Find the sum of all positive integers such that their expression in base $7$ digits is the reverse of their expression in base $16$ digits. Express your answer in base $10$.
When the number is a 1 digit number (1 digit number in base 7)- the numbers 0, 1, 2, 3, 4, 5, and 6
When the number is a 2 digit number=a*7+b=b*16+a / subtract (a+b)
a*6=15*b / divide by 3
2*a=5*b meaning a is divisible by 5 meaning a=5 (because if a=0 then b=0 and then the expression ab is not defined because the digit that represents the largest number in a certain number cannot be 0) meaning b=2. So there is only 1 2 digit number=52 (in base 7)=37 (in base 10)
When the number is a 3 digit number=49*a+7*b+c=256*c+16*b+a /subtract (a+c+7*b)
48*a=255*c+9*b / divide by 3
16*a=85*c+3*b. c cannot be 0 because it means the expression cba is not defined. if c=1-
16*a=85+3*b. there are no solutions to tht equation. if c>1- this means 16*a=85*c+3*b>=170 meaning a>6. But this is not possible! therefore there is no solution.
So the only numbers are 0, 1, 2, 3, 4, 5, 6 and 37. But Wait! didn't i miss something? what about 4 digit numbers? 5 digit numbers? 12 digit numbers? It is not possible for a number with more than 3 digits to maintain the attribute. How do i know? Well, i have discovered a truly remarkable proof of this theorem which this post is too small to contain. Can you prove it ;)?
(the sum of the numbers is 58)