AMC 10 · 2015 · #15
Grade 8 rate-ratioPick an answer.
Nobody wants to carry out this division. The way out is to rewrite the fraction so the denominator is a power of ten, because a fraction over 10^k wears its decimal digits on its face: the digits after the point are just the last k digits of the numerator. That single re-organisation converts a long-division question into a counting question. But the word asked for is minimum, and a minimum is two claims stacked together — one count must work, and every smaller count must fail. The working half comes free from the rewrite. The failing half is the part that actually needs proof, and it hinges on a fact easy to walk past: 123456789 is odd. A decimal can only be shortened by cutting trailing zeros, and an odd numerator leaves none to cut. So the plan is: rewrite, name the digit count as a variable, prove enough, then prove not fewer.
Turn digit count into an integer test
Digit count is a whole-number test.
Multiplying by 10^k slides the decimal point k places right, so if nothing is left past the point, k digits were enough.
5.NBT.A.2Introduce A VariableA minimum needs two proofs
Working once means working for more digits too.
Smallest is a claim about every value below it, so one successful example can never settle it alone.
6.EE.B.5Identify SubproblemsRebuild the denominator as a power of ten
The denominator rebuilds as a power of ten.
Whichever prime is in shorter supply, 2 or 5, is the one you top up to reach a power of ten — and the taller pile sets the exponent.
Whichever of the two primes is in shorter supply gets topped up, and the taller pile sets the exponent.
▸ Why?
Every number has one prime recipe, so the counts of twos and fives are fixed before anything is multiplied.
▸ Why?
A power of ten needs equal piles of both, so the smallest one that works matches the taller pile.
Twenty-six digits are enough
So 26 digits are enough.
Once the denominator is 10²⁶, the digits after the point are just the numerator's last 26 digits.
5.NBT.A.2Organize Information In More WaysThe final digit is 5, not 0
The final digit is odd, so one fewer fails.
The only way to shorten a decimal is to chop trailing zeros, and an odd numerator leaves none to chop.
4.OA.B.4Extreme PrincipleCollect the two halves
Both halves give 26, choice (D).
The digit count is set by whichever of 2 or 5 appears more often in the reduced denominator, never by their sum.
8.EE.A.1Eliminate PossibilitiesGive the fraction a power-of-ten denominator: the number of decimal digits is the exponent you needed, and if the top is odd, not a single digit can be trimmed off the end.
- Turn digit count into an integer test
- A minimum needs two proofs
- Rebuild the denominator as a power of ten
- Twenty-six digits are enough
- The final digit is 5, not 0
- Collect the two halves