Latest TCS Aptitude Question SOLUTION: what would be the remainder when (1111)^2222 + (2222)^3333 is divided by 9?

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(#M40030034) TCS question remainder theorem Keep an EYE

what would be the remainder when (1111)^2222 + (2222)^3333 is divided by 9?

Asked In TCS Moitri Paul (10 years ago)

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RAKESH  (10 years ago)

(1111)^2222 + (2222)^3333/9
=(9*123+4)^2222 +(9*247-1)^3333/9
=4^2222+(-1)^3333/9
={(4^3)^740}*4^2 +(-1)/9
=(9*7+1)^740*16 +(-1)/9
=16+(-1)/9
=15/9
so remainder=6

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saurabh agarwal  (10 years ago)

Ans 6.
((1107+4)^2222+(2223-1)^3333)/9
=(4^2222+(-1)^3333)/9
=(((4^3)^740)*4^2+(-1))/9
=((63+1)^740*4^2+(-1))/9
=(1^740*4^2+(-1))/9
=(16-1)/9
rem=6.

Is this solution Helpfull? Yes (6) | No (2)  

Anand Singh  (10 years ago)

Theorem: a^n+b^n is divisible by a+b when n is odd
So,
1111/9 remainder=4
2222/9 remainder=8
Hence the problem reduces to finding the remainder when (4^2222+8^3333)/9
Now using the above stated theorem for (4^2222+8^3333)/9
We get==== [4^2(1111)+8^3(1111)]
Again result we get,
4^2+8^3=528
If 528 had been a multiple of 9 the remainder would be 0,
Since it is not so divide 528/9
The remainder is 6.. ans

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soundarya  (10 years ago)

ans:2 find 2 digit num
(1111)^2222=units digit is 1 and tens digit 1*2 so 21
(2222)^3333=last 2 digits are 44
total 65 when divided by 9 gives 2

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Ankit Gupta  (10 years ago)

(4^3)^746*4^2/9=7
(-1)^3333=-1
7-1/9=rem 6

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