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.

Answer Link

Otras preguntas

What is 9 29/40 as a decimal
Round 936 to the nearest hundred
What is p over 2 equals 3 over 4 plus p over 3
decimals between 9.4 to 10.5 with an interval of 0.1 between each pair of decimals
example of a repeating decimal
Twenty two more than four times a number is less than 82
The Patels took out a 15-year mortgage. How many monthly payments will they have to make on this mortgage?
can a pair of numbers have more than one common factor?
What is 0.42 in a fraction in simplest form
why is radar useful when mapping areas that tend to be covered in clouds