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 $10$.

RektTheNoob  Sep 26, 2017

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)



Guest Sep 30, 2017
edited by Guest  Sep 30, 2017
edited by Guest  Sep 30, 2017

You know, blarney master, you really should cut it off, short --just use initials for your name. BM, seems more apropos, anyway. 

GingerAle  Sep 30, 2017

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