Numbers & Place Value

Depth · one concept from first principles · step 2 of 4

Make every prime exponent even

Find the smallest natural number A that makes 0.2‾0.\overline{2} × A equal to a perfect square of a natural number.
Number systemOperations
Your answer
How to solve
Strategy Organize Information in More Ways — A repeating decimal is hard to multiply, so change it to a fraction. Then write the product with letters: the denominator 9 has to cancel and the leftover factor 2 needs a partner to make an even exponent.
1STEP 1

Change the decimal to a fraction

One repeating digit after the point means divide by 9.

0.2‾0.\overline{2} = 2/9
2STEP 2

Write the square condition with letters

We need 2A/9 = m². So 9 divides A; with A = 9B, 2B = m² forces m even and B = 2 × (a square).

A = 9B, 2B = m² → B = 2k² → A = 18k²
3STEP 3

Take the smallest case and check

The smallest choice is k = 1, giving A = 18. Check the product.

2/9 × 18 = 4 = 2²
Answer
18
Smaller candidates fail: A must be a multiple of 9, and A = 9 gives 2/9 × 9 = 2, which is not a perfect square. So 18 is the first that works.
Takeaway

Turn the repeating decimal into a fraction, then pair up every prime factor.

  • Change the decimal to a fraction
  • Write the square condition with letters
  • Take the smallest case and check
Where next?
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▶ Practice — 12 problems