I am still not sure on this answer. I would like someone to help me see the solution to his question. I was working on it for a while and it is the only question that I looked at that I can not answer.

Find all pairs of positive integers $m$ and $n$ where $m<n$ such that the sum of $m$ and $n$ added to the product of $m$ and $n$ is equal to $2014$

I just thought about this question and wanted to know how would you solve something like this. Is there a formula that we can use? Is there a certain way we can do this? Out of curiosity, what would the solution look like? Can someone please show me how to do this?

I know the factors of $2014$ are $2*19*53$ and the factors of $2015$ are $5*13*31.$ I know that $mn+m+n$ $=$ $(m+1)(n+1).$ How would I figure out the integers for $m$ and $n$ though.

  • $\begingroup$ 3/5 of those tags don't apply here $\endgroup$ – ASKASK Sep 15 '14 at 4:19
  • $\begingroup$ I am still stuck on this question. Can someone please help me a little more so I can see it. $\endgroup$ – col Sep 15 '14 at 4:35

You (and Jimmy) have done the hard work, now just finish it off.

You know the factors of 2015, and you know $(m+1)(n+1) = 2015$

All you need to do now is rearrange the equation as follows:

$n+1 = \frac{2015}{m+1}$

$n = \frac{2015}{m+1} -1$

And substitute the factors of 2015 into $(m+1)$ to find $n$.

Eg. $n = \frac{2015}{m+1} -1,\ where\: (m+1) = 5$

$n = \frac{2015}{5} -1$

$n = 402,\ m=4$

Repeat for the other factors where $m<n$.

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Hint: If $mn+m+n = 2014$, then $2015 = mn+m+n+1 = (m+1)(n+1)$.

So $m+1$ and $n+1$ are complementary factors of $2015 = 5 \cdot 13 \cdot 31$.

Can you list all the factors of $2015$?

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  • $\begingroup$ The factors of 2015 are $5^1*13^1*31^1.$ It has $8$ divisors $\endgroup$ – col Sep 15 '14 at 3:50
  • $\begingroup$ The factors of 2014 are $2^1*19^1*53^1$ $\endgroup$ – col Sep 15 '14 at 4:05
  • $\begingroup$ @Col, you're looking for the number of pairs of cofactors of $2015$ which are there of the form $(m+1, n+1)$ where $0<m<n$. Hint: $(1, 2015)$ is not one of them. $\endgroup$ – Graham Kemp Sep 15 '14 at 4:12

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