AMC 10 · 2011 · #13

Grade 3 counting
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
digit-constraintspermutations-basicparity caseworksystematic-enumeration ↑ Prerequisites: permutations-basic
📏 Medium solution 💡 2 insights
📘 View easy version →
Problem
A three-digit whole number lies between 200 and 700, is even, has digits that are all different, and uses only digits from the set {1,2,5,7,8,9}. How many such numbers are there?

Pick an answer.

(A)
12
(B)
20
(C)
72
(D)
120
(E)
200

AMC 10 2011 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 I build the numbers digit by digit and count the valid choices at each place. The units digit is the tightest constraint (it decides evenness), so I split into cases by the hundreds digit, count each case as choices multiplied together, and add the cases.

1STEP 1

Pin down the hundreds digit

The range allows hundreds digit 2 through 6, but the set allows only 1,2,5,7,8,9 — the overlap is just 2 or 5.

hundreds ∈ {2,3,4,5,6} ∩ {1,2,5,7,8,9} = {2,5}
2STEP 2

Pin down the units digit

A number is even exactly when its last digit is even, and the set's only even digits are 2 and 8.

units ∈ {2,8}
3STEP 3

Case 1: hundreds digit is 2

With 2 used up front, the units digit must be 8, and the tens can be any of 1,5,7,9 — 4 numbers.

1 (units) × 4 (tens) = 4
4STEP 4

Case 2: hundreds digit is 5

With 5 in front, the units digit can be 2 or 8, and 4 digits remain for the tens — 2 × 4 = 8 numbers.

2 (units) × 4 (tens) = 8
5STEP 5

Add the cases

The two cases use different hundreds digits, so they cannot overlap: 4 + 8 = 12, which is choice (A).

4 + 8 = 12
Answer
12
The total 12 is small, which fits: the digit set is tight, the number must be even, and no digit may repeat, so most combinations are ruled out. A quick sanity check on Case 1 — hundreds 2 with units 8 and tens from {1,5,7,9} gives 218, 258, 278, 298, exactly 4 numbers — confirms the counting method, and it matches choice (A).
💡Key takeaway

Lock the digits that face hard rules first (range and even), then count the free choices left and multiply, adding up separate cases at the end.

  • Pin down the hundreds digit
  • Pin down the units digit
  • Case 1: hundreds digit is 2
  • Case 2: hundreds digit is 5
  • Add the cases

More like this

Same archetype — closest grade level first.