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
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asked Sep 14 “15 at 21:04
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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 $4cdot8cdot8+1cdot9cdot8=328$.
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 $4cdot4cdot8+5cdot5cdot8=328.$
answered Sep 14 “15 at 21:51
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