33Finally i got the proof.......
12 Answers
62A thought provoking one...
every single person should try this question... :)
62gr8 :)
actually i din answer this one cos i thought that this was for the users :)
Sorry buddy... but good that u figured out :)
Hint: Σak10k-Σak
33Oh frm now if i would have 2 ask the question as a doubt i will mention it.....
1a no is of the form
a_n10^n+a_{n-1}10^{n-1}+...+a_0=9k+(a_0+a_1+a_2+...+a_n)
for some integer k
so the number is divisible by 9 if the term in the bracket is a multiple of 9 ie if sum of digits is a multiple of 9
1heres the justification for that 9k
a_n10^n+a_{n-1}10^{n-1}+...+a_0=\sum_{i=1}^{n}{a_i(10^i-1)}+\sum_{1=0}^{n}{a_i}
10 \equiv 1mod(9)
so 10^i-1 \equiv0 mod(9)
so it is a multiple of 9
1We can also prove that in this way --
For the sake of simplicity , let the number be abcd , divisible by 9.
So abcd = 0 ( mod 9 )
But abcd = 103 a + 102 b + 10 c + d
= 999 a + 99 b + 9 c + ( a + b + c + d )
= ( a multiple of 9 ) + a + b + c + d
So if abcd is divisible by 9 , then a + b + c + d should be divisible by 9 also .
And the converse is also true .
1hope u know that if a≡b(mod)n implies P(a)≡P(b)(mod)n its easy to prove.,.,.,
a≡b(mod)n =>a^k≡b^k(mod)n multiply bothside by nCk
and take summation to get the result...P(a)≡P(b)(mod)n
now 10≡1(mod)9 =>P(10)≡P(1)mod 9
now in decimal presentation any number P(10)=an10^n+an-110^(n-1)+..........+a0
so P(10) to be divisible by 9 P(1)has to be 0(mod) 9
hence proved