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
mysticd:
check the problem again powers of xyz are not correct
no its correct
u see
lenvick has done it correctly
Answers
Answered by
73
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
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
tnks
what said is same
here powers of xyz are wrong
x^4*y^2*z^3 is correct
Answered by
36
x^4 * y^2 * z^3 =49392
= 2^4* 3^2* 7^3
Therefore
x=2,y=3, and z=7
= 2^4* 3^2* 7^3
Therefore
x=2,y=3, and z=7
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