Competition · AMC preparation · step 4 of 4
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.
Merge the three conditions
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.
A whole number is divisible by 15, 20, and 25 all at once exactly when it is a multiple of their least common multiple.
▸ Why?
Being divisible by 15, by 20, and by 25 means the number is a whole count of 15s, a whole count of 20s, and a whole count of 25s, so it appears on all three lists of multiples at the same time, which is exactly what being a common multiple of 15, 20, and 25 means.
▸ Why?
The numbers that sit on all three multiple lists at once are precisely the multiples of the single number where those lists first agree — their least common multiple — and after that first meeting the three lists keep coinciding again at every step of that same size.
Compute the LCM
Prime-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 SubproblemsList the multiples in range
Subproblem 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 ListCount the multiples
That 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.
- Merge the three conditions
- Compute the LCM
- List the multiples in range
- Count the multiples
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