Question number 4. How many 4-digit numbers are there that are divisible by 5 and all of whose digits are odd in base 10 notation? Let's begin by making an example number like that with 4 digits that I will label here as A, B, C, and D. We have several conditions here. Condition 1 says that all the digits are odd, so A, B, C, and D are elements in the set containing the odd digits, so 1, 3, 5, 7, and 9. And the second condition says that we have divisibility by 5, so that means the last digit has to be divisible by 5. D could be 0 or 5, but in fact it cannot be 0. 0 would make it an even number, so D has to be equal to 5. And then we count. Then we see that for A, B, and C, we have 5 choices, so each one of these 5 numbers here can be used with repetition for A, B, and C, and D can only be set to 5. So together, we multiply these, so 5 times 5 times 5 times 1, so that would be A, B, C, and D. 5 to the power 3rd, or 125 numbers are possible in this setup, and that is answer D, 125.

4-digit numbers

divisible by 5

odd digits

base 10

combinatorics