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June 4th, 2014, 10:46 PM  #1 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1  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 (a1) if and only if the sum of its digit are a multiple of (a1) when n is expressed in base a. Thanks in advance 
June 5th, 2014, 01:31 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
Let $n=k_ra^r + k_{r1}a^{r1} + \cdots + k_1a + k_0$. Since $a\equiv1\pmod{a1}$ we have $a^i\equiv1\pmod{a1}$ for all $i=0,1,\ldots,r$. Therefore $k_ia^i\equiv k_i\pmod{a1}$ for all $i=0,1,\ldots,r$. It follows that $n\equiv k_r+k_{r1}+\cdots+k_1+k_0\pmod{a1}$; hence $n$ is divisible by $a1$ if and only if the sum of its base$a$ coefficients is. 
June 5th, 2014, 07:50 AM  #3 
Member Joined: Apr 2014 From: Missouri Posts: 39 Thanks: 1 
Hi Olinguito Thanks...............It was quite easy but i didn't look in this way..............thanks again. 

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