Maths Olympiad Prep

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, 2026

Number theory Difficulty 1.6 Junior Find the answer Canada

Asha is asked to cut a 36 m36~\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 11
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 22 pieces, each with length 36 m2=18 m\dfrac{36\text{ m}}{2}=18~\text{m}.

She could make two cuts, cutting the ribbon into 33 pieces, each with length 36 m3=12 m\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 3636.

The positive divisors of 3636 that
are greater than 11 (since there
must be at least one cut and so at least 22 pieces) are 22, 33, 44, 66, 99, 1212, 1818, and 3636, and so there are 88 possibilities for the length of the
smaller ribbons.

The possible lengths of the smaller pieces are 18 m18~\text{m}, 12 m12~\text{m}, 36 m4=9 m\dfrac{36\text{ m}}{4}=9~\text{m}, 36 m6=6 m\dfrac{36\text{ m}}{6}=6~\text{m}, 36 m9=4 m\dfrac{36\text{ m}}{9}=4~\text{m}, 36 m12=3 m\dfrac{36\text{ m}}{12}=3~\text{m}, 36 m18=2 m\dfrac{36\text{ m}}{18}=2~\text{m}, and
$36 m36=1 m$.\$\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.