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During the sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and same number of pencils and crayons, how many of each would you need to buy?

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Hint- In order to find the same number of crayons and pencils, try to solve using L.C.M.

Number of colour pencils to be packed in a packet \[ = 24\]
Number of crayons to be packed in a packet \[ = 32\]
We have to find the L.C.M of $24$ and $32$.
\[
  24 = 2 \times 2 \times 3 \\
  32 = 2 \times 2 \times 2 \times 2 \times 2 \\
\]
L.C.M of $24$ and $ 32 $ \[{\text{ = }}2 \times 2 \times 2 \times 2 \times 2 \times 3 = 96\]
Capacity of $1$ packet of colour pencils \[ = 24\]
So, for 96 pencils, number of packets needed \[ = \dfrac{{96}}{{24}} = 4\]
$3$Now, capacity of $1$ packet of crayons \[ = 32\]
SO, for 96 crayons, number of packets needed \[ = \dfrac{{96}}{{32}} = 3\]
$\therefore $ In order to buy full packs of both and same number of pencils and crayons, we need to buy $4$ packets of colour pencils and $3$ packets of crayons.

Note- L.C.M stands for Lowest Common Multiple. For any two numbers a and b, L.C.M is the smallest positive integer that is divided by both a and b. Hence, whenever you see problems like these, L.C.M is the shortest way to find solutions.