What is the smallest whole number by which 3645 must be divided so that the quotient is a perfect square?

Class 8Squares and Square RootsSquares and Roots

Step-by-Step Solution

3645 =

3^{6} \times 5

. The six 3s already sit as three complete pairs, so only the lone 5 has to go:

3645 \div 5 = 729 = 2 7^{2}

. Dividing by 3, by 9 or by 45 strips out 3s that were already paired, so the quotient stops being a perfect square.

Answer: A) 5

Key idea: Properties of perfect squares

Useful patterns to spot a perfect square (and reject non-squares):

- **Last digit**: a perfect square ends in 0, 1, 4, 5, 6, or 9 — NEVER in 2, 3, 7, or 8. So 142 cannot be a perfect square (ends in 2).
- **Number of zeros at the end**: must be even. 100 is a perfect square

(1 0^{2})

; 1000 is not.
- **Squares of even numbers are even; squares of odd numbers are odd.**
- **Sum of consecutive odd numbers**: 1 + 3 + 5 + … +

(2 n - 1) = n^{2}

. Try it:

1 + 3 + 5 = 9 = 3^{2}

.

These let you reject many candidates without doing any heavy work.

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