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Answer: $170$
Starting with boxes of 12
${2005}\div{12}=167$ 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 $165+5=170$.
Starting with boxes of 5
$2005\div5=401$ 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
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Answer: $170$
Starting with boxes of 12
${2005}\div{12}=167$ 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 $165+5=170$.
Starting with boxes of 5
$2005\div5=401$ 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.
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