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Modulo Arithmetic and Congruence for CUET Applied Maths

About 5 min to learn · 97 questions

Modulo arithmetic opens Section B2 of CUET Maths, the Applied Mathematics section you answer if you choose Applied Mathematics instead of Section B1. Its questions are short and exact, which makes them quick marks once the rules are clear. Learn each rule here, from remainders and the mod operator to modular addition, subtraction and multiplication, congruence modulo n, solving congruences and residue classes, with worked examples, time-saving shortcuts and the traps CUET sets, then practise on free questions in CUET's own formats, each fully solved.

Syllabus area
Section B2: Applied Mathematics
On this page · 10 sections
  1. 01The modulo operation
  2. 02Modular addition, subtraction and multiplication
  3. 03Congruence modulo n
  4. 04Solving with congruences
  5. 05Residue classes
  6. 06How CUET asks this chapter
  7. 07Exam shortcuts
  8. 08Common mistakes
  9. 09Quick revision
  10. 10Check your understanding

The modulo operation

A shop's weekly sale starts on a Monday. Which day is it 45 days later? Nobody counts 45 days on a calendar: the days repeat every 7, so only the remainder of 45 on division by 7 matters. That remainder is all of modulo arithmetic, and by the end of the chapter we will answer far harder questions with it.

What does a mod n mean?

a mod na \bmod n is the remainder when the integer aa is divided by the positive integer nn. It is always one of 0,1,2,…,n−10, 1, 2, \dots, n - 1, even when aa is negative.

Example 1 · Remainders, positive and negative

Find 53 mod 753 \bmod 7 and (−17) mod 5(-17) \bmod 5.

  1. 53=7×7+453 = 7 \times 7 + 4, so 53 mod 7=453 \bmod 7 = 4.

  2. For −17-17, take the multiple of 5 just below it: −17=5×(−4)+3-17 = 5 \times (-4) + 3.

Answer

53 mod 7=453 \bmod 7 = 4 and (−17) mod 5=3(-17) \bmod 5 = 3.

Back at the shop: 45=7×6+345 = 7 \times 6 + 3, so 45 days on is 3 days after Monday, a Thursday.

Watch out

A remainder is never negative. (−17) mod 5(-17) \bmod 5 is not −2-2: go down to the multiple of 5 below −17-17, which is −20-20, and count up.

Modular addition, subtraction and multiplication

How do you add or multiply modulo n?

Reduce each number modulo nn first, then add, subtract or multiply the remainders, and reduce the result once more. The answer is the same as working with the full numbers, and much quicker.

Example 2 · Reducing first

Find (46+38) mod 7(46 + 38) \bmod 7 and (46×38) mod 7(46 \times 38) \bmod 7.

  1. 46 mod 7=446 \bmod 7 = 4 and 38 mod 7=338 \bmod 7 = 3.

  2. Sum: 4+3=74 + 3 = 7, and 7 mod 7=07 \bmod 7 = 0. Product: 4×3=124 \times 3 = 12, and 12 mod 7=512 \bmod 7 = 5.

Answer

(46+38) mod 7=0(46 + 38) \bmod 7 = 0 and (46×38) mod 7=5(46 \times 38) \bmod 7 = 5.

Powers work the same way: find a small power that leaves remainder 1, and the rest repeats. 32=9≡1(mod8)3^{2} = 9 \equiv 1 \pmod{8}, so every even power of 3 leaves 1 on division by 8, and 310 mod 8=13^{10} \bmod 8 = 1.

Key idea

The last digit of a number is its remainder modulo 10. The last digits of 3,32,33,343, 3^{2}, 3^{3}, 3^{4} are 3, 9, 7, 1 and then repeat, so 3223^{22}, with 22=4×5+222 = 4 \times 5 + 2, ends in 9.

Congruence modulo n

What does a ≡ b (mod n) mean?

a≡b(modn)a \equiv b \pmod{n}, read "aa is congruent to bb modulo nn", means that nn divides a−ba - b. Equivalently, aa and bb leave the same remainder when divided by nn.

So 29≡5(mod8)29 \equiv 5 \pmod{8}, because 29−5=24=8×329 - 5 = 24 = 8 \times 3, and both leave remainder 5.

Congruences behave like equations for adding, subtracting, multiplying and raising to powers: if a≡b(modn)a \equiv b \pmod{n}, then a+c≡b+ca + c \equiv b + c, ac≡bcac \equiv bc and ak≡bka^{k} \equiv b^{k}, all modulo nn.

Watch out

Dividing is different. 2×4≡2×1(mod6)2 \times 4 \equiv 2 \times 1 \pmod{6}, since both sides leave 2, but 4≢1(mod6)4 \not\equiv 1 \pmod{6}. Cancel a factor only when it shares no factor with nn.

Solving with congruences

How do you solve a linear congruence?

To solve ax≡b(modn)ax \equiv b \pmod{n}, find the multiplicative inverse of aa modulo nn, the number that multiplies aa to give 1, and multiply both sides by it. For small nn, trying x=0,1,…,n−1x = 0, 1, \dots, n - 1 is just as fast.

Example 3 · A linear congruence

Solve 2x≡3(mod5)2x \equiv 3 \pmod{5}.

  1. The inverse of 2 modulo 5 is 3, since 2×3=6≡12 \times 3 = 6 \equiv 1.

  2. Multiply both sides by 3: x≡9≡4(mod5)x \equiv 9 \equiv 4 \pmod{5}. Check: 2×4=8≡32 \times 4 = 8 \equiv 3.

Answer

x≡4(mod5)x \equiv 4 \pmod{5}: x=4,9,14,…x = 4, 9, 14, \dots

ax≡b(modn)ax \equiv b \pmod{n} has a solution exactly when gcd⁡(a,n)\gcd(a, n) divides bb, and then it has gcd⁡(a,n)\gcd(a, n) solutions among 0,1,…,n−10, 1, \dots, n - 1. So 4x≡2(mod6)4x \equiv 2 \pmod{6} has two, x=2x = 2 and x=5x = 5, while 4x≡3(mod6)4x \equiv 3 \pmod{6} has none.

Residue classes

What is a residue class?

The residue class [r][r] modulo nn is the set of all integers that leave remainder rr on division by nn. There are exactly nn of them, [0],[1],…,[n−1][0], [1], \dots, [n - 1]; they have no member in common and together they contain every integer.

Modulo 5, for example, [2]={…,−8,−3,2,7,12,… }[2] = \{\dots, -8, -3, 2, 7, 12, \dots\}. The five classes form Z5\mathbb{Z}_5, where classes are added and multiplied through their members: [3]+[4]=[7]=[2][3] + [4] = [7] = [2] and [3]×[4]=[12]=[2][3] \times [4] = [12] = [2].

How CUET asks this chapter

  • The mod operator: a remainder, a negative number's remainder, clock or calendar problems.
  • Modular arithmetic: a sum, difference or product modulo nn, the remainder of a power, a last digit.
  • Congruence: which congruence holds, which modulus works, which properties are always true.
  • Solving and classes: a linear congruence, an inverse, how many solutions, which numbers share a class.

Exam shortcuts

  • Reduce every number before you add, subtract or multiply.
  • For a power, find the first power that gives 1, then use the remainder of the exponent.
  • Last digit = remainder modulo 10; last digits of powers repeat in cycles of at most 4.
  • a≡b(modn)a \equiv b \pmod{n}: check that nn divides a−ba - b.
  • ax≡b(modn)ax \equiv b \pmod{n} solvable ⇔ gcd⁡(a,n)\gcd(a, n) divides bb.

Common mistakes

  • Giving a negative remainder, or the quotient instead of the remainder.
  • Forgetting to reduce a final answer such as 12 modulo 7.
  • Cancelling a factor that shares a factor with the modulus.
  • Mixing up the additive inverse (sum 0) and the multiplicative inverse (product 1).

Quick revision

Mod and arithmetic

  • a mod na \bmod n: remainder, 00 to n−1n - 1
  • Reduce, then add, subtract, multiply; powers repeat

Congruence

  • a≡b(modn)a \equiv b \pmod{n} ⇔ nn divides a−ba - b
  • ax≡bax \equiv b needs gcd⁡(a,n)\gcd(a, n) to divide bb
  • nn residue classes, disjoint, covering every integer

Next in Section B2: mixing quantities at different prices, in alligation and mixtures.

Next · 09Quick revision

Check your understanding

97 questions, grouped by type, easy to hard. Pick an answer and check it. The worked solution opens when you ask for it.

The modulo operationQ1 of 97Q1

The modulo operation

21 questions

Question 1

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What is 47 mod 647 \bmod 6?

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AnswerB · 55

  1. a mod na \bmod n is the remainder when aa is divided by nn.

  2. 47=6×7+547 = 6 \times 7 + 5, so 47 mod 6=547 \bmod 6 = 5.

Worth noting

7 is the quotient, not the remainder.

Question 2

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A 12-hour clock shows 9 o'clock. Which hour will it show 50 hours later?

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AnswerD · 1111

  1. The hour repeats every 12 hours, so work modulo 12.

  2. 9+50=599 + 50 = 59 and 59=12×4+1159 = 12 \times 4 + 11, so the clock shows 11.

Worth noting

2 is 50 mod 1250 \bmod 12 alone, forgetting the clock started at 9.

Question 3

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Today is Wednesday. Which day of the week will it be 100 days from today?

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AnswerB · Friday

  1. The days repeat every 7, so only 100 mod 7100 \bmod 7 matters: 100=7×14+2100 = 7 \times 14 + 2.

  2. Two days after Wednesday is Friday.

Worth noting

Saturday counts 3 days on, as if 100 left remainder 3.

Question 4

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It is 15:00 on a 24-hour clock. What hour will it show 100 hours later?

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AnswerB · 1919

  1. A 24-hour clock works modulo 24: 15+100=11515 + 100 = 115.

  2. 115=24×4+19115 = 24 \times 4 + 19, so it shows 19:00.

Worth noting

4 is 100 mod 24100 \bmod 24, forgetting the clock started at 15:00.

Question 5

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x mod 6=4x \bmod 6 = 4. What is (x+5) mod 6(x + 5) \bmod 6?

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AnswerB · 33

  1. x=6q+4x = 6q + 4, so x+5=6q+9=6(q+1)+3x + 5 = 6q + 9 = 6(q + 1) + 3.

  2. The remainder is 3.

Worth noting

9 is not a remainder modulo 6; reduce it: 9=6+39 = 6 + 3.

Question 6

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Read the two statements and choose the correct option.

I. 0 mod 5=00 \bmod 5 = 0.

II. If a mod n=0a \bmod n = 0, then nn divides aa.

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AnswerC · Both I and II

  1. I: 0=5×0+00 = 5 \times 0 + 0. True.

  2. II: a remainder of 0 means a=nqa = nq, a multiple of nn. True.

Question 7

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What is 125 mod 9125 \bmod 9?

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AnswerC · 88

  1. Divide by 9 and keep the remainder, which must be one of 0,1,…,80, 1, \dots, 8.

  2. 125=9×(13)+8125 = 9 \times (13) + 8, so the answer is 8.

Worth noting

13 is the quotient, not the remainder.

Question 8

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A factory runs round the clock. A machine is serviced every 30 hours, the first service at 06:00 on a Monday. The shifts run from 06:00 to 14:00, 14:00 to 22:00 and 22:00 to 06:00.

At what time of day is the second service?

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AnswerD · 12:00

  1. The time of day works modulo 24: 6+30=366 + 30 = 36, and 36=24+1236 = 24 + 12.

  2. So the second service is at 12:00, on Tuesday.

Worth noting

06:00 adds 24 hours instead of 30.

Question 9

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What is (−23) mod 5(-23) \bmod 5?

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AnswerC · 22

  1. A remainder modulo 5 is always one of 0, 1, 2, 3 and 4, even when the number divided is negative.

  2. −23=5×(−5)+2-23 = 5 \times (-5) + 2, so (−23) mod 5=2(-23) \bmod 5 = 2.

Worth noting

−3 takes the remainder of 23 and puts the sign back, but a remainder cannot be negative.

Question 10

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When a number aa is divided by 8, the remainder is 5. Which of these can aa be?

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AnswerA · 3737

  1. a mod 8=5a \bmod 8 = 5 means a=8q+5a = 8q + 5: 5, 13, 21, 29, 37, 45, …

  2. 37=8×4+537 = 8 \times 4 + 5. The others leave 3, 2 and 3.

Worth noting

43 is 5 more than 38, but 38 is not a multiple of 8.

Question 11

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Which of the following statements are true?

I. a mod na \bmod n is always one of 0,1,…,n−10, 1, \dots, n - 1.

II. (−1) mod n=n−1(-1) \bmod n = n - 1.

III. (−a) mod n=a mod n(-a) \bmod n = a \bmod n for every integer aa.

IV. a mod n=0a \bmod n = 0 exactly when nn divides aa.

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AnswerD · I, II and IV only

  1. I is the definition of a remainder, and IV says the remainder is 0. For II, −1=n×(−1)+(n−1)-1 = n \times (-1) + (n - 1).

  2. III is false: 3 mod 5=33 \bmod 5 = 3 but (−3) mod 5=2(-3) \bmod 5 = 2.

Question 12

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−38-38 is divided by 7 to give a quotient qq and a remainder rr with 0≤r<70 \le r < 7. What are qq and rr?

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AnswerA · q=−6, r=4q = -6,\ r = 4

  1. Take the multiple of 7 just below −38-38: 7×(−6)=−427 \times (-6) = -42.

  2. −38=7×(−6)+4-38 = 7 \times (-6) + 4, so q=−6q = -6 and r=4r = 4.

Worth noting

q=−5, r=−3q = -5,\ r = -3 does give −38-38, but its remainder breaks 0≤r<70 \le r < 7.

Question 13

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Match each expression in List I with its value in List II.

List IList II
A. 25 mod 725 \bmod 7I. 22
B. 36 mod 936 \bmod 9II. 44
C. (−10) mod 6(-10) \bmod 6III. 11
D. 100 mod 11100 \bmod 11IV. 00
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AnswerC · A-II, B-IV, C-I, D-III

  1. 25=7×3+425 = 7 \times 3 + 4 and 36=9×4+036 = 9 \times 4 + 0.

  2. −10=6×(−2)+2-10 = 6 \times (-2) + 2 and 100=11×9+1100 = 11 \times 9 + 1.

Question 14

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A train leaves at 22:00 on a 24-hour clock, and the journey takes 39 hours. At what hour does it arrive?

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AnswerB · 1313

  1. A 24-hour clock works modulo 24: 22+39=6122 + 39 = 61.

  2. 61=24×2+1361 = 24 \times 2 + 13, so it arrives at 13:00, two days later.

Worth noting

15 is 39 mod 2439 \bmod 24, the length of the journey alone.

Question 15

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What is 1,000 mod 131{,}000 \bmod 13?

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AnswerA · 1212

  1. Divide by 13 and keep the remainder, which must be one of 0,1,…,120, 1, \dots, 12.

  2. 1,000=13×(76)+121{,}000 = 13 \times (76) + 12, so the answer is 12.

Worth noting

76 is the quotient, 1000÷131000 \div 13 rounded down.

Question 16

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1 January 2027 is a Friday. On which day of the week does 1 March 2027 fall?

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AnswerD · Monday

  1. January has 31 days and February 2027 has 28, so 1 March is 59 days later.

  2. 59 mod 7=359 \bmod 7 = 3: three days after Friday is Monday.

Worth noting

Counting February as 29 days gives Tuesday; 2027 is not a leap year.

Question 17

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What is (−50) mod 7(-50) \bmod 7?

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AnswerC · 66

  1. Divide by 7 and keep the remainder, which must be one of 0,1,…,60, 1, \dots, 6.

  2. −50=7×(−8)+6-50 = 7 \times (-8) + 6, so the answer is 6.

Worth noting

1 is 50 mod 750 \bmod 7; for −50-50 the remainder is 7−1=67 - 1 = 6.

Question 18

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Today is Tuesday. Which day of the week was it 45 days ago?

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AnswerA · Saturday

  1. 45=7×6+345 = 7 \times 6 + 3, so going back 45 days lands where going back 3 days does.

  2. Three days before Tuesday is Saturday.

Worth noting

Friday counts 3 days forward; the question goes back.

Question 19

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A factory runs round the clock. A machine is serviced every 30 hours, the first service at 06:00 on a Monday. The shifts run from 06:00 to 14:00, 14:00 to 22:00 and 22:00 to 06:00.

When is the fifth service?

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AnswerA · Saturday at 06:00

  1. The fifth service comes 4×30=1204 \times 30 = 120 hours after the first.

  2. 120=24×5120 = 24 \times 5: exactly 5 days later, at 06:00, so Saturday at 06:00.

Worth noting

Saturday at 12:00 counts 5 gaps of 30 hours; from the 1st to the 5th service there are only 4.

Question 20

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A factory runs round the clock. A machine is serviced every 30 hours, the first service at 06:00 on a Monday. The shifts run from 06:00 to 14:00, 14:00 to 22:00 and 22:00 to 06:00.

After how many hours does a service next fall at 06:00?

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AnswerC · 120120

  1. A service falls at 06:00 when the hours passed are a multiple of both 30 and 24.

  2. LCM(30,24)=120\text{LCM}(30, 24) = 120 hours.

Worth noting

720 is 30×2430 \times 24, a common multiple but not the least.

Question 21

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A factory runs round the clock. A machine is serviced every 30 hours, the first service at 06:00 on a Monday. The shifts run from 06:00 to 14:00, 14:00 to 22:00 and 22:00 to 06:00.

In which shift does the third service fall?

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AnswerB · 14:00 to 22:00

  1. The third service is 2×30=602 \times 30 = 60 hours after the first: 6+60=666 + 60 = 66, and 66=24×2+1866 = 24 \times 2 + 18.

  2. 18:00 on Wednesday lies in the 14:00 to 22:00 shift.

Modular addition, subtraction and multiplication

23 questions

Question 22

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What is (58+37) mod 9(58 + 37) \bmod 9?

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AnswerC · 55

  1. Reduce each number first: 58 mod 9=458 \bmod 9 = 4 and 37 mod 9=137 \bmod 9 = 1.

  2. (4+1) mod 9=5(4 + 1) \bmod 9 = 5. Check: 95=9×10+595 = 9 \times 10 + 5.

Worth noting

4 is the remainder of 58 alone.

Question 23

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What is (23×19) mod 5(23 \times 19) \bmod 5?

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AnswerA · 22

  1. 23 mod 5=323 \bmod 5 = 3 and 19 mod 5=419 \bmod 5 = 4, so multiply the remainders: 3×4=123 \times 4 = 12.

  2. 12 mod 5=212 \bmod 5 = 2. Check: 437=5×87+2437 = 5 \times 87 + 2.

Worth noting

12 is the product of the remainders before it was reduced modulo 5.

Question 24

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What is the remainder when 115011^{50} is divided by 10?

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AnswerC · 11

  1. 11≡1(mod10)11 \equiv 1 \pmod{10}.

  2. So 1150≡150=111^{50} \equiv 1^{50} = 1: every power of 11 ends in 1.

Worth noting

0 treats 11 as a multiple of 10.

Question 25

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What is (89+76) mod 11(89 + 76) \bmod 11?

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AnswerA · 00

  1. 89 mod 11=189 \bmod 11 = 1 and 76 mod 11=1076 \bmod 11 = 10.

  2. (1+10) mod 11=0(1 + 10) \bmod 11 = 0. Check: 165=11×15165 = 11 \times 15.

Worth noting

11 is not a remainder modulo 11; it is 11≡011 \equiv 0.

Question 26

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A shop gives each product a 5-digit code. The first four digits name the product, and the fifth, the check digit, is the sum of the first four digits taken modulo 10.

What is the check digit for a product whose first four digits are 4728?

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AnswerA · 11

  1. 4+7+2+8=214 + 7 + 2 + 8 = 21.

  2. 21 mod 10=121 \bmod 10 = 1.

Worth noting

21 is the sum itself; a single digit is needed, its remainder by 10.

Question 27

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What is (17−30) mod 8(17 - 30) \bmod 8?

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AnswerC · 33

  1. 17−30=−1317 - 30 = -13, and −13=8×(−2)+3-13 = 8 \times (-2) + 3.

  2. So (17−30) mod 8=3(17 - 30) \bmod 8 = 3. Or reduce first: (1−6) mod 8=(−5) mod 8=3(1 - 6) \bmod 8 = (-5) \bmod 8 = 3.

Worth noting

5 is 13 mod 813 \bmod 8: the minus sign was dropped.

Question 28

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What is the last digit of 7357^{35}?

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AnswerB · 33

  1. The last digit is the remainder modulo 10. The last digits of 7,72,73,747, 7^{2}, 7^{3}, 7^{4} are 7, 9, 3, 1, and then they repeat.

  2. 35=4×8+335 = 4 \times 8 + 3, so 7357^{35} ends like 73=3437^{3} = 343: in 3.

Worth noting

9 is the last digit of 727^{2}, the second in the cycle, not the third.

Question 29

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What is the additive inverse of 5 modulo 12, written as one of 0,1,…,110, 1, \dots, 11?

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AnswerD · 77

  1. The additive inverse of aa is the number bb with a+b≡0(mod12)a + b \equiv 0 \pmod{12}.

  2. 5+7=12≡05 + 7 = 12 \equiv 0, so it is 7.

Worth noting

5 is the number itself: 5+5=105 + 5 = 10, not a multiple of 12.

Question 30

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Assertion (A): (45×46) mod 9=0(45 \times 46) \bmod 9 = 0.

Reason (R): 45 is a multiple of 9.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. 45≡0(mod9)45 \equiv 0 \pmod{9}, so 45×46≡0×46=045 \times 46 \equiv 0 \times 46 = 0.

  2. Both are true, and R is exactly why the product leaves no remainder.

Question 31

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What is (152+26) mod 4(15^{2} + 26) \bmod 4?

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AnswerD · 33

  1. Reduce before squaring: 15≡315 \equiv 3, so 152≡9≡1(mod4)15^{2} \equiv 9 \equiv 1 \pmod{4}; and 26≡226 \equiv 2.

  2. 1+2=31 + 2 = 3. Check: 225+26=251=4×62+3225 + 26 = 251 = 4 \times 62 + 3.

Worth noting

1 is 152 mod 415^{2} \bmod 4 alone, with 26 left out.

Question 32

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What is the remainder when 51005^{100} is divided by 4?

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AnswerB · 11

  1. 5=4+15 = 4 + 1, so 5≡1(mod4)5 \equiv 1 \pmod{4}.

  2. Then 5100≡1100=1(mod4)5^{100} \equiv 1^{100} = 1 \pmod{4}.

Worth noting

0 would need 4 to divide 51005^{100}, but 51005^{100} is odd.

Question 33

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What is (123×45) mod 7(123 \times 45) \bmod 7?

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AnswerB · 55

  1. Reduce first: 123 mod 7=4123 \bmod 7 = 4 and 45 mod 7=345 \bmod 7 = 3.

  2. (4×3) mod 7=12 mod 7=5(4 \times 3) \bmod 7 = 12 \bmod 7 = 5.

Worth noting

12 multiplies the remainders but does not reduce the product.

Question 34

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What is (40−73) mod 6(40 - 73) \bmod 6?

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AnswerA · 33

  1. 40−73=−3340 - 73 = -33, and −33=6×(−6)+3-33 = 6 \times (-6) + 3.

  2. So the remainder is 3.

Worth noting

−3-3 is not a remainder; add 6 to get 3.

Question 35

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What is the last digit of 320263^{2026}?

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AnswerD · 99

  1. Last digits of powers of 3 repeat in a cycle of 4: 3, 9, 7, 1.

  2. 2026 mod 4=22026 \bmod 4 = 2, so the last digit is the second in the cycle, 9.

Worth noting

7 is the third in the cycle: 2026=4×506+22026 = 4 \times 506 + 2, not +3+ 3.

Question 36

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What is (101×102×103) mod 100(101 \times 102 \times 103) \bmod 100?

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AnswerC · 66

  1. 101≡1101 \equiv 1, 102≡2102 \equiv 2, 103≡3(mod100)103 \equiv 3 \pmod{100}.

  2. 1×2×3=61 \times 2 \times 3 = 6.

Worth noting

306 is 101+102+103101 + 102 + 103, a sum, not reduced.

Question 37

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A shop gives each product a 5-digit code. The first four digits name the product, and the fifth, the check digit, is the sum of the first four digits taken modulo 10.

Which of these codes is valid?

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AnswerD · 3591835918

  1. For each code, add the first four digits and compare the remainder by 10 with the last digit.

  2. 3+5+9+1=183 + 5 + 9 + 1 = 18, which ends in 8: 35918 checks. The others give 8 against 7, 7 against 8 and 9 against 8.

Question 38

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A shop gives each product a 5-digit code. The first four digits name the product, and the fifth, the check digit, is the sum of the first four digits taken modulo 10.

A code reads 2, 9, 6, xx, 3, with the fourth digit smudged. What is xx?

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AnswerC · 66

  1. 2+9+6+x≡3(mod10)2 + 9 + 6 + x \equiv 3 \pmod{10}, that is 17+x≡317 + x \equiv 3.

  2. x≡3−17=−14≡6x \equiv 3 - 17 = -14 \equiv 6, and the only digit that works is 6.

Worth noting

4 drops the minus sign of −14-14 and takes its last digit.

Question 39

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What is 220 mod 72^{20} \bmod 7?

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AnswerD · 44

  1. 23=8≡1(mod7)2^{3} = 8 \equiv 1 \pmod{7}, so the powers of 2 repeat every 3.

  2. 220=(23)6×22≡1×4=4(mod7)2^{20} = (2^{3})^{6} \times 2^{2} \equiv 1 \times 4 = 4 \pmod{7}.

Worth noting

1 would be right for 2182^{18}; 20 leaves 2 over a multiple of 3.

Question 40

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What is the remainder when 1+2+3+⋯+201 + 2 + 3 + \dots + 20 is divided by 7?

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AnswerA · 00

  1. The sum is 20×212=210\frac{20 \times 21}{2} = 210.

  2. 210=7×30210 = 7 \times 30, so the remainder is 0.

Worth noting

6 is 20 mod 720 \bmod 7, the last term alone.

Question 41

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What is (210+35) mod 7\left(2^{10} + 3^{5}\right) \bmod 7?

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AnswerB · 00

  1. 23=8≡12^{3} = 8 \equiv 1, so 210=(23)3⋅2≡22^{10} = (2^{3})^{3} \cdot 2 \equiv 2. And 35=243=7×34+53^{5} = 243 = 7 \times 34 + 5, so 35≡53^{5} \equiv 5.

  2. 2+5=7≡02 + 5 = 7 \equiv 0.

Worth noting

2 and 5 are the two parts' remainders; their sum still has to be reduced.

Question 42

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What is the remainder when 71007^{100} is divided by 5?

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AnswerA · 11

  1. 7≡2(mod5)7 \equiv 2 \pmod{5}, and 24=16≡1(mod5)2^{4} = 16 \equiv 1 \pmod{5}.

  2. 7100≡2100=(24)25≡17^{100} \equiv 2^{100} = (2^{4})^{25} \equiv 1.

Worth noting

2 reduces the base but forgets the power.

Question 43

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What is the remainder when 1!+2!+3!+⋯+100!1! + 2! + 3! + \dots + 100! is divided by 5?

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AnswerD · 33

  1. From 5!5! on, every factorial contains the factor 5, so it leaves remainder 0.

  2. 1+2+6+24=331 + 2 + 6 + 24 = 33, and 33 mod 5=333 \bmod 5 = 3.

Worth noting

0 assumes every term is a multiple of 5; the first four are not.

Question 44

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A shop gives each product a 5-digit code. The first four digits name the product, and the fifth, the check digit, is the sum of the first four digits taken modulo 10.

Which mistake can this check digit NOT catch?

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AnswerB · Swapping two of the first four digits

  1. A swap leaves the sum of the first four digits unchanged, so the check digit still matches.

  2. Changing a single digit changes the sum by 1 to 9, which always changes its last digit.

Worth noting

This is why many real codes weight the digits differently, so that a swap changes the sum.

Congruence modulo n

16 questions

Question 45

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What does a≡b(modn)a \equiv b \pmod{n} mean?

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AnswerC · nn divides a−ba - b

  1. aa is congruent to bb modulo nn when their difference is a multiple of nn.

  2. For example, 17≡2(mod5)17 \equiv 2 \pmod{5} because 17−2=1517 - 2 = 15.

Worth noting

nn need not divide aa or bb: 5 divides neither 17 nor 2.

Question 46

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Which of these congruences is true?

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AnswerB · 38≡3(mod7)38 \equiv 3 \pmod{7}

  1. Check whether the modulus divides the difference.

  2. 38−3=35=7×538 - 3 = 35 = 7 \times 5. The others give 38, 22 and −13-13, which 4 and 6 do not divide.

Worth noting

−8≡5(mod6)-8 \equiv 5 \pmod{6} fails: −8−5=−13-8 - 5 = -13.

Question 47

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Which of these numbers is congruent to 100 modulo 7?

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AnswerB · 1616

  1. 100=7×14+2100 = 7 \times 14 + 2, so we need a number that leaves remainder 2.

  2. 16=7×2+216 = 7 \times 2 + 2. The others leave 3, 6 and 6.

Question 48

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For which of these values of nn is 47≡5(modn)47 \equiv 5 \pmod{n}?

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AnswerB · 66

  1. 47≡5(modn)47 \equiv 5 \pmod{n} means nn divides 47−5=4247 - 5 = 42.

  2. Of 5, 6, 8 and 9, only 6 divides 42.

Worth noting

5 is the remainder in the question, not a modulus that works.

Question 49

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a≡b(modn)a \equiv b \pmod{n} and cc is an integer. Which of the following are always true?

I. a+c≡b+c(modn)a + c \equiv b + c \pmod{n}

II. ac≡bc(modn)ac \equiv bc \pmod{n}

III. a≡b(mod2n)a \equiv b \pmod{2n}

IV. a2≡b2(modn)a^{2} \equiv b^{2} \pmod{n}

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AnswerD · I, II and IV only

  1. Adding, multiplying and raising to a power keep a congruence: if nn divides a−ba - b, it divides (a−b)c(a - b)c and a2−b2=(a−b)(a+b)a^{2} - b^{2} = (a - b)(a + b).

  2. III fails: 1≡4(mod3)1 \equiv 4 \pmod{3}, but 6 does not divide 4−1=34 - 1 = 3.

Question 50

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x≡4(mod9)x \equiv 4 \pmod{9}. What is 5x mod 95x \bmod 9?

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AnswerA · 22

  1. A congruence can be multiplied through: 5x≡5×4=20(mod9)5x \equiv 5 \times 4 = 20 \pmod{9}.

  2. 20=9×2+220 = 9 \times 2 + 2, so 5x mod 9=25x \bmod 9 = 2.

Worth noting

20 is right modulo 9 but is not reduced to a remainder below 9.

Question 51

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Assertion (A): 1000≡−1(mod7)1000 \equiv -1 \pmod{7}.

Reason (R): 7 divides 1001.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. 1000−(−1)=1001=7×1431000 - (-1) = 1001 = 7 \times 143.

  2. So both are true, and R is exactly the reason A holds.

Question 52

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Read the two statements and choose the correct option.

I. a≡b(modn)a \equiv b \pmod{n} exactly when aa and bb leave the same remainder on division by nn.

II. If a≡b(modn)a \equiv b \pmod{n}, then a≡b(mod2n)a \equiv b \pmod{2n}.

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AnswerA · Only I

  1. I: if a=nq1+ra = nq_1 + r and b=nq2+rb = nq_2 + r, then a−b=n(q1−q2)a - b = n(q_1 - q_2), and the reverse also holds. True.

  2. II: 1≡4(mod3)1 \equiv 4 \pmod{3}, but 4−1=34 - 1 = 3 is not a multiple of 6. False.

Question 53

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a≡2(mod6)a \equiv 2 \pmod{6} and b≡5(mod6)b \equiv 5 \pmod{6}. What is (a−b) mod 6(a - b) \bmod 6?

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AnswerA · 33

  1. a−b≡2−5=−3(mod6)a - b \equiv 2 - 5 = -3 \pmod{6}.

  2. −3+6=3-3 + 6 = 3.

Worth noting

−3-3 must be brought into 0,…,50, \dots, 5 by adding 6.

Question 54

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What is 4,567,891 mod 94{,}567{,}891 \bmod 9?

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AnswerD · 44

  1. 10≡1(mod9)10 \equiv 1 \pmod{9}, so every power of 10 is ≡1\equiv 1, and a number is congruent to the sum of its digits.

  2. 4+5+6+7+8+9+1=404 + 5 + 6 + 7 + 8 + 9 + 1 = 40, and 40 mod 9=440 \bmod 9 = 4.

Worth noting

1 is the last digit, which gives the remainder by 10, not by 9.

Question 55

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Which of these congruences is NOT true?

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AnswerD · 41≡6(mod8)41 \equiv 6 \pmod{8}

  1. a≡b(modn)a \equiv b \pmod{n} exactly when nn divides a−ba - b.

  2. 41−6=3541 - 6 = 35, not a multiple of 8. The others give 35, 48 and 24, multiples of 7, 8 and 6.

Question 56

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For how many integers n>1n > 1 is 23≡8(modn)23 \equiv 8 \pmod{n}?

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AnswerC · 33

  1. nn must divide 23−8=1523 - 8 = 15.

  2. The divisors of 15 greater than 1 are 3, 5 and 15: three values.

Worth noting

4 counts 1 as well, but the question asks for n>1n > 1.

Question 57

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Which of the following statements are true?

I. If a≡b(modn)a \equiv b \pmod{n}, then a2≡b2(modn)a^{2} \equiv b^{2} \pmod{n}.

II. If ac≡bc(modn)ac \equiv bc \pmod{n}, then always a≡b(modn)a \equiv b \pmod{n}.

III. If a≡b(modn)a \equiv b \pmod{n}, then b≡a(modn)b \equiv a \pmod{n}.

IV. If a≡ba \equiv b and b≡c(modn)b \equiv c \pmod{n}, then a≡c(modn)a \equiv c \pmod{n}.

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AnswerB · I, III and IV only

  1. I: congruences can be multiplied. III and IV: congruence is symmetric and transitive.

  2. II is false: 2×3≡4×3(mod6)2 \times 3 \equiv 4 \times 3 \pmod{6} (both are ≡0\equiv 0), but 2≢4(mod6)2 \not\equiv 4 \pmod{6}.

Worth noting

Cancelling cc is safe only when cc and nn have no common factor.

Question 58

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x≡3(mod7)x \equiv 3 \pmod{7}. What is (x2+1) mod 7(x^{2} + 1) \bmod 7?

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AnswerB · 33

  1. Replace xx by 3: x2+1≡9+1=10x^{2} + 1 \equiv 9 + 1 = 10.

  2. 10 mod 7=310 \bmod 7 = 3.

Worth noting

10 is right but not reduced: a remainder modulo 7 is below 7.

Question 59

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Assertion (A): 210≡1(mod11)2^{10} \equiv 1 \pmod{11}.

Reason (R): 25=32≡−1(mod11)2^{5} = 32 \equiv -1 \pmod{11}.

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AnswerA · Both A and R are true, and R is the correct explanation of A.

  1. R: 32=33−132 = 33 - 1, and 11∣3311 \mid 33. True.

  2. Squaring R: 210=(25)2≡(−1)2=12^{10} = (2^{5})^{2} \equiv (-1)^{2} = 1, so A is true and follows from R.

Question 60

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a≡3(mod5)a \equiv 3 \pmod{5} and b≡4(mod5)b \equiv 4 \pmod{5}. What is (a2+2b) mod 5(a^{2} + 2b) \bmod 5?

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AnswerA · 22

  1. Replace aa and bb by their remainders: a2+2b≡32+2×4=17(mod5)a^{2} + 2b \equiv 3^{2} + 2 \times 4 = 17 \pmod{5}.

  2. 17=5×3+217 = 5 \times 3 + 2, so the answer is 2.

Worth noting

7 squares nothing: 3+4=73 + 4 = 7 adds the remainders.

Solving with congruences

20 questions

Question 61

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What is the least positive xx with x+7≡3(mod10)x + 7 \equiv 3 \pmod{10}?

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AnswerC · 66

  1. Subtract 7 from both sides: x≡3−7=−4(mod10)x \equiv 3 - 7 = -4 \pmod{10}.

  2. −4+10=6-4 + 10 = 6. Check: 6+7=13≡36 + 7 = 13 \equiv 3.

Worth noting

4 drops the sign of −4-4; 4+7=11≡14 + 7 = 11 \equiv 1.

Question 62

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A number multiplied by 4 leaves remainder 1 on division by 7. What is the least positive such number?

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AnswerA · 22

  1. We need 4x≡1(mod7)4x \equiv 1 \pmod{7}.

  2. 4×2=8=7+14 \times 2 = 8 = 7 + 1, so x=2x = 2.

Worth noting

1 gives 4×1=44 \times 1 = 4, which leaves 4, not 1.

Question 63

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What is the least positive xx with x+9≡4(mod12)x + 9 \equiv 4 \pmod{12}?

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AnswerC · 77

  1. Subtract 9: x≡4−9=−5(mod12)x \equiv 4 - 9 = -5 \pmod{12}.

  2. −5+12=7-5 + 12 = 7. Check: 7+9=16≡47 + 9 = 16 \equiv 4.

Worth noting

5 drops the sign of −5-5: 5+9=14≡25 + 9 = 14 \equiv 2.

Question 64

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Republic Day, 26 January 2027, falls on a Tuesday. A school holds its annual day on 15 March 2027 and its sports day 100 days after Republic Day. 2027 is not a leap year.

How many days after Republic Day is the annual day?

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AnswerB · 4848

  1. From 26 January: 5 more days in January, 28 in February and 15 in March.

  2. 5+28+15=485 + 28 + 15 = 48 days.

Worth noting

49 counts 26 January itself; the count starts the day after.

Question 65

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Republic Day, 26 January 2027, falls on a Tuesday. A school holds its annual day on 15 March 2027 and its sports day 100 days after Republic Day. 2027 is not a leap year.

On which day of the week is the sports day?

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AnswerA · Thursday

  1. 100=7×14+2100 = 7 \times 14 + 2.

  2. Two days after Tuesday: Thursday.

Question 66

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What is the least positive xx with 3x≡4(mod7)3x \equiv 4 \pmod{7}?

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AnswerD · 66

  1. Try x=1,2,…,6x = 1, 2, \dots, 6: 3x3x leaves 3, 6, 2, 5, 1, 4 on division by 7.

  2. x=6x = 6 gives 18=7×2+418 = 7 \times 2 + 4.

Worth noting

4 copies the right-hand side; 3×4=12≡53 \times 4 = 12 \equiv 5.

Question 67

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What is the multiplicative inverse of 4 modulo 9?

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AnswerC · 77

  1. The inverse of 4 is the xx with 4x≡1(mod9)4x \equiv 1 \pmod{9}.

  2. 4×7=28=9×3+14 \times 7 = 28 = 9 \times 3 + 1, so the inverse is 7.

Worth noting

5 is the additive inverse: 4+5=9≡04 + 5 = 9 \equiv 0, not 1.

Question 68

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Which of these congruences has no solution?

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AnswerB · 6x≡4(mod9)6x \equiv 4 \pmod{9}

  1. ax≡b(modn)ax \equiv b \pmod{n} has a solution exactly when gcd⁡(a,n)\gcd(a, n) divides bb.

  2. gcd⁡(6,9)=3\gcd(6, 9) = 3 does not divide 4, so 6x≡4(mod9)6x \equiv 4 \pmod{9} has none: 6x6x only leaves 0, 3 or 6.

Worth noting

4x≡6(mod10)4x \equiv 6 \pmod{10} does have solutions: x=4x = 4 gives 16≡616 \equiv 6.

Question 69

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15 August 2026 is a Saturday. On which day of the week does 15 August 2027 fall?

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AnswerC · Sunday

  1. The days between are 365, as no 29 February falls in between, and 365=7×52+1365 = 7 \times 52 + 1, so 365≡1(mod7)365 \equiv 1 \pmod{7}.

  2. So the day moves on by one: Sunday.

Worth noting

Monday counts 366 days, but 2027 is not a leap year.

Question 70

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Which numbers among 0,1,…,90, 1, \dots, 9 satisfy 2x≡4(mod10)2x \equiv 4 \pmod{10}?

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AnswerD · 2 and 7

  1. 2x2x must end in 4, so 2x2x is 4 or 14 for xx up to 9.

  2. That gives x=2x = 2 and x=7x = 7: two solutions, as gcd⁡(2,10)=2\gcd(2, 10) = 2 promises.

Worth noting

Dividing both sides by 2 gives only x=2x = 2; dividing a congruence also divides its modulus, to 5, and x≡2(mod5)x \equiv 2 \pmod{5} has two values below 10.

Question 71

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Chairs set out in rows of 3 leave 2 over, and in rows of 5 leave 3 over. What is the least number of chairs?

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AnswerB · 88

  1. We need x≡2(mod3)x \equiv 2 \pmod{3} and x≡3(mod5)x \equiv 3 \pmod{5}.

  2. Numbers ≡3(mod5)\equiv 3 \pmod{5}: 3, 8, 13, … The first that leaves 2 on division by 3 is 8.

Worth noting

23 works too but is not the least; solutions repeat every 15.

Question 72

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What is the multiplicative inverse of 7 modulo 10?

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AnswerA · 33

  1. We need 7x≡1(mod10)7x \equiv 1 \pmod{10}: a multiple of 7 ending in 1.

  2. 7×3=217 \times 3 = 21, so the inverse is 3.

Worth noting

9 gives 7×9=63≡37 \times 9 = 63 \equiv 3, not 1.

Question 73

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Which of these congruences has no solution?

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AnswerD · 6x≡4(mod9)6x \equiv 4 \pmod{9}

  1. ax≡b(modn)ax \equiv b \pmod{n} has a solution exactly when gcd⁡(a,n)\gcd(a, n) divides bb.

  2. gcd⁡(6,9)=3\gcd(6, 9) = 3 does not divide 4. The others: 3 divides 3; gcd⁡(2,9)=1\gcd(2, 9) = 1; gcd⁡(4,10)=2\gcd(4, 10) = 2 divides 6.

Question 74

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1 March 2027 is a Monday. On which day of the week does 1 March 2028 fall?

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AnswerD · Wednesday

  1. From 1 March 2027 to 1 March 2028 there are 366 days, because 29 February 2028 falls in between.

  2. 366 mod 7=2366 \bmod 7 = 2, so the day moves two places on: Wednesday.

Worth noting

A one-day shift is right for an ordinary year; this span includes 29 February.

Question 75

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What is the least positive xx with 5x≡3(mod8)5x \equiv 3 \pmod{8}?

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AnswerD · 77

  1. The inverse of 5 modulo 8 is 5, since 5×5=25≡15 \times 5 = 25 \equiv 1.

  2. Multiply both sides by 5: x≡15≡7x \equiv 15 \equiv 7. Check: 5×7=35=32+35 \times 7 = 35 = 32 + 3.

Worth noting

3 copies the right side; 5×3=15≡75 \times 3 = 15 \equiv 7, not 3.

Question 76

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Republic Day, 26 January 2027, falls on a Tuesday. A school holds its annual day on 15 March 2027 and its sports day 100 days after Republic Day. 2027 is not a leap year.

On which day of the week is the annual day?

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AnswerD · Monday

  1. 48=7×6+648 = 7 \times 6 + 6, so 48 days later is 6 days on in the week.

  2. Six days after Tuesday is Monday.

Worth noting

The day before counts 26 January itself among the 48 days; whole weeks change nothing, only the remainder 6 matters.

Question 77

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Republic Day, 26 January 2027, falls on a Tuesday. A school holds its annual day on 15 March 2027 and its sports day 100 days after Republic Day. 2027 is not a leap year.

On which day of the week will Republic Day 2028 fall?

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AnswerB · Wednesday

  1. 26 January 2027 to 26 January 2028 is 365 days: 29 February 2028 comes later.

  2. 365=7×52+1365 = 7 \times 52 + 1: one day on, Wednesday.

Worth noting

A two-day shift would need 29 February to fall between the two dates; it comes after 26 January 2028.

Question 78

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How many solutions does 4x≡8(mod12)4x \equiv 8 \pmod{12} have among x=0,1,…,11x = 0, 1, \dots, 11?

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AnswerD · 44

  1. gcd⁡(4,12)=4\gcd(4, 12) = 4, and 4 divides 8, so there are exactly 4 solutions modulo 12.

  2. They are x=2,5,8,11x = 2, 5, 8, 11: each gives 4x4x = 8, 20, 32 or 44, all ≡8\equiv 8.

Worth noting

1 would be right if 4 had an inverse modulo 12, but 4 and 12 share the factor 4.

Question 79

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Eggs in a basket are counted in fours, leaving 3 over, and in fives, leaving 4 over. What is the least number of eggs the basket can hold?

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AnswerB · 1919

  1. x≡3(mod4)x \equiv 3 \pmod{4} and x≡4(mod5)x \equiv 4 \pmod{5} both say x≡−1x \equiv -1: one more egg would make whole fours and whole fives.

  2. So x+1x + 1 is a multiple of 20, and the least is x=19x = 19.

Worth noting

23 leaves 3 in fours but also 3 in fives.

Question 80

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How many solutions does 6x≡12(mod18)6x \equiv 12 \pmod{18} have among x=0,1,…,17x = 0, 1, \dots, 17?

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AnswerC · 66

  1. gcd⁡(6,18)=6\gcd(6, 18) = 6 divides 12, so there are exactly 6 solutions modulo 18.

  2. Dividing through by 6: x≡2(mod3)x \equiv 2 \pmod{3}, so x=2,5,8,11,14,17x = 2, 5, 8, 11, 14, 17.

Worth noting

1 is the count when gcd⁡(a,n)=1\gcd(a, n) = 1; here it is 6.

Residue classes

17 questions

Question 81

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How many residue classes are there modulo 8?

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AnswerC · 88

  1. Every integer leaves one of the remainders 0,1,…,70, 1, \dots, 7 on division by 8.

  2. Each remainder is a class, [0],[1],…,[7][0], [1], \dots, [7]: 8 classes.

Worth noting

7 leaves out the class [0][0] of the multiples of 8.

Question 82

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In Z6\mathbb{Z}_6, the residue classes modulo 6, what is [4]+[5][4] + [5]?

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AnswerD · [3][3]

  1. Add the representatives and reduce: 4+5=94 + 5 = 9.

  2. 9 mod 6=39 \bmod 6 = 3, so [4]+[5]=[3][4] + [5] = [3].

Worth noting

[9][9] is the same class as [3][3] but is not written with a representative from 0 to 5.

Question 83

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In Z9\mathbb{Z}_9, what is [4]×[7][4] \times [7]?

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AnswerB · [1][1]

  1. [4]×[7]=[28][4] \times [7] = [28].

  2. 28=9×3+128 = 9 \times 3 + 1, so it is [1][1].

Worth noting

[2][2] is 11 mod 911 \bmod 9: it adds the classes instead of multiplying.

Question 84

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In Z10\mathbb{Z}_{10}, with classes written [0][0] to [9][9], which class is the additive inverse of [3][3]?

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AnswerD · [7][7]

  1. The additive inverse of [a][a] is the class that adds to [0][0].

  2. [3]+[7]=[10]=[0][3] + [7] = [10] = [0].

Worth noting

[1][1] is the multiplicative idea; the additive inverse must add to [0][0]: [3]+[1]=[4][3] + [1] = [4].

Question 85

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In Z8\mathbb{Z}_8, the residue classes modulo 8 written [0][0] to [7][7], what is [5]+[6][5] + [6]?

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AnswerC · [3][3]

  1. [a]+[b]=[a+b][a] + [b] = [a + b], reduced modulo 8.

  2. 5+6=11≡35 + 6 = 11 \equiv 3, so the sum is [3][3].

Worth noting

Add, then take away 8 when the sum is 8 or more: 11−8=311 - 8 = 3.

Question 86

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A school numbers its 60 students from 1 to 60 and puts each in a house by the remainder of the roll number on division by 5: remainder 1 is Red, 2 Blue, 3 Green, 4 Yellow and 0 White.

Which house is roll number 47 in?

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AnswerD · Blue

  1. 47=5×9+247 = 5 \times 9 + 2.

  2. Remainder 2: Blue.

Question 87

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A school numbers its 60 students from 1 to 60 and puts each in a house by the remainder of the roll number on division by 5: remainder 1 is Red, 2 Blue, 3 Green, 4 Yellow and 0 White.

How many students are in Green house?

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AnswerB · 1212

  1. Green holds 3, 8, 13, …, 58: the numbers 5k+35k + 3 up to 60.

  2. k=0k = 0 to 11: 12 students.

Worth noting

Each of the 5 houses gets exactly 605=12\frac{60}{5} = 12, since 60 is a multiple of 5.

Question 88

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Which of these numbers lies in the same residue class as 17 modulo 5?

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AnswerA · −3-3

  1. 17 leaves remainder 2, so its class is [2]={…,−8,−3,2,7,12,17,… }[2] = \{\dots, -8, -3, 2, 7, 12, 17, \dots\}.

  2. −3=5×(−1)+2-3 = 5 \times (-1) + 2, so −3-3 is in it. 8, 21 and 14 leave 3, 1 and 4.

Worth noting

−3-3 looks far from 17, but 17−(−3)=2017 - (-3) = 20 is a multiple of 5.

Question 89

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Which set is the residue class [3][3] modulo 7?

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AnswerB · {…,−11,−4,3,10,17,… }\{\dots, -11, -4, 3, 10, 17, \dots\}

  1. [3][3] is every integer that leaves 3 on division by 7: 7k+37k + 3 for every integer kk, negative ones included.

  2. k=−2,−1,0,1,2k = -2, -1, 0, 1, 2 give −11,−4,3,10,17-11, -4, 3, 10, 17.

Worth noting

{3,10,17,24,… }\{3, 10, 17, 24, \dots\} stops at 3 and leaves out −4-4 and −11-11.

Question 90

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Read the two statements and choose the correct option.

I. Two different residue classes modulo nn have no integer in common.

II. Every integer lies in one of the residue classes modulo nn.

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AnswerC · Both I and II

  1. An integer has exactly one remainder on division by nn, so it is in exactly one class: I is true.

  2. And it does have a remainder, so it is in some class: II is true. Together the classes split the integers into nn parts.

Question 91

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In Z7\mathbb{Z}_7, what is [3]×[5][3] \times [5]?

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AnswerB · [1][1]

  1. Multiply the representatives: 3×5=153 \times 5 = 15.

  2. 15=7×2+115 = 7 \times 2 + 1, so [3]×[5]=[1][3] \times [5] = [1]: 5 is the inverse of 3 modulo 7.

Worth noting

[8][8] adds the representatives instead of multiplying them.

Question 92

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Read the two statements and choose the correct option.

I. In Z5\mathbb{Z}_5, every class except [0][0] has a multiplicative inverse.

II. In Z6\mathbb{Z}_6, [2]×[3]=[0][2] \times [3] = [0].

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AnswerC · Both I and II

  1. I: 5 is prime, so 1, 2, 3, 4 each share no factor with it; the inverses are [1],[3],[2],[4][1], [3], [2], [4]. True.

  2. II: 2×3=6≡0(mod6)2 \times 3 = 6 \equiv 0 \pmod{6}. True: two non-zero classes can multiply to [0][0].

Question 93

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Which of these numbers belongs to the residue class [2][2] modulo 6?

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AnswerC · 3838

  1. [2][2] modulo 6 is {…,−4,2,8,14,20,26,32,38,… }\{\dots, -4, 2, 8, 14, 20, 26, 32, 38, \dots\}: numbers 6k+26k + 2.

  2. 38=6×6+238 = 6 \times 6 + 2. 40, 21 and 27 leave 4, 3 and 3.

Question 94

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A school numbers its 60 students from 1 to 60 and puts each in a house by the remainder of the roll number on division by 5: remainder 1 is Red, 2 Blue, 3 Green, 4 Yellow and 0 White.

Which pair of roll numbers are in the same house?

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AnswerA · 23 and 38

  1. Two numbers share a house when they are congruent modulo 5, that is, their difference is a multiple of 5.

  2. 38−23=1538 - 23 = 15: both are Green. The other differences are 27, 27 and 19.

Worth noting

14 and 41 use the same digits, but 41−14=2741 - 14 = 27 is not a multiple of 5.

Question 95

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A school numbers its 60 students from 1 to 60 and puts each in a house by the remainder of the roll number on division by 5: remainder 1 is Red, 2 Blue, 3 Green, 4 Yellow and 0 White.

A Blue student and a Yellow student add their roll numbers. Which house matches the sum?

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AnswerC · Red

  1. In classes modulo 5: [2]+[4]=[6]=[1][2] + [4] = [6] = [1].

  2. Remainder 1 is Red, whatever the two roll numbers are.

Worth noting

The sum can be checked with any pair, such as 2+4=62 + 4 = 6 or 7+9=167 + 9 = 16: both leave 1.

Question 96

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Which class in Z10\mathbb{Z}_{10} has no multiplicative inverse?

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AnswerD · [4][4]

  1. [a][a] has an inverse modulo 10 exactly when aa and 10 have no common factor.

  2. 4 shares the factor 2 with 10, so 4x4x is always even and never ≡1\equiv 1. The others: 3×7=213 \times 7 = 21, 7×3=217 \times 3 = 21, 9×9=819 \times 9 = 81, all ≡1\equiv 1.

Worth noting

[9][9] is its own inverse, since 81≡1(mod10)81 \equiv 1 \pmod{10}.

Question 97

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How many classes in Z12\mathbb{Z}_{12} have a multiplicative inverse?

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AnswerA · 44

  1. [a][a] has an inverse modulo 12 exactly when gcd⁡(a,12)=1\gcd(a, 12) = 1.

  2. Those are [1],[5],[7],[11][1], [5], [7], [11]: 4 classes (each is its own inverse).

Worth noting

11 counts every non-zero class, but [2][2], [3][3], [4][4] and others share a factor with 12.

NTA's Mathematics / Applied Mathematics syllabus has three sections: Section A1, which every candidate answers, then Section B1 (Mathematics) or Section B2 (Applied Mathematics), of which a candidate answers one. Each section lists units and topics but no chapters, so the chapters here are ours. Where both B sections list a topic that Section A also lists, its questions sit once, in Section A.