AMC 10 · 2016 · #25
Grade 7 number-theoryPick an answer.
AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
An lcm condition on whole numbers is really three separate conditions hiding inside one, because lcm works prime by prime: the power of 2 in the answer depends only on the powers of 2 in the inputs, and likewise for 3 and 5. So tool #7 (Identify Subproblems) splits the hard count into three independent prime puzzles. Tool #4 (Introduce a Variable) names the exponent of each prime in x, y, z and turns each lcm into a clean 'the larger of two exponents equals this' statement. For each prime the allowed exponents form a tiny set, so tool #2 (Make a Systematic List) counts the valid exponent triples directly. The three counts multiply because the primes are chosen independently.
Factor the three targets
Prime-factor each target: 72=2³·3², 600=2³·3·5², 900=2²·3²·5². The only primes anywhere are 2, 3, and 5.
A number is fully described by how many of each prime it contains, so factoring lays out exactly what we are counting.
4.OA.B.4Identify SubproblemsTurn each lcm into a max of exponents
Write each prime's exponents in x,y,z; since an lcm exponent is the larger of the two inputs, primes 2, 3, 5 solve independently.
lcm stacks each prime to the tallest tower present, so 'lcm' just means 'take the bigger exponent' on every prime separately.
6.NS.B.4Use Matrix LogicCount the powers of 2
For prime 2, e_x=3 is forced, and max(e_y,e_z)=2 with both ≤2 gives pairs (2,0),(2,1),(2,2),(1,2),(0,2) — 5 choices.
Once one variable is pinned to the top value, you just list the small pairs whose larger entry hits the required height.
6.EE.B.5Make A Systematic ListCount the powers of 3
For prime 3, e_y=2 is forced, and max(e_x,e_z)=1 with both ≤1 gives pairs (1,0),(0,1),(1,1) — 3 choices.
The same pattern repeats: a low cap pins one variable to the top, then you list the small pairs reaching the smaller height.
6.EE.B.5Make A Systematic ListCount the powers of 5
For prime 5, max(e_x,e_y)=0 forces e_x=e_y=0, then e_z=2 — everything is pinned, so exactly 1 choice.
A max of zero means nobody carries that prime, which immediately freezes every exponent.
6.EE.B.5Make A Systematic ListMultiply the independent counts
The three prime choices are independent, so by the multiplication principle the total is 5×3×1=15 — answer (A).
Independent choices stack by multiplying, so the three prime counts combine into one product.
7.SP.C.8Identify SubproblemsBecause lcm just takes the bigger exponent on each prime, split the problem into separate prime puzzles, count 5, 3, and 1 ways, then multiply to get 15.
- Factor the three targets
- Turn each lcm into a max of exponents
- Count the powers of 2
- Count the powers of 3
- Count the powers of 5
- Multiply the independent counts