AMC 10 · 2005 · #14

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

digit-constraintsparity caseworkconvert-to-algebra ↑ Prerequisites: place-valueparitymean-median-mode-range

📏 Medium solution 💡 2 insights

Problem

A three-digit number has a first digit, a middle digit, and a last digit. Here the middle digit is the average of the first digit and the last digit. How many such three-digit numbers are there?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

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.

1STEP 1

Turn the average into a sum

Call the digits $a$ , $b$ , $c$ . 'Middle is the average' says $b=\frac{a+c}{2}$ , so doubling clears the fraction: $2b=a+c$ .

b = (a+c)/2 ⟹ 2b = a + c

2STEP 2

The middle digit takes care of itself

Pick $a$ and $c$ and $b$ is forced. Since $a+c$ is at most 18, $b$ always lands in 0 to 9; it only fails when $a+c$ is odd.

0 ≤ a + c ≤ 18 → 0 ≤ (a+c)/2 ≤ 9

3STEP 3

Even sum means matching parity

$a+c$ is even only when $a$ and $c$ are both odd or both even, so the task is just counting matching-parity pairs.

4STEP 4

Case 1 — both digits odd

The odd digits 1, 3, 5, 7, 9 are all nonzero, so both ends have 5 choices — that is 25 numbers.

5 × 5 = 25

5STEP 5

Case 2 — both digits even

Even digits are 0, 2, 4, 6, 8, but the front cannot be 0 — 4 choices there, 5 at the end, so 20 numbers.

4 × 5 = 20

6STEP 6

Add the cases

The two cases never overlap, so adding them gives 45 three-digit numbers — choice (E).

25 + 20 = 45

Answer

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.

💡Key takeaway

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.

Turn the average into a sum
The middle digit takes care of itself
Even sum means matching parity
Case 1 — both digits odd
Case 2 — both digits even
Add the cases