Months and years

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).