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.