AMC 10 · 2006 · #9

Grade 7 counting
combinations-basicdigit-constraintssystematic-enumeration casework ↑ Prerequisites: combinations-basic
📏 Medium solution 💡 2 insights
Problem
A three-digit even number has digits that strictly increase from left to right. Count how many such numbers there are.

Pick an answer.

(A)
21
(B)
34
(C)
51
(D)
72
(E)
150
How to solve
Strategy Make a Systematic List

A "how many" question with a small finite universe calls for Tool #2 (Make a Systematic List), but listing all 450 even three-digit numbers is not the plan — the plan is to list the right objects. Tool #3 (Eliminate Possibilities) first shrinks the alphabet: no digit is 0, and the units digit is one of only three values. Then Tool #15 (Organize Information in More Ways) does the load-bearing work. A strictly increasing number carries no information beyond which three digits it uses, because their order is already forced, so a number can be re-seen as a set of digits. That re-seeing is the whole problem, and it must be argued in both directions: every valid number gives a digit set, and — the part that is usually asserted rather than shown — every allowed digit set gives back a valid number, exactly one of them. The second direction is where "no digit is 0" earns its keep, because it is what guarantees the rebuilt number really has three digits. Tool #7 (Identify Subproblems) then splits the count by units digit into three non-overlapping cases and adds them.

1STEP 1

Rule zero out of every slot

Strictly increasing digits push zero out of every slot.

a ≠ 0 and a < b < c → 1 ≤ a < b < c ≤ 9
2STEP 2

Pin down the units digit

The last digit must be even and leave room, so only three remain.

c even, c ≠ 0, and c=2 leaves no room for a < b < c → c∈{4,6,8}
3STEP 3

Match numbers with digit pairs

Order being fixed, each number matches a pair of smaller digits.

{abc : 1 ≤ a < b < c} ⟷ { {p,q}⊆{1,2,…,c-1} }
4STEP 4

Count the pairs below each c

Counting the pairs below each gives 3, 10 and 21.

C(n, 2)=(n(n-1))/2; C(3, 2)=3, C(5, 2)=10, C(7, 2)=21
5STEP 5

Add the three cases

Adding gives 34, choice (B).

3+10+21=34 → (B)
Answer
34
Recount by pivoting on the middle digit instead of the last one — a route that uses no subset counting at all. Fix b. The hundreds digit can be any of 1,2,…,b-1, which is b-1 choices, and the units digit can be any even digit larger than b; the two choices are independent, so multiply. b=2: 1 × 3=3 (with c=4,6,8). b=3: 2 × 3=6. b=4: 3 × 2=6 (c=6,8). b=5: 4 × 2=8. b=6: 5 × 1=5 (c=8). b=7: 6 × 1=6. Nothing else survives: b=1 would force a=0, and b ≥ 8 leaves no even digit above b. The total is 3+6+6+8+5+6=34, matching. The choices can also be attacked directly. Every strictly increasing three-digit number, even or odd, corresponds to a three-element subset of {1,…,9}, and there are C(9, 3)=84 of those — so 150 is larger than the entire universe and (E) is impossible on sight. Among those 84, pair each even ending with the odd ending just above it: c=4 against c=5 gives 3 < 6, c=6 against c=7 gives 10 < 15, c=8 against c=9 gives 21 < 28, and c=3 adds 1 more to the odd side with no partner. So the even-ending numbers are strictly fewer than half of 84, that is fewer than 42, which rules out (C) 51 and (D) 72 as well. That leaves (A) 21 and (B) 34, and 21 is exactly the c=8 subtotal on its own — the answer of someone who counted the largest case and stopped.
💡Key takeaway

When digits are forced to increase, the order is already decided — so counting these numbers is just counting which digits to use.

  • Rule zero out of every slot
  • Pin down the units digit
  • Match numbers with digit pairs
  • Count the pairs below each c
  • Add the three cases