# Months and Years

The product of Mary's age at her last birthday and her age in complete months is 1800. How old is Mary?

## Problem

The product of Mary's age in years on her last birthday and her age now in complete months is 1800.

How old was Mary on her last birthday?

If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.

## Student Solutions

**Answer**: 12

**Using factors of $1800$**

Let $x$ be Mary's age in years last birthday

Mary's age in months now is $12x+k$ where $k\lt12$

$\therefore x(12x+k)=1800$

So need a factor pair of $1800$ with one factor a bit more than $12$ times the other

$\begin{align} 1800&=2\times900\quad&2\times12=24\\

&=10\times180\quad&10\times12=120\\

&=12\times150\quad&12\times12=144\end{align}$

So it is possible with $x=12$ (and $m=6$)

Check there are no other posibilities:

$1800=15\times120$ (next factor) $15\times12=180$ too big

So Mary is $12$ years old (and $6$ months)

**Using full inequalities**

Let Mary's age now be $y$ years and $m$ months, where $0\le m<12$ (so her age on her latest birthday was $y$ years).

We are told that Mary's age now in months times her age in years on her latest birthday is $1800$. So $$y(12y+m)=1800.$$So, dividing both sides by $12$, $$y\bigg( y+\frac{m}{12}\bigg) =150.$$Expanding, $$y^2 +\frac{ym}{12}=150.$$Therefore, since $\frac{ym}{12}\ge 0$, $y^2\le 150$, so, since $y>0$, $y\le \sqrt{150}=12.247...$. Because $y$ must be an integer, $y\le 12$.

Also, since $\frac{m}{12}<1$, $$y\bigg( y+\frac{m}{12}\bigg) <(y+1)(y+1)=(y+1)^2.$$

$y\bigg( y+\frac{m}{12}\bigg) =150$, so $$(y+1)^2>150.$$

So $$y+1>\sqrt{150}=12.247...$$Since $y$ is an integer, $$y+1\ge 13$$so $$y\ge 12.$$

Therefore since $12\le y\le 12$, Mary was $12$ years old on her latest birthday.

(Since $y(12y+m)=1800$ and $y=12$, $12(12\times 12+m)=1800$, $144+m=150$ so $m=6$ and Mary is now $12$ years and $6$ months old).