calyntanqh calyntanqh
  • 11-07-2020
  • Mathematics
contestada

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

Respuesta :

PollyP52 PollyP52
  • 11-07-2020

Answer:

Smallest possible x = 98.

Step-by-step 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.

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