MBA
Exam
Remainder when (2222^5555 + 5555^2222) divided by 7 ? 1) 3 2) 2 3) 1 4) 0
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- Sol:
When 5555 and 2222 are divided by 7 we get the remainders 4 and 3 respectively.
so, problem reduces to find the remainder when 42222 + 35555 is divided by 7
so, 42222 + 35555 = (42)1111 + (35)1111
since, an + bn is divisible by (a + b) when n is ODD
therefore, it is divisible by 42 + 35 = 16 + 243 = 259
since, 259 is divisible by 7,
therefore, 7 will divide the 55552222 + 22225555
so, Remainder = 0
- 10 years agoHelpfull: Yes(2) No(2)
- 2222= 3 mod 7
5555 = 4 mod 7
2222^5555 + 5555^2222 = (3^5)^1111 + (4^2)^ 1111
as it is a^n + b^n and n is odd so it is divisible by
3^5 + 4^ 2 = 259
hence by 7 as 7 is a factor of 259 - 9 years agoHelpfull: Yes(0) No(0)
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