Question - A proof of divisibility by the number 9 with a assumed number. The aim is to prove two assertions state within divisibility rule of 9 -(adsbygoogle = window.adsbygoogle || []).push({});

1. the sum of the digits of the number should be divisible by 9 and vice-versa

Attempt-

Let the number be N

N=a_{0}a_{1}a_{2}a_{3}..........a_{n-1}a_{n}(They do not represent the product but denotes the digits)

As in decimal number system -

N = a_{n}.10^{n}+a_{n-1}.10^{n-1}+......+a_{2}.10^{2}+a_{1}.n^{1}

Where 10^{n}= 9999.....9(n times)+1, and [itex]n\geq1[/itex]

Then,

N= [a_{n}.99999....9(n times)+a_{n-1}.99....9(n-1 times)+.....+a_{2}.99+a_{1}.9]+(a_{0}+a_{1}+a_{2}+a_{3}+..........+a_{n-1}+a_{n})

The first summand is definitely divisible by 9 but i am stuck over here to prove the remaining assertions, either verbally or i'm missing some computation too.

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# Proof of divisibility(9)

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