Hi! Thank you guys for responding. I solved it! I did it like this:
The base 7 number can be "abc_7", and a, b, and c are all digits from 1 to 6, except for b which can also be 0. Then you can expand it to be 49a + 7b + c. Then when it is reversed, you get cba. expanding in base 9, cba_9= 81c+ 9 b + a. The two expressionss are equal, so you can use the equation 49a + 7b + c = 81c + 9b + a. Then you move everything to one side and get 80c + 2b - 48a = 0, and you can simplify and get 40 c+ b- 24a = 0. Then we can solve for b and get b = 24a - 40c = 8(3a - 5c). This means that b is divisible by 8, but it is also a base 7 digit so it has to be 0. then you have 3 a = 5c sp a is divisible by 5 and c is divisible by 3. They cant be 0 because they are both left digits, so they are 5 and 3. This means that the answer is 503_7.
Hooray! Thank you again for helping me :)
Hello! I am new to this website and I am not the best at math, but I like helping people so I will try to help. Please tell me if I get it wrong, and don't be mad. Thanks!
My thinking is that there are exactly five possible even digits, and we can first take a look at the possible 4-digit numbers where the first digit is even. There are 4 choices for the first one because it can't be 0, and 10 for each one following. This gives 4*10*10*10=4000. We can do the same for when the even digit is in the hundereds, tens, and ones place, then add them all up: 4000+5000+5000+5000=19000. So the answer is 19000. Again, I might not be correct. :)