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# 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.

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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
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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)

~blarneymaster~

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