# Solution: How Many Positive Integers Less Than 1000 (Counting)

I am not looking for an answer on this. Just need to clarify why my approach is failing –

\$N_1 + N_2 + N_3\$, i.e. single digit, double digit, 3 digit

single \$= 2, 4, 6, 8\$, i.e 4

double = X non-zero \$= 8 cdot 4 = 32\$ X zero \$= 9 cdot 1 = 9\$

Now the confusing part three digit, breaking into 4 cases

X zero zero \$= 9\$

X nz nz \$= 7 cdot 8 cdot 4 = 224\$

X z nz \$= 8 cdot 1 cdot 4 = 32\$

X nz z \$= 8 cdot 9 cdot 1 = 72\$

Three digit total comes to \$= 9 + 224 + 32 + 72 = 337\$. This answer is wrong and it should be \$328\$. What am I missing in the logic? Please suggest.

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edited May 6 “18 at 21:57

N. F. Taussig
asked Sep 14 “15 at 21:04

totalmokshtotalmoksh
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(I was going to add this as a comment, but it turned out to be too long.)

Here are two alternate ways to handle the 3-digit case:

a) If the last digit is not zero, there are 4 choices for the last digit, 8 choices for the first digit, and 8 choices for the middle digit.

b) If the last digit is zero, there are 9 choices for the first digit and 8 choices for the middle digit.

This gives a total of \$4cdot8cdot8+1cdot9cdot8=328\$.

Alternatively,

1) If the first digit is even, there are 4 choices for this digit, 4 choices for the last digit, and 8 choices for the middle digit.

2) If the first digit is odd, there are 5 choices for this digit, 5 choices for the last digit, and 8 choices for the middle one.

This gives a total of \$4cdot4cdot8+5cdot5cdot8=328.\$

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answered Sep 14 “15 at 21:51

user84413user84413
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