Competition · AMC preparation · step 4 of 4
AMC 8 · 2008 · #22
Grade 6 number-theoryPick an answer.
AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The problem stacks two separate conditions onto the same n, so Tool #7 (Break Into Subproblems) says: handle each condition by itself, then combine. Condition 1 (n/3 is a three-digit whole number) pins down a range for n and forces n to be a multiple of 3. Condition 2 (3n is a three-digit whole number) pins down another range for n. Intersecting the two ranges leaves a short list, and Tool #13 (Count Carefully) gives the final count of multiples of 3 in that overlap.
Bound n from the first clue
Turn condition 1 into a range: 100 ≤ ≤ 999, then multiply each part by 3 to get 300 ≤ n ≤ 2997.
Writing the size constraint as an inequality is the Grade 6 "write an inequality for a real-world constraint" move.
6.EE.B.8Identify SubproblemsNote n is a multiple of 3
The same condition also makes a whole number, so n must be a multiple of 3.
Grade 4 multiples: n/3 is a whole number exactly when n is in the list 3, 6, 9, 12, …
4.OA.B.4Identify SubproblemsBound n from the second clue
Turn condition 2 into a range: 100 ≤ 3n ≤ 999, then divide each part by 3 to get 34 ≤ n ≤ 333.
Since n is an integer and 100/3 ≈ 33.3, the smallest integer that works is 34. (3n is automatically a whole number when n is.)
6.EE.B.8Identify SubproblemsIntersect the two ranges
Intersect the two ranges: 300 ≤ n ≤ 2997 and 34 ≤ n ≤ 333 overlap in 300 ≤ n ≤ 333.
Grade 6 inequality reasoning: keep only the values that satisfy every condition.
6.EE.B.5Identify SubproblemsCount the multiples of 3
Count the multiples of 3 in [300, 333]: (333 - 300)/3 + 1 = 12.
Counting evenly-spaced terms is the Grade 4 "generate and analyze a pattern" idea: 12 multiples of 3 fit in this stretch.
The number of multiples of 3 from 300 to 333 is found by taking 333 - 300, dividing by the step of 3, and then adding 1.
▸ Why?
Counting these evenly spaced multiples is the same as counting how many jumps of 3 fit from the first to the last, and that jump-count is (333 - 300) ÷ 3; the extra +1 is there because the very first multiple must be counted too, not only the jumps that come after it.
▸ Why?
Every multiple of 3 in the stretch is 300 plus a whole number of jumps of 3, and matching each multiple to its own jump-count pairs the multiples one-to-one with the whole numbers 0, 1, 2, and on up.
▸ Why?
Starting at 300, which is itself a multiple of 3, and adding 3 again and again lands on each next multiple in turn, and adding 3 a total of k times is just 3 times k, so the multiple is exactly 300 + 3k.
▸ Why?
Each multiple pairs with exactly one jump-count and each jump-count with exactly one multiple, a perfect one-to-one match, so counting the multiples is the same as counting the jump-counts.
▸ Why?
A run of whole numbers that starts at 0 and ends at the top jump-count holds one more number than that top value, because relabeling 0, 1, 2, up to the top as 1, 2, 3, up to one past the top pairs them one-to-one with a plain count that leaves out no endpoint.
Two rules on the same n? Turn each one into its own range, then keep only the n values that fit both — and remember the divisibility rule from the fraction. After that, this AMC 8 problem is just counting multiples of 3 from 300 to 333.
- Bound n from the first clue
- Note n is a multiple of 3
- Bound n from the second clue
- Intersect the two ranges
- Count the multiples of 3
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