AMC 8 · 2003 · #19
Grade 6 number-theoryPick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Three divisibility rules at once feels like three problems, but Tool #7 (Break into Subproblems) splits it cleanly: first collapse the three rules into one by finding lcm(15, 20, 25); then count how many multiples of that LCM land in the open interval (1000, 2000). Once the LCM is in hand, Tool #2 (Make a Systematic List) finishes the job — there are only a few candidates, so listing them is faster and safer than algebra.
Subproblem 1: a number divisible by 15, 20, and 25 is a multiple of their lcm(15, 20, 25) — three checks become one.
One LCM rule replaces three separate divisibility checks. That is the Grade 6 "least common multiple" idea in action.
6.NS.B.4Identify SubproblemsPrime-factorize and take the highest power of each prime: 2² · 3 · 5² = 300.
Highest power of 2 is 2² (from 20); of 3 is 3¹ (from 15); of 5 is 5² (from 25). Multiply: 4 · 3 · 25 = 300.
6.NS.B.4Identify SubproblemsSubproblem 2: the multiples of 300 strictly inside (1000, 2000) are 1200, 1500, 1800.
300 · 3 = 900 is below the range and 300 · 7 = 2100 is above it. Only the values strictly between 1000 and 2000 count.
4.OA.B.4Make A Systematic ListThat is 3 integers inside the interval, so the answer is (C).
A short systematic list makes the count obvious: three multiples fit.
4.OA.A.3Make A Systematic ListSeveral divisibility rules at once is really one LCM rule in disguise. Find the LCM, then list its multiples inside the range and count.