**Question:**

Find the least number which when divided by 35, 56 and 91 leaves the same remainder 7 in each case.

**Solution:**

Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91.

Prime factorization of 35, 56 and 91 is:

35 = 5 × 7

$56=2^{3} \times 7$

91 = 7 × 13

LCM = product of greatest power of each prime factor involved in the numbers $=2^{3} \times 5 \times 7 \times 13=3640$

Least number which can be divided by 35, 56 and 91 is 3640.

Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.

Thus, the required number is 3647.

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