AMC 10 · 2017 · #25
Grade 4 countingPick an answer.
AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Checking all 900 numbers one by one is hopeless, so use Tool #16 (Change Focus): instead of testing each number, start from the multiples of 11 and count outward to every number that shares a multiple's digits. There are only 81 such multiples. Tool #2 (Make a Systematic List) then counts, for each multiple, how many valid three-digit rearrangements its digits produce. Tool #7 (Identify Subproblems) sorts the 81 multiples by how many of their digits repeat, because repeats and zeros change the permutation count, and it flags the double-counting from the reversal rule.
Count the multiples of 11
The three-digit multiples of 11 run 110 to 990, giving 81 targets; a number qualifies exactly when its digits shuffle into one of them.
Counting from the rare multiples outward is far easier than scanning all 900 numbers.
4.OA.B.4Count The ComplementA reversal rule that doubles up
Divisibility by 11 tests A-B+C, unchanged when ABC flips to CBA, so a multiple of 11 reversed is still a multiple of 11, sharing its digits.
Flipping a multiple of 11 front-to-back lands on another multiple of 11 with the same digits.
4.OA.A.3Identify SubproblemsSort the 81 by repeated digits
Sort the 81 by repeats: all-equal is impossible (111 isn't a multiple of 11), leaving 17 with a repeated digit and 64 all different.
Repeated digits and zeros each shrink how many real numbers a set of digits can spell.
4.OA.A.3Identify SubproblemsGroup II — a repeated digit, no zero
The no-zero repeats are the 8 palindromes aba (121, 242, …, 979); each spells 3!/2!=3 valid numbers and is its own reverse, giving 24.
Two matching digits cut the six orderings down to three.
4.NBT.B.5Make A Systematic ListGroup II — a repeated digit with a zero
The 9 sets {a,a,0} (110, 220, …, 990) order as aa0, a0a, 0aa, but 0aa is illegal, leaving 2 each for 18.
A leading zero is illegal, so a doubled digit plus a zero gives only two real numbers.
4.NBT.B.5Make A Systematic ListGroup III — all digits different, with a zero
The 8 all-different multiples with a 0 (209, 308, …, 902) give 3!=6 orders minus 2 zero-led = 4 each, halved for reversal pairs: 16.
Three different digits with a zero make four legal numbers, but reversal counts each pair twice.
4.NBT.B.6Make A Systematic ListGroup III — all digits different, no zero
The remaining 56 multiples have three different nonzero digits, each spelling 3!=6 valid numbers, halved for reversal pairs: 168.
Three distinct nonzero digits make six numbers, halved because each multiple's reverse is counted too.
4.NBT.B.6Make A Systematic ListAdd the groups
Every qualifying number lands in exactly one disjoint case, so add: 24+18+16+168 = 226, choice (A).
Separate cases with no overlap simply add up.
3.NBT.A.2Make A Systematic ListWhen few numbers fit a rule, start from those and count outward, adjusting for leading zeros and for shuffles you count twice.
- Count the multiples of 11
- A reversal rule that doubles up
- Sort the 81 by repeated digits
- Group II — a repeated digit, no zero
- Group II — a repeated digit with a zero
- Group III — all digits different, with a zero
- Group III — all digits different, no zero
- Add the groups