This is another great math challenge for those who love critical thinking challenges. This puzzle doesn’t require advanced *mathematical* knowledge, just logical reasoning. Let’s see how smart you are!

## Challenge:

How many positive integers less than 1000 do not have any 5 as any digits?

**A-** 700

**B-** 728

**C-** 736

**D-** 770

**E-** 819

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The correct answer is B.

Let’s consider 1 digit, 2 digits, and 3 digits numbers separately:

One digit number: From 9 1-digit numbers, eight numbers don’t have any digit as 5.

Two digits numbers: Put A for^{ }\(10^{th}\) place and B for unit place. Then, we have AB as a 2 digits number. We can put 8 digits for A (all digits except 0 and 5) and 9 digits for B (except 5). So, we’ll have 8 × 9 = 72 2-digits numbers that don’t have 5 as any digits.

For three digits numbers, ABC: For A, we can put 8 digits (not 0 or 5), for B, we have 9 digits (not 5) and for C, we have 9 digits (not 5). So, there are 648 three digits numbers that don’t have 5 as any digits. 8 × 9 × 9 = 648

In total: 8 + 72 + 648 = 728