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find the remainder when 2222^(5555)+5555^(2222)is divided
by 7
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Sol: the remainder of a^c + b^d when it is divided by x is
same as the sums of rems of a,b to the power c and d.
2222/7 has a remainder of 3 and 5555/7 has a remainder of 4.
So (3)^5555 mod 7 + (4)^2222 mod 7 is the same as
(3)^5 mod 7 + (4)^2 mod 7 = (5 + 2)mod 7 = 0
| Is This Answer Correct ? | 38 Yes | 13 No |
Answer / prakash kumar ojha
3^5555 +4^2222
=3^5*1111 +4^2*1111
=(3^5)^1111 +(4^2)^1111
=243^1111 +16^1111
Which is divisible by 243+16=259
[x^n + y^n always divisible by x+y if n is odd]
But 259 is divisible by 7
Therefore 3^5555 + 4^2222 is divisible by 7
| Is This Answer Correct ? | 18 Yes | 3 No |
Answer / gaurav
kindly explain the last step.
(3)^5 mod 7 + (4)^2 mod 7 = (5 + 2)mod 7 = 0
| Is This Answer Correct ? | 14 Yes | 3 No |
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