Question

x^4.y^4.z^3 equal to 49392

then find the values of X Y and Z where X Y and Z are different positive primes

Answer 1

I think it's x^4.y^2.z^3
multiples of prime it can divide also
prime no between 1 to 10
2,3,7,5
there was unit digit 2 so not multiple of 5 because multiple of 5 unit digit 0 and 5
unit digit 2 so 49392 can divided by 2 in 4 time we get 3087
Sum of all digit of 3087 is 18 , 18 is divide by 3 in 2 time we get 343
we know that 7^3=343
so it's not x^4y^4z^3
x^4.y^2.z^3=49392
prime no. are 2,3,7

Answer 2

x^4 * y^2 * z^3 =49392
= 2^4* 3^2* 7^3
Therefore
x=2,y=3, and z=7