This cool puzzle (and solution) can help you challenge your students. It only requires basic knowledge of mathematics. Let’s see who can solve it!

## Challenge:

Find the highest four-digit number that is divisible by each of the numbers 16, 36, 45 and 80.

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The correct answer is9360.

For a positive integer to be divisible by those numbers, it must be divisible by their Least common multiple:

Factorize the numbers to find their Least Common Multiple.

\(16 = 2^4\)

\(36 = 2^2 × 3^2\)

\(45 = 3^2 × 5\)

\(80 = 2^4 × 5\)

Least Common Multiple \(= 2^4 × 3^2 × 5 = 720\)

Every multiple of 720 is divisible by all those above.

Every multiple of 720 can be written as 720n, where n is an integer. The largest 4-digit integer is 9999. Therefore:

\(720n ≤ 9999 → n ≤ \frac{9999}{720} → n ≤ 13.8875 → n = 13 →

720n = 9360\)