AMC 10 · 2018 · #15

Grade 7 number-theory
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
divisibility-rulesdigit-sumdigit-constraintsfundamental-counting-principlemodular-arithmetic identify-subproblemscaseworksystematic-enumeration ↑ Prerequisites: divisibility-rulesfundamental-counting-principle
📏 Medium solution 💡 3 insights
Problem
Count the whole numbers from 100 to 999 that satisfy three conditions at once: the number is odd, the number is a multiple of 3, and none of its three digits is the digit 3.

Pick an answer.

(A)
96
(B)
97
(C)
98
(D)
102
(E)
120
How to solve
Strategy Identify Subproblems

Two of the three conditions live in a single digit slot: 'odd' only touches the ones digit, and 'no digit 3' is one separate ban per slot. Only 'divisible by 3' ties the slots together, and rewriting the number turns that condition into a statement about the digit sum. So I plan to treat each digit slot as its own small counting problem, handle the two picky slots first, and let the last slot absorb the divisibility condition. If that last slot always has the same number of legal digits, the whole count collapses into one multiplication instead of a long case list.

1STEP 1

Name the three digits

Give each digit a name.

N = 100A + 10B + C, A ∈ {1,…,9}, B ∈ {0,…,9}, C ∈ {1,5,7,9}
2STEP 2

Turn divisibility into a digit sum

Turn the multiple rule into a digit sum.

100A + 10B + C = (99A + 9B)_multiple of 3 + (A + B + C)
3STEP 3

Count the hundreds and ones digits

Count the hundreds and ones digits.

A ∈ {1,2,4,5,6,7,8,9} (8 choices), C ∈ {1,5,7,9} (4 choices)
4STEP 4

Sort the tens digits by remainder

The tens digit always leaves three choices.

{0,6,9}, {1,4,7}, {2,5,8} → 3 choices for B in every case
5STEP 5

Multiply the three slot counts

Multiplying the three counts gives 96.

8 · 4 · 3 = 96
Answer
96
Two checks agree. First, the count must be a multiple of 3, because every legal pair (A, C) contributes exactly 3 numbers; that instantly rules out 97 and 98, and it is consistent with 96. Second, compare against the unrestricted count: the odd 3-digit multiples of 3 are 105, 111, …, 999, spaced 6 apart, so there are (999 - 105)/6 + 1 = 150 of them. Banning the digit 3 keeps 8/9 of hundreds digits, 9/10 of tens digits, and 4/5 of odd ones digits, and 150 · 8/9 · 9/10 · 4/5 = 96. Both routes land on 96, which is smaller than 150 as it must be, and 102 and 120 are too big to be a proper subset of a 150-element list after three separate bans.
💡Key takeaway

Fill the fussy digit slots first and save the evenly balanced one for last, because then every case has the same number of options and the whole count becomes one multiplication.

  • Name the three digits
  • Turn divisibility into a digit sum
  • Count the hundreds and ones digits
  • Sort the tens digits by remainder
  • Multiply the three slot counts

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