



 
Hi Peter, Let me ask a different question.
You can answer this by looking at the prime factorization of 15 × 48.
The square root of 2^{4} is 2^{2} and the square root of 3^{2} is 3 but the square root of 5^{1} is not an integer so the square root of 15 × 48 = 2^{4} × 3^{2} × 5^{1} is not an integer. Now what about this question?
Now I can see how to do this. In the prime factorization of 15 × n × 48 all the primes need to have an even power. 15 × n × 48 = 2^{4} × 3^{2} × 5^{1} × n so if n = 5 then 15 × n × 48 = 2^{4} × 3^{2} × 5^{2} and the square root of 15 × n × 48 is an integer. Penny  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 