<p>Okay, so I'm looking to bump my SAT Math Score from a 630 to at least 660 and hopefully a 760 (go big or go home!!!) I ran into an especially tricky "Level 5" question in prepping for the SAT's Math Section:</p>
<p>A perfect square is an integer that is the square of an integer. Suppose that m and n are positive integers such that mn>15. If 15mn is a perfect square, what is the least possible value of mn?</p>
<p>Now, I know that the answer is 60, but I have no clue how to get it. Any help at all is greatly appreciated!</p>
<p>was this problem a multiple choice question or a fill in? if it was a fill in question, it would be incredibly difficult…</p>
<p>i’m asking because obviously, if it was multiple choice, this question would boil down to some simple calculator crunching.</p>
<p>uh…i guess its a matter of finding perfect squares until one of them is divisible by 15.</p>
<p>16, 25, 36, 49, 64, … goes on until you reach 900 and that’s how you would get 60 for the answer.
this is a really inefficient way of solving it but I can’t think of anything else right now to solve it :</p>
<p>IMO, this question is too hard (or at least too time consuming) and if it shows up on the test, i would save it for last</p>
<p>This question would likely never show up on an SAT test, and it definitely would not be on a grid-in, so you could test out the answers and it would be easy.</p>
<p>15mn = p²
==> sqrt(15mn) = p ∈ ℕ where mn > 15</p>
<p>mn must be a multiple of 15 (or how would it ever be a perfect square?), so check the multiples and find that mn = 60. If the condition was mn ≥ 15, the answer would be mn = 15. Since it has to be >15, the answer is mn = 60, which means 2sqrt(15*15) = p. You can see that it is the least possible alternative.</p>
<p>Honors calculus in university asks us to do stuff like this. It’s annoyingly difficult, but it makes me feel so smart!</p>
<p>Since 15mn is a perfect square and 15 > mn, we can write 15mn = 15²α² Where α > 1. To get the smallest value for mn, make α the smallest integer greater than 1; that is, 2. Then 15mn = 15²<em>2² = 15(15</em>2²), so mn = 15*2² = 60.</p>
<p>15mn is a perfect square, note that 15 = 5*3. Therefore mn must have a factor of 5 and a factor of 3 to make it a perfect square, so 15 divides mn (note that letting mn = 15 results in 15^2).</p>
<p>However mn > 15, so we should multiply by a perfect square so that the resulting number is still a perfect square. The smallest perfect square other than 1 is 4, and 15*4 = 60, which is our answer. Of course we can find integers m and n such that mn = 60 (m = 1, n = 60 for example).</p>
<p>This seems pretty appropriate for a Level 5 problem – it can be solved easily by noting that mn must contain one factor of 3, one factor of 5, and then noting that we must multiply by a perfect square other than 1 to satisfy the mn > 15 constraint.</p>