I took a number, $2012^{2012}$.
Then I cancelled all the $9$'s which appeared in its expansion. I also cancelled all the digits which added up to $9$ viz $1 + 3 + 5 , 2 + 7$ etc. What will be the sum of remaining digits?
Please help.
I took a number, $2012^{2012}$.
Then I cancelled all the $9$'s which appeared in its expansion. I also cancelled all the digits which added up to $9$ viz $1 + 3 + 5 , 2 + 7$ etc. What will be the sum of remaining digits?
Please help.
The answer of your problem is $2012^{2012}\pmod {9}$. Since $2012^{6}=1\pmod 9\implies 2012^{6(335)+2}\pmod 9=2012^2\pmod9=7$. Thus the sum of the remaining digits is $7$.
Here's a method for computing $2012^{2012} \bmod 9$ that avoids doing any divisions (with remainder) of "large" numbers. First, modular arithmetic tells us
$2012^{\displaystyle 2012} \equiv (2012 \bmod 9)^{\displaystyle (2012 \bmod \phi(9))} \equiv (2012 \bmod 9)^{\displaystyle (2012 \bmod 6)} \pmod 9.$
For $2012 \bmod 9$, you can add the digits to get
$2012 \equiv 2 + 0 + 1 + 2 \equiv 5 \pmod 9$
For $2012 \bmod 6$, let us use the Chinese Remainder Theorem on $2012 \bmod 2,3$. Obviously:
$2012 \equiv 0 \pmod 2.$
Modulo $3$ you can also add digits to get
$2012 \equiv 2 + 0 + 1 + 2 \equiv 5 \equiv 2 \pmod 3.$
Combining them with CRT, you get
$2012 \equiv 2 \pmod 6.$
So in total we get
$2012^{\displaystyle 2012} \equiv 5^2 \equiv 25 \equiv 2 + 5 \equiv 7 \pmod 9.$
We know a≡b(mod m)=>$a^n≡b^n(mod\ m)$
$2012≡5(mod\ 9)$ (as in base n, any natural number (mod n-1) => digit sum(mod n-1))
=>$2012^{2012}≡5^{2012}(mod\ 9)$
Using Euler's Totient theorem, $a^{\phi(m)}≡1(mod\ m)$ where (a.m)=1
=> $a^{b\phi(m)+c}≡a^c(mod\ m)$ where b is any natural number.
$\phi(9)=6$ and 2012=2+6.350
$2012^{2012}≡5^{2012}(mod\ 9)≡5^2(mod\ 9)=7$