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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?

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8  
Hint: $1575=3^2\cdot 5^2 \cdot 7$. –  Chris Eagle Jul 30 '12 at 20:38
    
MistyD, I think that most of your questions are homework. Please tag them accordingly, if so. That's what the homework tag is for. Don't worry, there are always people around here, that will help you anyway. Thank you! –  draks ... Jul 30 '12 at 20:49
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My choice for the answer is $0$, unless "possible" is a typo for "positive." –  André Nicolas Jul 30 '12 at 21:03
    
@draks I will thanks –  MistyD Jul 30 '12 at 22:46
    
then do so..... –  draks ... Jul 31 '12 at 14:44
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4 Answers 4

up vote 4 down vote accepted

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.

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Thanks for the great tip –  MistyD Jul 30 '12 at 22:47
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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? –  Michael Hardy Jul 30 '12 at 23:43
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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.

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+1 for determine the parity of the power of any one of the primes –  draks ... Jul 31 '12 at 14:42
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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 ;-)

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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.

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