What is the smallest number which, when divided by 6, 8 and 15, leaves a remainder of 2 in each case?

हल

1. 01

   A number that leaves the same remainder 2 under all three divisors is 2 more than a common multiple of all three.

2. 02

   So find the LCM first: $6 = 2 \times 3$, $8 = 2^3$, $15 = 3 \times 5$, giving $\mathrm{LCM} = 2^3 \times 3 \times 5 = 120$.

3. 03

   The smallest such number is therefore $120 + 2 = 122$.

4. 04

   Check: $122 = 6 \times 20 + 2 = 8 \times 15 + 2 = 15 \times 8 + 2$.