Challenge Level

${2005}\div{12}=167$ remainder $1$.

So, if $167$ of the larger boxes are used $1$ bar will remain.

If $166$ larger boxes are used there will be $13$ bars left (not a mulitple of $5$).

If $165$ larger boxes are used this will leave $25$ bars, which can be packed into $5$ of the smaller boxes.

Total number of boxes is $165+5=170$.

$2005\div5=401$ so could use $401$ small boxes

$12$ boxes of $5$ contain the same number as $5$ boxes of $12$

$401\div12 = 33$ remainder $5$

So $401$ boxes of $5 = 33\times(12$ boxes of $5)$ remainder $5$ boxes of $5$

$ =33\times(5$ boxes of $12)$ remainder $5$ boxes of $5$

$ = 33\times5 + 5 =170$ boxes

This problem is taken from the UKMT Mathematical Challenges.

You can find more short problems, arranged by curriculum topic, in our short problems collection.