Competition · AMC preparation · step 4 of 4

AMC 10 · 2007B · #8

Grade 7 arithmetic

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

digit-constraintsparitysystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: parity

📏 Long solution 💡 3 insights

Problem

A five-digit number has digits of the form bbcac, where the digits satisfy 0 ≤ a ＜ b ＜ c ≤ 9 and b is the average of a and c. Count how many different five-digit numbers fit all of these conditions.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

The question is a pure 'how many' count, so Tool #2 (Make a Systematic List) is the engine: once the real freedom in the problem is found, list and count the valid cases. Tool #4 (Introduce a Variable) is already handed to us as a,b,c; the key is reading b = (a+c)/2 as a rule that ties the three together. Tool #16 (Change Focus) is the crux move: instead of hunting for triples (a,b,c), notice that b is completely decided by a and c, so the whole problem collapses to counting the allowed pairs (a,c) — and the average condition quietly forces a and c to share the same parity.

1STEP 1

Read the digit pattern and its rules

Only three values fill bbcac: a, b, c, climbing strictly, with b the average of a and c. The pattern is fixed; only the digits are free.

digits = b b c a c, 0 ≤ a ＜ b ＜ c ≤ 9, b = (a+c)/2

2STEP 2

See b as the midpoint of a and c

The average (a+c)/2 sits exactly halfway between a and c, so b is their midpoint — pick a and c, and b is forced.

b = (a+c)/2 ⟺ a+c = 2b

3STEP 3

The midpoint forces a and c to match in parity

a + c = 2b is even, and a sum is even only if both terms share the same parity — and then a ＜ b ＜ c and b ≥ 1 come free.

a + c = 2b is even → a ≡ c (mod 2)

4STEP 4

Reframe: just count valid pairs (a, c)

Since b is decided by a and c, the whole count equals the number of same-parity pairs (a, c) with a ＜ c.

#{bbcac} = #{(a,c): a ＜ c, a≡ c (mod 2)}

5STEP 5

Count the both-even pairs

Among the even digits 0, 2, 4, 6, 8, fixing the smaller one in turn gives 4+3+2+1 = 10 pairs.

4+3+2+1 = 10 even pairs

6STEP 6

Count the both-odd pairs and add

The odd digits 1, 3, 5, 7, 9 tally the same 10, so 10 + 10 = 20 numbers — choice (D); ignoring parity leaves the trap 25.

10 (even) + 10 (odd) = 20 → (D)

Answer

Cross-check the total a different way: among all digits 0–9 there are C(10, 2)=45 pairs a ＜ c. A pair is same-parity (a+c even) or opposite-parity (a+c odd). Same-parity pairs are exactly the 20 we counted, leaving 45-20=25 opposite-parity pairs. Only the same-parity pairs give a whole-number average b, so 20 is right — and the leftover 25 is precisely the trap choice (E) you'd land on by ignoring the even-sum requirement. The count is comfortably in the range of the answer choices, confirming (D).

💡Key takeaway

When one value is forced to be the average of two others, their sum must be even, so both are even or both are odd — then you just count the pairs.

Read the digit pattern and its rules
See b as the midpoint of a and c
The midpoint forces a and c to match in parity
Reframe: just count valid pairs (a, c)
Count the both-even pairs
Count the both-odd pairs and add

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