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Geometry Difficulty 4.7 AIME Find the answer Canada

$PQRS$ is a square with side length 60 and centre $C$ . Point $W$ lies on $PS$ so that $WS=53$ . Point $X$ lies on $SR$ so that $XR=40$ . The midpoint of $QR$ is $Y$ . Point $Z$ lies on $PQ$ .

What is the length of $ZQ$ so that the total area of the shaded regions is equal to the total area of the non-shaded regions?

Pick one

Solution

Each of 14 and 21 is a divisor of $n$ .

Since $14=2\times7$ , then each of 2 and 7 is also a divisor of $n$ .

Since $21=3\times7$ , then 3 is a divisor of $n$ (as is 7 which we already noted).

So far, the positive divisors of $n$ are: $1,2,3,7,14$ , and 21.

Since 2 and 3 are divisors of $n$ , then their product $2\times3=6$ is a divisor of $n$ .

Since $2,3$ and 7 are divisors of $n$ , then their product $2\times3\times7=42$ is a divisor of $n$ .

The positive divisors of $n$ are: $1,2,3,6,7,14,21$ , and 42.

We are given that $n$ has exactly 8 positive divisors including 1 and $n$ , and so we have found them all, and thus $n=42$ .

The sum of these 8 positive divisors is $1+2+3+6+7+14+21+42=96$ .

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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.