I have a fun thought to ponder. I thought of this one sleepless night. In base 10, a number is divisible by 9 if the sum of the digits add up to a multiple of 9. This also holds true for 3 which is a factor of 9.

Examining the other number bases it can be found that a number whose digits add up to a multiple of (base-1) are also divisible by that numbers. This also holds true for it's factors.

Question

I have a fun thought to ponder. I thought of this one sleepless night. In base 10, a number is divisible by 9 if the sum of the digits add up to a multiple of 9. This also holds true for 3 which is a factor of 9.

Examining the other number bases it can be found that a number whose digits add up to a multiple of (base-1) are also divisible by that numbers. This also holds true for it's factors.

anonymous · Answer

An example would be base 8. 70 in base 8 is 106. The sum of the digits is 7 which is a multiple of 7. For base 16, 150 is 96 which is a multiple of 15.

Another example in base 16 to prove the factor perspective is 25. This is not divisible by 15. The hex value of 25 is 19. The sum of these digits is 10 which is divisible by 5 which is a factor of 15, i.e. (base-1).

This can be further validified by testing 21 which is not divisible by 5 or 15. 21 in hex is 15. The sum of the digits is 6 which is a multiple of 3.

anonymous · Answer

The question here is why does the rule seem to hold, i.e. is there a proof that shows why this works?