# What is smallest possible integer k such that $1575 \times k$ is perfect square?

I wanted to know how to solve this question:

What is smallest possible integer $k$ such that $1575 \times k$ is a perfect square?

a) 7, b) 9, c) 15, d) 25, e) 63. The answer is 7.

Since this was a multiple choice question, I guess I could just put it in and test the values from the given options, but I wanted to know how to do it without hinting and testing. Any suggestions?

• Hint: $1575=3^2\cdot 5^2 \cdot 7$. Jul 30 '12 at 20:38
• My choice for the answer is $0$, unless "possible" is a typo for "positive." Jul 30 '12 at 21:03
• @draks I will thanks Jul 30 '12 at 22:46

Hint $\rm\, 1575 = 15\!\cdot\! 100\! +\! 75 = 25\,(15\!\cdot\! 4\! +\! 3) = 5^2\cdot 3^2\cdot\color{#C00} 7\:$ lacks only one $\,\color{#C00}{??}\,$ to become square.

Remark $\$ Suppose, instead, that $\rm\:n\:$ is not $\,1575\,$ but is a bigger integer that is difficult to factor completely, but we are given that it is not square. First, we can rule out $\rm\:k = 9\,$ and $\,25,\,$ since multiplying $\rm\:n\:$ by a square does not change the squareness of $\rm\:n.\:$ Since $\,63 = 7\cdot 3^2,\:$ we infer that if $\rm\:63\,n = 3^2\!\cdot\! 7\,n\:$ is a square then so too is the smaller $\rm\:7n.\:$ This leaves only the possibilities $\rm\: k = 7\:$ or $\rm\:15 = 3\!\cdot\! 5.\:$ To determine which $\rm\:kn\:$ is square, we need only determine the parity of the power of any one of the primes $\,3,5,7\,$ in the factorization of $\rm\:n.\:$ For example, in your case, once we have determined that $\rm\:n = 5^2\,j,\ 5\nmid j,\:$ then we know $\rm\:k\:$ must have have an even power of $5$, which excludes $\rm\:k=3\!\cdot\!5,\:$ leaving $\rm\:k = 7\:$ as the only possible solution.

• +1 for determine the parity of the power of any one of the primes Jul 31 '12 at 14:42

Consider the prime factorization of $1575$: $1575=3^2 \cdot 5^2 \cdot 7$. Then $\sqrt{1575}=\sqrt{3^2 \cdot 5^2 \cdot 7}=\sqrt{3^2}\sqrt{5^2}\sqrt{7}=3\cdot 5 \cdot \sqrt{7}$. Clearly the problem here is that $7$ is not a perfect square. So what's the smallest possible integer $k$ that we can multiply $1575$ by so that $1575k$ is a perfect square? Well, we have to fix the problem of $7$ not being a perfect square, so let's multiply by $7$. Is $7$ the smallest number that makes $1575k$ a perfect square? Well, multiplying by $1$ clearly doesn't help, multiplying by $2$ doesn't help since we'll end up with a $\sqrt{2}$ which isn't an integer, etc. Multiplying by $3$ or $5$ won't help since we'd end up with a $\sqrt{3^3}$ or $\sqrt{5^3}$, which does not yield an integer. So we can safely conclude that $7$ is the least integer $k$ such that $1575k$ is a perfect square.

In short, since we can split the square root function apart across the prime decomposition of a number, each distinct prime power must be a perfect square in order for the whole number to be a perfect square. Hope this helps.

• So we want $\sqrt{3^2\cdot5^2\cdot7\cdot k}$ $=\sqrt{3^2\cdot5^2}\cdot\sqrt{7k\ {}}$. The question is then: What is the smallest integer $k$ such that $7k$ is a square? Jul 30 '12 at 23:43

If $1575\times k$ is a perfect square, we can write it as $\prod_k p_k^2$. Since $1$ is not among the choices given, let's assume that $1575$ is not a square from the beginning.

Therefore $9=3^2$ and $25=5^2$ are ruled out, since multiplying a non square with a square doesn't make it a square. $63=9\cdot 7$ is ruled out, since $7$ would be the smaller possible choice.

We are left with $7$ and $15$ and it jumps to the eye that $1575$ is dividible by $15$ twice: $$1575=1500+75=100\cdot 15+5\cdot 15=105\cdot 15=(75+30)\cdot 15=(5+2)\cdot 15^2.$$ So what are you left with...? Good luck ;-)

If you take the prime factorisation of a perfect square, all the exponents come out even, because if $n=p_1^{a_1}\cdot p_2^{a_2} \cdots p_m^{a_m}$, then $n^2=p_1^{2a_1}\cdot p_2^{2a_2} \cdots p_m^{2a_m}$.

Therefore, if you want to multiply $1575 = 3^2\cdot 5^2 \cdot 7$ by some number $k$ to make a perfect square, then the prime factorisation of $k$ will have to include an odd exponent for 7, and an even exponent for any other primes. Clearly, the smallest such $k$ (apart from the trivial $k=0$) would have no primes other than 7 in its prime factorisation, and an exponent of 1 for the prime 7; in other words, $k=7$ is the smallest possible.