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June 4th, 2014, 10:46 PM   #1
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natural number multiple of another number if its digit sum equal to that number

Hi

I have been trying to do this. What is the algorithm here?

Show that a natural number n is a multiple of (a-1) if and only if the sum of its digit are a multiple of (a-1) when n is expressed in base a.

Thanks in advance
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June 5th, 2014, 01:31 AM   #2
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Math Focus: Abstract algebra
Let $n=k_ra^r + k_{r-1}a^{r-1} + \cdots + k_1a + k_0$.

Since $a\equiv1\pmod{a-1}$ we have $a^i\equiv1\pmod{a-1}$ for all $i=0,1,\ldots,r$.

Therefore $k_ia^i\equiv k_i\pmod{a-1}$ for all $i=0,1,\ldots,r$.

It follows that $n\equiv k_r+k_{r-1}+\cdots+k_1+k_0\pmod{a-1}$; hence $n$ is divisible by $a-1$ if and only if the sum of its base-$a$ coefficients is.
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June 5th, 2014, 07:50 AM   #3
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Hi Olinguito

Thanks...............It was quite easy but i didn't look in this way..............thanks again.
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