Sum of factors problem (I think)

Asked by PhiNotPi (12681

) February 21st, 2011

What is L+M+N, if
1) L^3*M^2*N = A+B
2) A^2*B = 1259
3) A, B, L, M, and N are all natural numbers
4) N has to be the lowest possible number that will make the equation work.

There seems to be many conditions on the problem, and I don’t know how to work it out.
This is not homework.

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4 Answers

1259 is prime, so a^2*b has no natural number solutions except 1259,1, right?

a+b = 1259+1 = 1260

I get 9^3+23^2+2 = 1260

n==2

6rant6 (13700

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@6rant6 Looks like you did L^3+M^2+N = 1260, not L^3*M^2*N = 1260. It should be multiplication, not addition. Also, finding N is part of the problem, but I am looking for L+M+N.

PhiNotPi (12681

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let’s see. 1260 factors to 2*2*3*3*5*7, so L must be 1. Largest possible M= 36, N=35?

6rant6 (13700

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@6rant6 Actually, you forgot to square root 36. 36*35 = 1260, so 6^2*35 = 1260, and M = 6. It was only a very simple math error, so GA anyway. That leaves L = 1, M = 6, N = 35, so L+M+N = 42 (the answer to everything).

PhiNotPi (12681

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