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Kartik.
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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 -
1. the sum of the digits of the number should be divisible by 9 and vice-versa
Attempt-
Let the number be N
N=a0a1a2a3...an-1an (They do not represent the product but denotes the digits)
As in decimal number system -
N = an.10n+an-1.10n-1+...+a2.102+a1.n1
Where 10n = 9999...9(n times)+1, and [itex]n\geq1[/itex]
Then,
N= [an.99999...9(n times)+an-1.99...9(n-1 times)+...+a2.99+a1.9]+(a0+a1+a2+a3+...+an-1+an)
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.
1. the sum of the digits of the number should be divisible by 9 and vice-versa
Attempt-
Let the number be N
N=a0a1a2a3...an-1an (They do not represent the product but denotes the digits)
As in decimal number system -
N = an.10n+an-1.10n-1+...+a2.102+a1.n1
Where 10n = 9999...9(n times)+1, and [itex]n\geq1[/itex]
Then,
N= [an.99999...9(n times)+an-1.99...9(n-1 times)+...+a2.99+a1.9]+(a0+a1+a2+a3+...+an-1+an)
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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