AMC 10 · 2019 · #14
Grade 6 number-theoryPick an answer.
Tool #16 (Complement): list ALL candidate values (divisors of 10¹⁰, which is 11 · 11 = 121) and subtract the few that cannot be written as a product of two distinct divisors of 10⁵. Tool #7 (Subproblems): a product of two divisors has the form 2^a · 5^b with 0 ≤ a, b ≤ 10 — so the question splits into "which (a, b) pairs are reachable?" Tool #9 (Easier Problem): the same kind of question on 100 = 2² · 5² first, then generalize.
Count the candidates
Two exponent ranges give the candidates.
Products live in a 11 × 11 grid of (a, b) exponent sums — start with the whole grid.
6.NS.B.4Identify SubproblemsState the condition in exponents
The two elements must differ.
Translate "distinct divisors" to "distinct (a, b) pairs".
Distinct divisors are exactly distinct pairs of exponents.
▸ Why?
Every number has exactly one prime recipe, so a divisor is nothing but a choice of exponents.
▸ Why?
Each exponent pair names one divisor and each divisor names one pair, so counting either counts both.
Check the smallest product
One needs the same element twice.
Only way to multiply to 1 is 1 × 1, but that's the same divisor twice.
6.EE.B.6Change Focus Count The ComplementCheck the largest product
The largest fails for the same reason.
Top corner of the grid has only one factorization — twin divisors.
6.EE.B.6Change Focus Count The ComplementCheck the extreme exponents
A single extreme exponent is blocked too.
Same forced-pair trap, just on the 2-axis.
6.EE.B.6Change Focus Count The ComplementCheck the other extreme
The other side is blocked by symmetry.
Mirror image of the previous case.
6.EE.B.6Change Focus Count The ComplementCollect the impossible cases
Only four cases are blocked.
Force-the-pair fails only when BOTH coordinates have a unique factorization — and that's only at the four extreme corners.
6.EE.B.6Solve An Easier Related ProblemSubtract to finish
Subtracting gives 117.
Universe minus exceptions — done.
4.OA.A.3Change Focus Count The ComplementThis AMC 12 problem only needs Grade 6 prime-factorization reasoning you already know — every divisor of 10⁵ is 2^a · 5^b, so products live among the 121 divisors of 10¹⁰, and only 4 "corner" values fail the distinct-pair rule. 121 - 4 = 117. The answer is (C).