How many 3 digit numbers contain the digit 3 at least once?

Answer 1

To find the number of 3-digit numbers that contain the digit 3 at least once, we can use the concept of complementary counting. There are a total of 900 three-digit numbers (ranging from 100 to 999). To find the numbers that do not contain the digit 3, we calculate the numbers with digits 0, 1, 2, 4, 5, 6, 7, 8, and 9 in each place. There are 8 choices for each place, so the total number of three-digit numbers without the digit 3 is 8 x 9 x 9 = 648. Therefore, the number of three-digit numbers that contain the digit 3 at least once is 900 - 648 = 252.

Answer 2

172

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Another contributor was amazed:

Wow ! That's a neat little problem.

Let's see . . . . .

-- There are three ways to have a single 3 in the number: 3xx, x3x, and xx3.

For 3xx, there can be 81 different numbers. (2nd and 3rd digits can't be a 3. 9x9=81.)

For x3x, there can be 72 numbers. (1st can't be 0 or 3. 3rd can't be a 3. 8x9=72.)

For xx3, there are also 72. (1st can't be 0 or 3. 2nd can't be a 3. 8x9=72.)

So far, we have 81+72+72 = 225 possibilities, with a single 3 in each one.

-- There are three ways to have two 3s in the number: 33x, 3x3, and x33.

For 33x, there can be 9 different numbers. (3rd digit can't be a 3.)

For 3x3, there can be 9 numbers. (2nd digit can't be a 3.)

For x33, there are 8 numbers. (1st digit can't be a 0 or a 3.)

So we have 9+9+8 = 26 more possibilities.

-- There is only one way to have three 3s in the number: 333.

That's one more possibility.

So the total number of 3-digit numbers where at least one of the digits is a 3 is

225 + 26 + 1 = 252 all together.