How can I find integers which satisfy $\frac{150+n}{15+n}=m$?

Here are some facts about myself:

1. In 2017, I was $15$ years old.
2. Canada, my country, was $150$ years old.

When will be the next time that my country's age will be a multiple of mine?

I've toned this down to a function. With $n$ being the number of years before this will happen and $m$ being any integer,

$$\frac{150+n}{15+n}=m$$

How would you find $n$?

First thing I would do is say that Canada is $135$ years older than you.

That gives you a simpler

$$\frac {135+n}{n} = k\\ \frac {135}{n} = k-1\\$$

It will happen every time your age is a factor of $135.$ It last happened when you were $15.$ It will next happen when you are $27$

You want $\frac{150+n}{15+n}=m$, and clearing denominators gives us $$150+n=(15+n)m.$$ Subtracting $15+n$ from both sides give us $$135=(15+n)(m-1).$$ Now you are looking for the smallest $n>1$ for which such an $m$ exists, so the smallest $n>1$ for which $15+n$ divides $135$.

We want the smallest positive integer $n$ such that there is some (positive) integer $k$ such that $$\frac{150+n}{15+n}=k.$$ Note that $k=1$ can never work, so we can assume $k-1\neq0.$ Now we rearrange the above equation: multiplying both sides by $15+n,$ we get $150+n=15k+nk;$ now rearrange and factorize to get $15(10-k)=(k-1)n;$ and now divide both sides by $k-1,$ to get $$n=\frac{15(10-k)}{k-1}.$$ Since we want the smallest positive integer $n,$ we can just try values of $k\in\{2,3,\ldots,9\},$ starting from the largest and working our way down (because the function of $k$ on the right-hand side is decreasing in this range), until we arrive at an integer value of $n.$* When $k=9,$ $8$ or $7$ we get non-integer values of $n;$ when $k=6$ we find $n=15\times4/5=12.$

So last year, $n$ was $0,$ and the ratio of Canada's age to your age was $k=150/15=10;$ and $11$ years from now, $n$ will be $12,$ and the ratio of Canada's age to your age will be $k=162/27=6.$

*Incidentally, it is not obvious in advance that we will ever get an integer value of $n;$ if this were the case then the problem would simply have no answer. As it happens, the problem does have answers, namely $(n,k)\in\{(0,10),(12,6),(30,4),(120,2)\}.$