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2005 AMC 10A Problem 16

2005 AMC 10A Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Bases & Digits · difficulty 2 of 5 · about 3 min · position P16-20

The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is $6$ . How many two-digit numbers have this property?

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

Answer

Key insight

Subtracting the digit sum from 10a + b leaves 9a, independent of b; the only multiple of 9 up to 81 ending in 6 is 36, so a = 4.

Solution

Write the number as $10a + b$ with tens digit $a$ ( $1 \leq a \leq 9$ ) and units digit $b$ . Subtracting the digit sum:
$(10a + b) - (a + b) = 9a.$
The units digit $b$ drops out entirely, so the result depends only on $a$ .

The possible results are $9, 18, 27, 36, 45, 54, 63, 72, 81$ . Exactly one of these ends in $6$ : $36$ , which needs $a = 4$ .

Every number with tens digit $4$ works, whatever its units digit: $40, 41, \dots, 49$ . That is $10$ numbers.

The answer is $\boxed{\textbf{(D)}\ 10}$ .

Why this works

"Number minus its digit sum" is always $9 \times$ (tens digit) for two-digit numbers, a special case of the fact that a number and its digit sum agree mod $9$ . Because the units digit cancels, the condition constrains only one digit, and the other digit is free, which is where the count of $10$ comes from.

The trap

Stopping at the single tens digit a = 4 and answering 1, or forgetting that all ten units digits 0 through 9 work.

Common mistakes

Stopping at the single tens digit a = 4 and answering 1, or forgetting that all ten units digits 0 through 9 work.
Testing a few numbers like $46$ or $56$ by hand and guessing the pattern without deriving $9a$ .

Techniques

Set up the equation/formula and compute; no special trick needed