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The answer key on the back of Combinatorics A Problem Based Approach by Pavle Mladenovic says it's 512 but I don't think that's right.

Here's my thought process:

A 5 digit number divisible by 5 means the ones digit is 0 or 5

Case last Digit is 5:

1st Digit: (1-9) choices but can't have 5 because it would repeat the last digit so possible choices are 8

2nd Digit: (0-9) choices but can't have 5 and can't have the previous digit so 8 choices

3rd Digit: (0-9) Choices but can't have 5 and the previous digit so 7 choices

4th Digit: (0-9) choices but can't have 5 and previous 3 digits so 6 choices

5th Digit: Must have 5 so 1 choice

Product: 8 * 8 * 7 * 6 * 1 = 2688 possible numbers

Case last Digit is 0:

1st Digit: (1-9) choices so 9

2nd Digit: (0-9) choices but last digit is 0 and cant repeat previous digit so 8 choices

3rd Digit: (0-9) choices but last digit is 0 and can't repeat previous 2 so 7 choices.

4th Digit: (0-9) choices but last digit is 0 and can't repeat previous 3 choices so only 6 choices.

5th Digit Last Digit is 0 so only 1 choice.

Product: 9 * 8 * 7 * 6 = 3024

Total Choices is 2688 + 3024 = 5712.

How do you get 512 from this question?

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    $\begingroup$ It might be just a typo. Accidentally forgetting the 7. I could be wrong, though. $\endgroup$
    – user1135823
    Apr 17, 2023 at 2:58
  • $\begingroup$ To the original poster: Your analysis looks good. $\endgroup$ Apr 17, 2023 at 4:48
  • $\begingroup$ Thanks!! I thought something was strange $\endgroup$ Apr 17, 2023 at 23:48

2 Answers 2

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Let $a_i$ be the $i$th digit of a $5$-digit number divisble by $5$.
If $a_5=5$, then $a_1$ can be $1,\dots,4,6\dots,9$, so there are $8$ choices for $a_1$ in this case.
If $a_5=0$, then $a_1$ can be $1,\dots,9$, so there are $9$ choices for $a_1$ in this case.
Therefore, there are $8+9=17$ chocies for $(a_1,a_5)$ pairs.

Now, $(a_2,a_3,a_4)$ can be any permutation constructed from any digits that are not $a_1$ and not $a_5$.
Therefore, we have to determine the number of permutations of $3$ digits from $10-2=8$ digits, which is $8\cdot 7\cdot 6=336$.

Finally, any $5$-digit number divisible by $5$ can be obtained from any combination of $(a_1,a_5)$ and $(a_2,a_3,a_4)$, and any such combination also results in a $5$-digit number divisible by $5$, so the answer is $17\cdot 336=5712$.

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First and last digit combined: 17 choices. 2nd, 3rd and 4th digits: 8, 7 and 6 choices. 17 times 8 times 7 times 6 = 5712 choices.

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