Question

The HCF of 56, x and 154 is 14 and their LCM is 4312. Find the smallest possible integer x

Answer 1

Smallest possible x = 98.

Explanation:

4312 = 2 * 2 * 2 * 7 * 7 * 11

56 = 2 *2 * 2 * 7

154 = 2 * 7 * 11

Because the HCF is 14:-

x = = 2 * 7 OR 2 * 7 * other prime number(s).

As there are 2 7's in the prime factors of 4312 there must also be another 7 in the prime factors of x.

x = 2 * 7 * 7 = 98.