A number leaves a remainder of 4 when divided by 6, by 9 and by 15, and it is exactly divisible by 11. What is the least such number?

94Correct answerYour answer364Correct answerYour answer814Correct answerYour answer904Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$\operatorname{LCM}(6, 9, 15) = 90$ , so the number has the form $90k + 4$ .
Try $k = 1$ , 2, 3, …: 94, 184, 274, 364, 454, 544, 634, 724, 814, …
The first one divisible by 11 is $814 = 11 \times 74$ $(k = 9)$ .
Check: $814 = 90 \times 9 + 4$ , so it leaves 4 when divided by 6, 9 or 15.

Worth noting

94 satisfies the remainder condition but $94 \div 11$ is not whole. Shortcut: 90 leaves 2 on division by 11, so you need $2k + 4$ to be a multiple of 11, first true at $k = 9$ .

Concept: Least number with a given remainder and an extra divisibility condition
Typical timeTypical: 120 s
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A number leaves a remainder of 4 when divided by 6, by 9 and by 15, and it is exactly divisible by 11. What is the least such number?

Explanation