AMC 10 · 2005 · #14
Grade 6 arithmeticHow many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Count the three-digit numbers where the middle digit equals the average of the first digit and the last digit.
Givens: The number has three digits: a hundreds digit, a tens digit, and a units digit.; The tens (middle) digit is the average of the hundreds (first) digit and the units (last) digit.
Unknowns: How many such three-digit numbers exist.
Understand
Restated: Count the three-digit numbers where the middle digit equals the average of the first digit and the last digit.
Givens: The number has three digits: a hundreds digit, a tens digit, and a units digit.; The tens (middle) digit is the average of the hundreds (first) digit and the units (last) digit.
Plan
Primary tool: #2 Make a Systematic List
Secondary: #13 Convert to Algebra, #7 Identify Subproblems, #5 Look for a Pattern
The phrase 'middle is the average' hides a cleaner rule. Turn it into an equation first, discover that only the first and last digits are free, then split the count into two tidy cases by parity and add them.
Execute — Answer: E
6.SP.A.3 Step 1 Turn the average into a sum
- Let the digits be a (first), b (middle), c (last).
- 'The middle is the average of the first and last' means b = (a + c) / 2.
- Multiply both sides by 2 to clear the fraction: 2b = a + c.
💡 The average of two numbers is just their sum split in half, so doubling it gives the sum back.
2.OA.C.3 Step 2 The middle digit takes care of itself
- Once you know a and c, the middle digit is forced: b = (a + c) / 2.
- Since a and c are each between 0 and 9, their sum is between 0 and 18, so b = (a + c)/2 always lands between 0 and 9 — a legal digit.
- The only thing that can go wrong is b not being a whole number, which happens exactly when a + c is odd.
💡 Half of a number is a whole number only when that number is even, so a + c must be even.
2.OA.C.3 Step 3 Even sum means matching parity
- A sum a + c is even only when a and c are both odd or both even.
- So the whole problem reduces to counting pairs (a, c) where the first and last digits share the same parity.
- The middle digit then comes along for free.
💡 Odd plus odd and even plus even both give even; a mismatch gives an odd sum.
3.OA.A.1 Step 4 Case 1 — both digits odd
- The odd digits are 1, 3, 5, 7, 9.
- The first digit can be any of these 5 (all are nonzero, so all are allowed for the leading spot).
- The last digit can also be any of these 5.
- That gives 5 x 5 = 25 numbers.
💡 Every odd first digit can be paired with every odd last digit, so multiply the choices.
4.NBT.A.2 Step 5 Case 2 — both digits even
- The even digits are 0, 2, 4, 6, 8.
- The first digit cannot be 0 (or the number would not be three digits), leaving 4 choices: 2, 4, 6, 8.
- The last digit can be any of the 5 even digits, including 0.
- That gives 4 x 5 = 20 numbers.
💡 Zero is fine at the end but banned at the front, so the first digit loses one option.
4.NBT.B.4 Step 6 Add the cases
- The two cases never overlap (a pair is either both-odd or both-even, not both).
- Add them: 25 + 20 = 45 three-digit numbers.
- The answer is (E).
💡 Separate, non-overlapping cases just get added together for the grand total.
6.SP.A.3 Let the digits be a (first), b (middle), c (last). 'The middle is the average of 2.OA.C.3 Once you know a and c, the middle digit is forced: b = (a + c) / 2. Since a and 2.OA.C.3 A sum a + c is even only when a and c are both odd or both even. So the whole pr 3.OA.A.1 The odd digits are 1, 3, 5, 7, 9. The first digit can be any of these 5 (all are 4.NBT.A.2 The even digits are 0, 2, 4, 6, 8. The first digit cannot be 0 (or the number wo 4.NBT.B.4 The two cases never overlap (a pair is either both-odd or both-even, not both). Review
Reasonableness: There are 9 x 10 = 90 ways to choose a first and last digit (first from 1-9, last from 0-9). Exactly the ones with an even sum work, and even sums are close to half of all pairs — 45 out of 90 is exactly half, which fits and matches the count from the two cases.
Alternative: Instead of parity cases, loop over the middle digit b from 0 to 9 and count valid (a, c) with a + c = 2b and a from 1-9, c from 0-9. Summing those counts also gives 45, confirming the answer.
CCSS standards used (min grade 6)
6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Reading 'the middle is the average of the first and last' as b = (a+c)/2 and rewriting it as 2b = a + c.)2.OA.C.3Determine whether a group of objects has an odd or even number (Seeing that a + c must be even, which forces the first and last digits to share parity.)3.OA.A.1Interpret products of whole numbers as total number of objects in groups (Counting digit pairs by multiplying the choices for the first digit by the choices for the last digit.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Applying the place-value rule that a three-digit number's first digit cannot be 0.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the two case counts, 25 + 20 = 45, for the final total.)
⭐ The middle digit being the average just means the first and last digits add to an even number, so count the matching-parity pairs and add the cases.
⭐ The middle digit being the average just means the first and last digits add to an even number, so count the matching-parity pairs and add the cases.
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