AMC 10 · 2005 · #11

Grade 6 arithmetic
digit-constraintsparity caseworkconvert-to-algebra ↑ Prerequisites: place-valueparitymean-median-mode-range
📏 Medium solution 💡 2 insights
Problem
In a three-digit number the middle digit equals the average of the first and last digits. Count how many such numbers there are.

Pick an answer.

(A)
41
(B)
42
(C)
43
(D)
44
(E)
45
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

Clearing the fraction turns the condition into a sum.

b = (a+c)/2 ⟹ 2b = a + c
2STEP 2

The middle digit takes care of itself

The middle digit is then forced and always legal.

0 ≤ a + c ≤ 18 → 0 ≤ (a+c)/2 ≤ 9
3STEP 3

Even sum means matching parity

It is whole exactly when the two outer digits share parity.

4STEP 4

Case 1 — both digits odd

Both odd gives 25 numbers.

5 × 5 = 25
5STEP 5

Case 2 — both digits even

Both even gives only 20, since the lead cannot be zero.

4 × 5 = 20
6STEP 6

Add the cases

The cases do not overlap, so the count is 45, choice (E).

25 + 20 = 45
Answer
45
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