What is the smallest positive integer k such that the product 1575 * k is a perfect square?
How does one solve this problem ?
Hello,
First find the prime decomposition of 1575, then multiply by some prime numbers in order to make the final product a product of even powers of prime numbers...
Example :
smallest k such that 450*k is a perfect square.
$\displaystyle 450=2*3^3*5^2$
In order to have an even (hence a square) power of 2, we have to multiply by 2.
In order to have an even power of 3, we have to multiply by 3.
There's no need to do it for 5, because it's already an even power.
So k=2*3=6 is the number we're looking for.
Now try to do it for 1575.
the prime factorisation of 1575 is $\displaystyle 1575 = 3^2 \times 5^2 \times 7$
so
$\displaystyle 1575 \times k = 3^2 \times 5^2 \times 7 \times k$
square numbers have only even powers in their prime factorisation. We need to find the smallest factor k which makes all the powers even. This is (by inspection) 7.
$\displaystyle 1575 \times 7 = 3^2 \times 5^2 \times 7 \times 7 = 3^2 \times 5^2 \times 7^2 = 105^2 $
Edit: Didn't see previous post.
It follows from Fundamental theorem of arithmetic - Wikipedia, the free encyclopedia.