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A set of consecutive natural numbers beginning with 1 is written on the board. A student came along and wrote one of these numbers one more time and found the average of the numbers written on the board as 40 20/27.What was the number that was written twice on the board?

27

53

26

60

27

53

26

60

Find the remainder when 37^(1257) *28^3512 is divided by 5.

2

6

8

4

2

6

8

4

2222^7777 + 7777^2222

=(2222^7)^1111 + (7777^2)^1111

The number has to be divisible by 2222^7 + 7777^2( a^n+b^n is divisible by a+b if n is odd)

2222^7 + 7777^2

=1111^2( 2^7*1111^5 + 7^2 )

If we check for 88 and get it right then 44 should also be right, hence 88 cannot be the answer.

We are left with 44, 77 and 99. .

11 divides 1111^2.

So the remaining term should be divisible by either 7,4 or 9

2^3 = -1(mod 9) => 2^7 = 2(mod 9)

1111 = 4(mod 9) => 1111^5 = 4^5(mod 9) = 2^10(mod 9) = -1*2

= -2(mod 9)

2^7*1111^5 = -4(mod 9) and 7^2 = +4( mod 9 )

9 divides ( 2^7*1111^5 + 7^2 )

Hence,99 divides the given number

Let me know if you have any problem

did not get this step.

plz rahulmn.............

My take

2) 99

it's ryt. bt cn u xplain it 4 me?

options are

1)999

2)99

3)88

4)77

5)44

1)999

2)99

3)88

4)77

5)44

2222's power 7777 + 7777's 2222

is divisible by

?

is divisible by

?

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