AMC 10 · 2016 · #22
Grade 7 number-theoryPick an answer.
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
All three targets factor into the same primes.
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
Each condition is a maximum of exponents.
lcm stacks each prime to the tallest tower present, so 'lcm' just means 'take the bigger exponent' on every prime separately.
A least common multiple stacks each prime to the tallest tower present, so it just means taking the bigger exponent.
▸ Why?
A common multiple must contain every prime at least as often as either number does.
▸ Why?
Each number has exactly one prime recipe, so the primes never interfere with one another.
Count the powers of 2
One prime allows 5 arrangements.
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
The next allows 3.
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
The last is forced completely.
A max of zero means nobody carries that prime, which immediately freezes every exponent.
6.EE.B.5Make A Systematic ListMultiply the independent counts
Multiplying gives 15, choice (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