 My Math Forum natural number multiple of another number if its digit sum equal to that number

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 June 4th, 2014, 11:46 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 (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 June 5th, 2014, 02:31 AM #2 Senior Member   Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra Let $n=k_ra^r + k_{r-1}a^{r-1} + \cdots + k_1a + k_0$. 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. June 5th, 2014, 08:50 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. Tags digit, equal, multiple, natural, number, sum ### number have multiple and sum equal

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