If m and n are positive integers such that sqrt((50+m)n) is

Question

If m and n are positive integers such that sqrt((50+m)n) is an integer, what is the smallest possible value of m+n ?

6 8 10 12 14

eschn3am · Answer

I don't know any short way to do this. List of perfect squares:

100
121
144
169
196
225
256
289
324
361
400
441

Now we're looking for a number that's close to a multiple of 50, and the difference between the number and the multiple of 50 is divisible by whatever you need to multiply by to get to the number. I know this is the most confusing sentence written in the English language, so here's an example.

324 is 24 away from 300
300/50 = 6
324-300 = 24 which is divisible by 6 (24/6=4)

SO...54*6=324

4+6=10

This eliminates 12 and 14 as being correct, and I don't see any obvious examples for 6 or 8...so I'd put down C and move on

I'd love to have a faster approach if anyone has one though 8-)

GMATBLACKBELT · Answer

sqrt((50+m)n) I just used the answer choices. I did these in my head so it wasnt too hard.

Could find any for 6 or 8. I then just tried values for 10. 9,1 Nope 8,2 no 7,3 no 6,4 --> (54)6) --> 6*9*6

we have 6^2 and 3^2 --> this is an integer.

There is def a much better approach, I may have just gotten lucky here and quickly found a value. Id like to see a more systematic approach.

C

maratikus · Answer

obviously, n doesn't have a square as its divisor. For example, n = 12 can't give the smallest possible value of m+n because if n = 12 works, then n = 3 also works.

Therefore, 50+m = n*s^2

n = 1: 50 + m = s^2: s = 8 -> m = 14 //m+n = 15
n = 2: 50 + m = 2*s^2: s = 6 -> m too large (m+n >15)
n = 3: 50 + m = 3*s^2: s = 5 -> m too large (m+n > 15)
n = 4: skip - non-optimal
n = 5: 50 + m = 5*s^2: s = 4 -> m too large
n = 6: 50 + m = 6*s^2: s = 3 -> m = 4 (m+n = 10)
n = 7: 50+m= 7*s^2: s = 3 -> m too large (m+n > 10)
n = 8: skip - non-optimal
n = 0: skip - non-optimal
n = 10: m+n > 10

C is the answer.

incognito1 · Answer

(C)

This is a little plug and play, though I'm sure theres a proof to it.

We're looking for a number that will make   a perfect square, and want (m+n) to be minimal at the same time. Let's ignore n for now.

What will make (50+m) closest to a perfect square? ('n' will later complete the perfect square, btw)

If you plug in the first few values for m, you see 54 = 2 * 3 * 3 * 3
Multiplying this by 2*3 yields a perfect square (324 = 18*18). Resulting in (m+n) = (4+6) = 10.

As you increment the value of m, you realize nothing beats a total of 10. When you're at 59 and it doesn't work, you select (C) as your answer and submit.

srp · Answer

used the same approach.

sqrt(50n+mn) = sqrt(50*6 + (4*6))= sqrt(324)

4+6=10

If m and n are positive integers such that sqrt((50+m)n) is

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