Question

If x, y, and z are positive integers and 3x=4y=7z, then the least possible value of x+y+z is?

A. 33
B. 40
C. 49
D. 61
E. 84

Answer 1

D

Explanation:

3x, 4y, and 7z must be equal to the LCM of 3, 4, and 7 in order to be the smallest value. The LCM is 84 which means x = 28, y = 21 and z = 12. 28 + 21 + 12 = 61.

Answer 2

61

Explanation:

3x=4y=7z

x =4/3 y

x = 7/3 z

Since they have to be integers

y and z must be multiples of 3

y = 7/4 z

Since they have to be integers

z must be multiple of 4

Z must be a multiple of 12

Let z = 12

Then

y = 7/4 *12

y = 21

x = 7/3 *12

x = 28

x+y+z

28+ 21+12

61