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So, I tried to solve like this:

One's place: 5 (since the number is a multiple of 5 and odd); Ten's place: the rest 4 odd/even numbers + a zero; Hundred's place: the remaining 4 odd/even numbers

So the required number of 3 digit natural odd numbers, which are also a multiple of 5, should be: 4 * 5 * 1 = 20

But the book I use says it should be 12.

and a quick google search also told me it should be 12. (which I am not convinced is the right answer)

Like this:

number of ways of filling unit place is only one i.e. 5. Now, four odd digits are left, hence ten’s place can be filled in four ways and hundred’s place in three ways. number of required three-digit natural numbers is 1 x 4 x 3 = 12

So now, I understand up to the part that it says filling up the ten's place with the four remaining odd digits. After that it says that "hundred's place can be filled in three ways." But there are five digits remaining? (4 even numbers + a zero)

Now, to my question. Did I get the right answer(20)? Or am I missing something here?

Also, I have seen the other related posts here but they seem to lead into a completely different answer, like 64 [8(hundred's) * 8(ten's) * 1(one's)]

Thank you, in advance.

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  • $\begingroup$ Since 5 has to be the last digit, there are 4 digits left 1,3,7,9. All the combinations from these numbers are $\binom{4}{2}$=6. Every combination has two possible permutations, so the total number we want is 2x6=12 $\endgroup$
    – user1054388
    Commented Jun 30, 2022 at 17:12

1 Answer 1

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You can think of it as a permutation problem:

A three digit number that is a multiple of 5 and is odd must end with 5, since it can't end in 0. So you need all possible 2-permutations for digits (1,3,7,9). In this way you get the first 2 digits of the number we are intested in, sticking the 5 in the last position, as you said.These are 4 *3 possible permutation, In this way you are sure you use only odd digits for the permutations, as the ex. requires

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    $\begingroup$ Oh! Now I get it, the question asked all the digits to be odd and not the number itself (which is by default odd with 5 being in the one's place) well that was kinda dumb of me... Thank you so much, been banging my mind for too long on this. $\endgroup$
    – Rex
    Commented Nov 19, 2021 at 8:20
  • $\begingroup$ It is just (excluding five, last digit) $ \,2\binom{4}{2}$=12 $\endgroup$
    – user1054388
    Commented Jun 30, 2022 at 19:24

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