If p is a prime factor of a perfect square,
then the maximum k for which pk is a factor,
then k MUST be an even number.
We prime factor 2352 with a factor tree:
2352
/ \
2 1176
/ \
2 588
/ \
2 294
/ \
2 147
/ \
3 49
/ \
7 7
So 2352 = 24∙3∙72
The primes 2 and 7 already have even powers,
so we must multiply by 3 so that the 3 will
also have an even power.
So we must multiply by 3:
2352∙3 = 24∙32∙72
The answer is 3. Then the perfect square will be
7056, which is 24∙32∙72
and (22∙3∙7)2 = (4∙3∙7)2 = 842
Edwin