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 and . 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.
= 5/9 = 50/90, = 8/9 = 80/90
So the numerator x must satisfy 50 ＜ x ＜ 80.
8.NS.A.1Organize Information In More Ways2STEP 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)
If any 3 is left in the denominator, the decimal repeats forever.
8.NS.A.1Identify Subproblemsis a terminating decimal exactly when is a multiple of .
Why?
A decimal that stops after places is a whole number over .
🧱Place-value groupingTen ones make one ten and ten tens make one hundred — each place bundles ten of the last.
Why?
If , then ; the left side has at least two s and has none, so itself must hold .
🧱Every number has one prime factorizationKeep splitting a number into factors and you always land on the same bag of primes, no matter which split you tried first.
Why?
If , then , which does stop.
🧱Scaling top and bottom keeps a fraction's valueMultiplying or dividing a fraction's top and bottom by the same number only renames it — the pieces change size but the amount you hold does not.
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
The next ones, 45 and 81, fall outside the range, so there are 3.
4.OA.B.4Make A Systematic ListAnswer
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 = 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
Where next?
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