Numbers & Place Value

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

Count numerators that cancel the denominator's 3s

Look at fractions x/90 that sit strictly between 0.5‾0.\overline{5} and 0.8‾0.\overline{8}. Count how many of them are terminating decimals.
Number systemOperations
Your answer
How to solve
Strategy Organize Information in More Ways — The bounds are repeating decimals, so first rewrite them as fractions over 90; that turns the question into a range of whole-number numerators. Then factor 90 to see which numerators remove the troublesome prime 3, and list them.
1STEP 1

Rewrite the bounds over 90

A single repeating digit d after the point is d/9. Scale both bounds up to denominator 90.

0.5‾0.\overline{5} = 5/9 = 50/90, 0.8‾0.\overline{8} = 8/9 = 80/90
2STEP 2

Find which numerators terminate

Factor the denominator: 90 = 2 × 3² × 5. The 2 and 5 are harmless; the factor 3² = 9 must cancel completely, so x must be a multiple of 9.

x/90 = x/(2 × 3² × 5)
3STEP 3

List the multiples of 9 in range

The multiples of 9 strictly between 50 and 80 are 54, 63, 72.

9 × 6 = 54, 9 × 7 = 63, 9 × 8 = 72
Answer
3
Check each: 54/90 = 3/5 = 0.6, 63/90 = 7/10 = 0.7, 72/90 = 4/5 = 0.8. All terminate and all lie between 0.555… and 0.888…, while a neighbour such as 60/90 = 2/3 = 0.6‾0.\overline{6} does not terminate.
Takeaway

Put everything over one denominator, then just ask which numerators cancel the 3s.

  • Rewrite the bounds over 90
  • Find which numerators terminate
  • List the multiples of 9 in range
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