Qfyilyi

Here are some facts about myself:

1. Last year, 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 $mathbb{Z}$ being any integer,

$$frac{150+n}{15+n}=mathbb{Z}$$

How would you find $n$?

You want $frac{150+n}{15+n}inBbb{Z}$, so there is some $minBbb{Z}$ such that $frac{150+n}{15+n}=m.$ Then
$$150+n=(15+n)m=15m+mn,$$
and hence
$$0=mn+15m-n-150=(m-1)(n+15)-135,$$
or equivalently $135=(m-1)(n+15)$. Now you are looking for the smallest $n>1$ for which such an $m$ exists, so the smallest $n>1$ for which $n+15$ divides $135$.

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 $27$

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 $k-1neq0.$ 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 $kin{2,3,ldots,9},$ 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 $n=15times4/5=12.$

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 $(n,k)in{(0,10),(12,6),(30,4),(120,2)}.$

Note that if $k mid a$ and $k mid b$, then $k mid a - b$. In this particular instance, we have $15 + n mid 15 + n$ and $15 + n mid 150 + n$, so $15 + n mid 135$. In other words, we are looking for $n+15$ to be the next factor of $135$ which is larger than $15$. Can you continue from here?

Alternatively:

$frac {150 + n}{15 + n} = frac {150+ 10n}{15+n} +frac {-9 n}{15+n}$

$=10 -frac {9 n}{15+n}$ (which is an integer for $n=0$ but when next?)

$=10 - frac {9n + 9*15}{15+n} + frac {9*15}{15+n}=$

$=10 - 9 + frac{3^3*5}{15+n}= 1 + frac{3^3*5}{15+n}$

which is an integer if $15+n$ is one of the factors of $3^3*5$.

And the factors of $3^3*5$ are $1, 3,9, 27, 5,15, 45, 135$.

So this will occur when $n = -14,-12, -10, -6,0, 12,30, 120$

When you are $1, 3, 5, 9, 15, 27, 45, 135$ and canada is $136, 138, 140, 144, 150, 162, 180, 270$ and canada is exactly $136,46, 28, 16, 10,6,4, 2$ as old as you are.

(Enjoy your $45$ birthday when your country annexes my country after we collapse from the thirty year aftermath of the unrecoverable mistakes of the last two years.)

That's a fun problem. It's nice to see other people like to think about these things.