2006 AMC 10A Problem 21

Question

How many four-digit positive integers have at least one digit that is a

2

or a

3?

2439

4096

4903

4904

5416

Count the complement: four-digit numbers with no

2

and no

3

The leading digit has

7

choices; each other digit has

8

There are

9000

four-digit integers. For those avoiding

2

and

3,

the leading digit is one of

\{1, 4, 5, 6, 7, 8, 9\}

(

7

choices) and each remaining digit is one of

\{0, 1, 4, 5, 6, 7, 8, 9\}

(

8

choices):

7 \cdot 8^3 = 3584.

So

9000 - 3584 = 5416

have at least one

2

or

3.

Thus, the correct answer is E.

Problem 21 in other years