What is the remainder when (25^99 x 4^99)^99 is divided by 11?

What is the remainder when (25^99∗4^99)^99 is divided by 11?

A. 1
B. 3
C. 7
D. 9
E. 10

Binomial expansion of
\((x+y)^n = x^n + C_1^n x^n-1 y + .... C_k^n x^n-k y^k +.....y^n\)

(25^99∗4^99)^99 = (100^99)^99 = ((9*11+1)^99)^99
(9*11+1)^99 = multiples of 11 + 1^99 = 11k+1
((9*11+1)^99)^99 = (11k+1)^99 = multiples of 11 + 1^99 = 11y+1

=> Remainder when (25^99∗4^99)^99 is divided by 11 = 1

IMO A

What is the remainder when (25^99∗4^99)^99 is divided by 11?

\((25^{99}∗4^{99})^{99}\)
= \([(25*4)^{99}]^{99}\)
=\([(100)^{99}]^{99}\)

100=1 [MOD 11]

\(100^{99}\)= \((1)^{99}\) (Mod 11)
\(100^{99}\)= (1) (Mod 11)

\([(100)^{99}]^{99}\)= (1)^{99} (Mod 11)
\([(100)^{99}]^{99}\)= (1) (Mod 11)

Remainder is 1 when \((25^{99}∗4^{99})^{99}\) is divided by 11.

IMO A

What is the remainder when \((25^{99}∗4^{99})^{99}\) is divided by 11?
99*99=9801

So, \(25^{9801}*4^{9801}\)

\(25^1\) /11 => Reminder 3
\(25^2\) /11 => Reminder = 3*3= 9
\(25^3\) /11 => Reminder = 9*3/11 = 5
\(25^4\) /11 => Reminder = 5*3/11 = 4
\(25^5\) /11 => Reminder = 4*3 / 11 = 1
\(25^6\) /11 => Reminder = 1*3 / 11 = 3

Pattern has formed.

Reminder cyclicity will be : 3, 9, 5, 4, 1

9801/5 => Reminder =1 =>\(25^{9801}\)/11 => Reminder should be 3

Similarly,

\(4^1\) /11 => Reminder 4
\(4^2\) /11 => Reminder = 4*4/11= 5
\(4^3\) /11 => Reminder = 5*4/11 = 9
\(4^4\) /11 => Reminder = 9*4/11 = 3
\(4^5\) /11 => Reminder = 3*4 / 11 = 1
\(4^6\) /11 => Reminder = 1*4 / 11 = 4

Pattern has formed.

Reminder cyclicity will be : 4, 5, 9, 3, 1

9801/5 => Reminder = 1 =>\(4^{9801}\)/11 => Reminder should be 4

Finally,
\(25^{9801}/11*4^{9801}/11\) => Reminder = 3*4 = 12/11

So, Ans should be 1.
That's option (A)

\((25^{99}∗4^{99})^{99}\) = \((100^{99})^{99}\)

When 100 is divided by 11 the remainder = 1

i.e. when \((100^{99})^{99}\) is divided by 11 the remainder = \((1^{99})^{99}\) = 1

Answer: Option A

The answer should be 1. Ans A
Explanation:
25 = 5 ^ 2
4 = 2 ^ 2
So, for every 5 there is one 2 available. Hence there multiplication should be 10 raised to power of some number regardless of the exponent used on 25 and 4.
Reducing the scale of the problem from 99 to just 1, 25 * 4 = 100 and Remainder of 100/11 = 1

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