Working in other bases

apple2357

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Something i have been thinking about..

So take the multiple of 9 : 9,18,27,36 etc.. and one thing we notice is that if we keep adding digits up repeatedly they come to 9, so you have a simple divisibility test. Presumably, this works because 9 is one less than the base we are working in which is 10.

Does it follow if we work in other bases, e.g. base 5 and look at multiples of 4 that the digits add up to 4?
It appears so, upon a few test cases? Is this generally going to be true for other bases? Is this a theorem?
 
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Jomo

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Something i have been thinking about..

So take the multiple of 9 : 9,18,27,36 etc.. and one thing we notice is that if we keep adding digits up repeatedly they come to 9, so you have a simple divisibility test. Presumably, this works because 9 is one less than the base we are working in which is 10.

Does it follow if we work in other bases, e.g. base 5 and look at multiples of 4 that the digits add up to 4?
It appears so, upon a few test cases? Is this generally going to be true for other bases? Is this a theorem?
1st of all what you say about digits on the nine table always adding up to 9 is simply not true. Are you saying that the only number in the 900's is 900? After all, any other number on the nine table in the 900s will have the sum of the digits adding up to more than 9. Consider 909, 918, 927,.... They are all on the nine table but their digits do not add up to 9. The divisibility test for 9 is that the sum adds up to a multiple of 9.

Why is this? Consider the integer x = an10n + an-110n-1 + ... + a1101 + a0100. Now we can write ak10k as ak*999...9 + ak. Clearly 9 goes evenly into ak*999...9 for each k. So 9 goes evenly into x precisely when 9 goes in an+an-1+...a0.

Now you try to see if this will work for base 5? Does it work for an arbitrary base b?
 

apple2357

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Ah ok Thanks for that clarification, When i said 'repeatedly' i sort of meant that but it wasn't very clear!
 

Jomo

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Ah ok Thanks for that clarification, When i said 'repeatedly' i sort of meant that but it wasn't very clear!
So did you prove this result in base 5? in base n?
 

apple2357

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apple2357

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Not yet! Still thinking about how to put it together!



Is it as simple as replacing 10n with 5n and breaking up 5nwith 4 and the extra one? in the way you have with 9 above?
In which this will always work?
 

HallsofIvy

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Let b be a positive integer. A number written in "base b" is of the form \(\displaystyle a_nb^n+ a_{n-1}b^{n-1}+ \cdot\cdot\cdot+ a_2b^2+ a_1b^1+ a_0b ^0\). Repeat what Jomo did,
Consider the integer \(\displaystyle x = a_n10^n + a_{n-1}10^{n-1} + ... + a_110^1 + a_010^0\). Now we can write \(\displaystyle a_k10^k\) as \(\displaystyle a_k*999...9 + a_k\). Clearly 9 goes evenly into \(\displaystyle a_k*999...9\) for each k. So 9 goes evenly into x precisely when 9 goes in \(\displaystyle a_n+a_{n-1}+...a_0\)
using b- 1 instead of 9.
 
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