I disagree a little with Rich. I think there are 8 solutions.
Here's how I did it.
I am 4-digit number
Let the 4-digit number be "ABCD" where
A = the thousands digit, B = the hundreds digit, C = the tens digit, and D = the
one's or units digit.
If you reverse my digits, I am divisible by 2.
That says DCBA is even. So A is even, either 2,4,6, or 8
My tens digit is three times as great as my thousands digit.
So C = 3A
The only even digit that A can be of these (2,4,6,or 8) that we can multiply
by 3 to get another digit C is 2. So A = 2 and C=6
Therefore the number has the form:
2B6D
and sum of my digit is 15.
2+B+6+D = 15, so
B+D = 7
For that we can have any of these:
0+7 = 1+6 = 2+5 = 3+4 = 4+3 = 5+2 = 6+1 = 7+0 = 7
So we have eight solutions:
1. 2067
2. 2166
3. 2265
4. 2364
5. 2463
6. 2562
7. 2661
8. 2760
Let's see if any of the other clues that we did not use
narrow the list down:
divisible by 3
A number is divisible by 3 if the sum of its digits is divisible by 3. So
"divisible by 3" is redundant (unnecessary) since the sum of the digits is 15
which tells us it is divisible by 3.
If you reverse my digits, I am divisible by 2 as well as by 6.
We used the "divisible by 2" part above. However the "as well as by 6" is
redundant (unnecessary) because the sum of the digits is not changed by
reversing the digits. We already know that ABCD is divisible by 3, so since its
reverse is both even and divisible by 3, so it is automatically divisible by 6.
So all the other information given was unnecessary. Therefore there are 8
solutions.
Edwin
