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CEMC Gauss (Grade 7), 2026

Number theory Difficulty 1.6 Junior Find the answer Canada

Asha is asked to cut a $36~\text{m}$ length of ribbon into smaller
pieces so that each piece has equal length, and the ribbon width does
not change. The length of each of the pieces must be a whole number of
metres. If she must make at least $1$
cut, how many possibilities are there for the length of the smaller
ribbons?

Pick one

Solution

Asha could make one cut, cutting the ribbon into $2$ pieces, each with length $\dfrac{36\text{ m}}{2}=18~\text{m}$ .

She could make two cuts, cutting the ribbon into $3$ pieces, each with length $\dfrac{36\text{ m}}{3}=12~\text{m}$ .

Since the length of each of the equal pieces must be a whole number of
metres, then the number of pieces must be a positive divisor of $36$ .

The positive divisors of $36$ that
are greater than $1$ (since there
must be at least one cut and so at least $2$ pieces) are $2$ , $3$ , $4$ , $6$ , $9$ , $12$ , $18$ , and $36$ , and so there are $8$ possibilities for the length of the
smaller ribbons.

The possible lengths of the smaller pieces are $18~\text{m}$ , $12~\text{m}$ , $\dfrac{36\text{ m}}{4}=9~\text{m}$ , $\dfrac{36\text{ m}}{6}=6~\text{m}$ , $\dfrac{36\text{ m}}{9}=4~\text{m}$ , $\dfrac{36\text{ m}}{12}=3~\text{m}$ , $\dfrac{36\text{ m}}{18}=2~\text{m}$ , and
$\dfrac{36\text{ m}}{36}=1~\text{m}$ .

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.