Algebra/number theory question How do you solve it

Find the smallest positive or zero integer z that makes the quantity
(3 * (z+ 7)^25 )/121 a positive integer.

The answer is 4, but i don't know how to get it, can someone explain? Thanks!

The answer appears to come from the fact that the denominator is the same as 11^2. So, in order for the expression to be a positive integer, the numerator has to somehow produce a value that has 11^2 as part of its prime factorization.

Obviously z = 0 doesn't work (since numerator becomes 3*7^25)
Obviously z = 1 doesn't work (since numerator becomes 3*2^75)
Obviously z = 2 doesn't work (since numerator becomes 3^51)
Obviously z = 3 doesn't work (since numerator becomes 3*(2^25)*(5^25))

But, z = 4 does work, giving us a numerator of 3*(11^25), which is then divisible by 11^2.

121 = 11*11

3 (z+7)^25
---------------
...11*11

must be an integer.

Since 3 and 11 don't have any common factors, we can ignore the 3

(z+7)^25
------------
11*11

If P ia a prime number and it divides into abc...z,
then P must divide one of a, b, c, ... z.

11 is a prime number and it divides into (z+7)(z+7)...(z+7).
So 11 must divide into z + 7.
The easiest solution would be z + 7 = 11,
which gives us z = 4

Letting z = 4, we get (3 * (z+ 7)^25 )/121
= (3 * 11^25 )/11^2
= 3 * 11^23
= 2686290729765712116739593
which is a big integer.