AMC 10 · 2007 · #16

Grade 7 countingnumber-theory

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

paritydigit-constraintscombinations-basicpermutations-basic caseworkcomplementary-countingsystematic-enumeration ↑ Prerequisites: combinations-basic

📏 Long solution 💡 4 insights

Problem

A three-digit number has all different digits and one digit equals the average of the other two. Count how many such numbers there are.

Pick an answer.

(A)

(B)

104

(C)

112

(D)

120

(E)

256

How to solve

Strategy Organize Information in More Ways

Chasing all 900 three-digit numbers is hopeless, and checking three separate "is this one the average?" equations for every digit set is nearly as bad. Tool #15 (Organize Information in More Ways) is the spine: re-describe a qualifying number not by its three digits but by its two outer digits, because once those are chosen the middle one is forced. Getting there needs Tool #13 (Convert to Algebra) to turn the words "is the average of" into an equation, and Tool #14 (Extreme Principle) to test that equation at the smallest and largest digit — which kills two of the three cases and, just as importantly, proves a set can never qualify in two different ways, so nothing gets counted twice. Tool #2 (Make a Systematic List) then counts the outer pairs, and counts them a second time by a different slicing as a check. Finally the count is of numbers, not sets, so Tool #16 (Change Focus / Count the Complement) handles the leading zero by counting all arrangements and deleting the illegal ones, and Tool #3 (Eliminate Possibilities) reads the distractors to confirm which slip each one corresponds to.

1STEP 1

Sort the digits, then write the equation

Sorting first shows only the middle digit can be the average.

0 ≤ x ＜ y ＜ z ≤ 9, and one of x=(y+z)/2, y=(x+z)/2, z=(x+y)/2

2STEP 2

Only the middle digit can be the average

The two outer digits must share a parity.

(y+z)/2 ＞ (x+x)/2=x and (x+y)/2 ＜ (z+z)/2=z ⟹ the condition is exactly y=(x+z)/2

3STEP 3

The two outer digits decide everything

Counting those pairs gives 20 digit sets.

x≡ z (mod 2) ⇔ x+z even ⇔ y=(x+z)/2 is a digit, and then x ＜ y ＜ z

4STEP 4

Count the outer pairs, twice

Each set arranges six ways, minus the ones starting with zero.

C(5, 2)+C(5, 2)=10+10=20, and Σ_m=0⁹min(m, 9-m)=0+1+2+3+4+4+3+2+1+0=20

5STEP 5

Turn sets into numbers, then delete the leading zeros

A second count the same way confirms 112.

20 · 3!-4 · 2!=120-8=112

6STEP 6

Total, and what each wrong choice is

So the count is 112, choice (C).

16 · 6+4 · 4=96+16=112 ⟹ (C)

Answer

112

Two independent tallies of the same numbers agree: 20 · 6-8=112 and 16 · 6+4 · 4=112. Scale is sensible too — three-digit numbers with all-distinct digits number 9 · 9 · 8=648, and only 20 of the C(10, 3)=120 possible digit sets qualify, one set in six, so a total near 648/6≈ 108 is exactly what to expect; 112 sits right there, while 256 does not. Hand-check the two kinds of set: {0,1,2} gives 102, 120, 201, 210 — four numbers, not six, just as the zero rule says; {1,3,5} gives 135, 153, 315, 351, 513, 531 — six. Then test the whole method on a shrunken version of the problem where only the digits 0,1,2,3 are allowed. Same-parity pairs are {0,2} and {1,3}, so the method predicts the sets {0,1,2} and {1,2,3} and a total of 4+6=10 numbers. Listing every three-digit number with distinct digits drawn from 0 to 3 confirms it: 102, 120, 201, 210 and 123, 132, 213, 231, 312, 321 — ten. The method reproduces a case small enough to check completely.

💡Key takeaway

Pick the two outer digits with the same parity and the middle digit is decided for you, so counting these numbers is really just counting pairs, shuffling each set, and dropping the arrangements that start with zero.

Sort the digits, then write the equation
Only the middle digit can be the average
The two outer digits decide everything
Count the outer pairs, twice
Turn sets into numbers, then delete the leading zeros
Total, and what each wrong choice is