AMC 10 · 2013 · #13

Grade 6 number-theory
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
digit-constraintsdivisibility-rulesdigit-sum caseworksystematic-enumeration ↑ Prerequisites: digit-constraints
📏 Medium solution 💡 2 insights
Problem
Count the three-digit whole numbers that satisfy all three conditions at once: the number is not a multiple of 5, the sum of its three digits is less than 20, and its first (hundreds) digit equals its third (units) digit.

Pick an answer.

(A)
52
(B)
60
(C)
66
(D)
68
(E)
70

AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

The question asks 'how many', so the natural tool is #2 (Make a Systematic List): sort the valid numbers into a few clean cases and count each. To set that up, first use Tool #4 (Introduce a Variable) to name the digits: since the hundreds digit equals the units digit, every number looks like aba, controlled by just two digits a and b. Then Tool #3 (Eliminate Possibilities) throws out the outer digits that break the 'not divisible by 5' rule, leaving a short list of a values. Finally the digit-sum condition becomes a bound on b, and counting the allowed b for each a and adding gives the total.

1STEP 1

Name the digits: the number is aba

The hundreds and units digits must match, so every number we want has the shape aba — fixed by the outer digit a and middle digit b.

aba = 100a + 10b + a, a = hundreds = units, b = tens
2STEP 2

Apply the not-divisible-by-5 rule to a

Divisibility by 5 is read off the units digit, which here is a, so a is neither 0 nor 5: 8 values left, and b is still free.

a ∈ {1,2,3,4,6,7,8,9} (8 values), b ∈ {0,1,…,9}
3STEP 3

Turn the digit-sum condition into a bound on b

The digits a, b, a sum to 2a + b, so the sum condition becomes b ＜ 20 - 2a, with b still capped at 9.

a + b + a = 2a + b ＜ 20 → b ＜ 20 - 2a
4STEP 4

Count the allowed b for each a

For a = 1,2,3,4 the bound clears 9, so all 10 digits b work; a = 6,7,8,9 allow only 8, 6, 4, 2.

a=1,2,3,4: 10 each; a=6:8; a=7:6; a=8:4; a=9:2
5STEP 5

Add the cases to get the total

Add the cases: 4 × 10 = 40 from the small outer digits plus 8 + 6 + 4 + 2 = 20 from the rest, giving 60 numbers.

4 × 10 + (8 + 6 + 4 + 2) = 40 + 20 = 60 → (B)
Answer
60
The eight outer digits split into two groups: a=1,2,3,4 each allow all 10 middle digits, and a=6,7,8,9 allow 8,6,4,2. The second group forms an even, decreasing pattern (8,6,4,2) that makes sense because a larger outer digit eats more of the budget of 20, leaving less room for b. The maximum possible if no sum condition applied would be 8 × 10 = 80 numbers, and our count 60 is below that, which is expected since the sum condition removes some. 60 matches choice (B).
💡Key takeaway

When the first and last digits must match, the number is really just aba; pin down the outer digit with the divisibility rule, turn the sum condition into a limit on the middle digit, then count case by case.

  • Name the digits: the number is aba
  • Apply the not-divisible-by-5 rule to a
  • Turn the digit-sum condition into a bound on b
  • Count the allowed b for each a
  • Add the cases to get the total

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