Wow - really nice method, rupsk! And great question. I did it similarly to you but in a slightly different way so I might as well post.

Like you, I knew that the evens were all out, and those ending in 5 were all out, but that some even numbers and some multiples of 5 would also be multiples of 3. Since the 3s were the toughest, I did the 5s and 2s first:

1) There are 101 total numbers in this set, so I'll start with that

2) 51 of them are even (using the property of inclusive sets just like in step 1)

3) I've already counted the multiples of 5 that end in 0, so I only need the 10 that end in 5. So I'm already up to 61 numbers that don't count.

4) NOW it gets pretty interesting. But I've already eliminated the evens and those that end in 5, so I'm looking for the ODD multiples of 3 that don't end in 5. Well, I want to find a pattern so I'll start with:

201 --> and note that, by adding 6, I'll hit all the odd multiples of 3, so
206
213
219
225 ---BUT this one doesn't count...
231
237
243
249 ---> and here I notice a pattern. I go for 4 terms and then skip one because then I'm at a units digit of 5 when I get to:
255 (which doesn't count)

The pattern (which you could easily continue just writing out if you want but if this were a larger set you might want to just use the pattern) is:

2 sets of numbers that count in the first ten
2 set that count in the second ten
NONE in the third
start again

So I know I'll have: 2 + 2 + none (that's 225) + 2 + 2 + none (that's 255) + 2 + 2 + none (that's 285) + 2 (and that's it...that gets us to the 290s). So there are 14 multiples of 3 I need to eliminate.

So my totals are:

51 evens
10 that end in 5
14 that are odd multiples of 3 that don't end in t (using the pattern above)

75 numbers I can't use out of 101, so there are 26 numbers I CAN use, and the answer is E.


I think that's pretty similar to what you did, but maybe a different take if that clicks for people. With multiples/divisibility it's often convenient to look for patterns, so I've trained my mind to work that way and it's really helped...

You had almost the same approach as I did:

Out of the possible unit digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 we can knock out 0, 2, 4, 5, 6, and 8 because they are divisible by 5 or by 2.

This leaves unit digits 1, 3, 7, 9, which would give 4*10 = 40 integers without considering multiples of 3 that end with these digits.

The first digit will always be 2, since we knocked out 300 on the first step.

So, for 200-299, we want to find how many combinations of 2+x+y are divisible by 3, where x is a digit 1-10 and y is 1, 3, 7, or 9.

For units digit 1, 2+x+1 = 3+x is to a multiple of 3 for x = 0, 3, 6, 9
For units digit 3, 2+x+3 = 5+x is to a multiple of 3 for x = 1, 4, 7
For units digit 7, 2+x+7 = 9+x is to a multiple of 3 for x = 0, 3, 6, 9
For units digit 9, 2+x+9 = 11+x is to a multiple of 3 for x = 1, 4, 7

4 + 4 + 3 + 3 = 14 integers between 200 and 299 with unit digits of 1, 3, 7, or 9 are multiples of 3, so removing these integers we have 40 - 14 = 26.