AMC 10 · 2005 · #16

Grade 6 arithmetic
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
digit-decompositionplace-value convert-to-algebrasystematic-enumeration ↑ Prerequisites: place-valuemulti-digit-arithmetic
📏 Short solution 💡 2 insights
Problem
Take a two-digit number and subtract the sum of its two digits from it. The result must have units (ones) digit 6. How many two-digit numbers do this?

Pick an answer.

(A)
5
(B)
7
(C)
9
(D)
10
(E)
19

AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Introduce a Variable

Testing all 90 two-digit numbers by hand is slow and easy to botch. Tool #4 names the tens digit a and units digit b and writes the number as 10a+b; then "subtract the digit sum" becomes (10a+b)-(a+b), which collapses to 9a. That single simplification is the whole problem: the result never depends on b at all, only on the tens digit. Tool #2 then just lists the nine multiples of 9 to see which tens digit makes the result end in 6, and the count of matching numbers falls out immediately.

1STEP 1

Name the digits

Let a be the tens digit and b the units digit, so the number is 10a+b and the digit sum is a+b.

number = 10a + b, digit sum = a+b
2STEP 2

Subtract the digit sum

Subtracting gives (10a+b)-(a+b)=9a. The b cancels, so the result depends only on the tens digit a.

(10a+b)-(a+b) = 10a + b - a - b = 9a
3STEP 3

Which tens digit ends in 6

So 9a must end in 6. Among 9,18,27,36,45,54,63,72,81 only 36 does, which needs a=4.

9·1=9, 9·2=18, 9·3=27, 9·4=36, 9·5=45,…,9·9=81
4STEP 4

Count the numbers

The tens digit is pinned to 4 while b stays free, so 40 through 49 all work: 10 numbers, choice (D).

1 tens digit × 10 units digits = 10 → (D)
Answer
10
Spot-check both ends of the decade: 40 → 4+0=4, 40-4=36 (ends in 6), and 49 → 4+9=13, 49-13=36 (ends in 6). Both land on 36, matching the claim that the result is always 9a=36. A different tens digit fails, e.g. 50 → 50-5=45 ends in 5. Because the result ignores the units digit, the answer must be a whole block of ten, so 10 (D) is exactly right; the tempting 9 (C) would come from forgetting one of the ten units digits, and 19 (E) from wrongly thinking two tens digits qualify.
💡Key takeaway

Subtracting a two-digit number's digit sum always leaves 9 times the tens digit, so the units digit is forgotten — find the one tens digit that works and every number in that decade of ten counts.

  • Name the digits
  • Subtract the digit sum
  • Which tens digit ends in 6
  • Count the numbers

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