How many 3-digit numbers containing at least one digit of 7?

Answer 1

-4

Answer 2

Answer is 199.

The set of three-digit numbers starts at 000 and goes to 999. So we have to find each number with at least one "7" in this set.

In the first hundred (000 to 099), instances of one or more 7s occur 11 times, for:

007

017

027

037

047

057

067

070

077

087

097

This same sequence happens for the 100s, 200s, 300s, 400s, 500, 600s, 800, and 900s (notice I left out 700s). So far, this totals 11 times 9 = 99 instances.

Then we add the number in 700 through 799. Every one of these has at least one digit of 7, totaling 100 numbers.

So, in total, there are 11 * 9 + 100 = 199 instances.

000-099 has 11

100-199 has 11

200-299 has 11

300-399 has 11

400-499 has 11

500-599 has 11

600-699 has 11

700-799 has 100

800-899 has 11

900-999 has 11

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TOTAL = 199

Answer 3

2331