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What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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Updated on: 20 Jul 2019, 05:31
(100^99)^99 divided by 11 yields what remainder?
\(\frac{10^1}{11}\) remainder 10
\(\frac{10^2}{11}\) remainder 1
\(\frac{10^3}{11}\) remainder 10
\(\frac{10^4}{11}\) remainder 1
Cyclicity is 2. When exponent is odd, remainder is 10, when exponent is even, remainder is 1. 10^2*99*99 will yield even number, thus remainder will be 1 (A)
Originally posted by mira93 on 19 Jul 2019, 22:11.
Last edited by mira93 on 20 Jul 2019, 05:31, edited 1 time in total.



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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19 Jul 2019, 23:33
What is the remainder when (25^99∗4^99)^99 is divided by 11?5 to any power is always 5 in units. 4 to odd power is 64 at the end. 64*5=320 320/11 ~ reminder is 10. A. 1 B. 3 C. 7 D. 9E. 10E is the answer.
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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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19 Jul 2019, 23:39
What is the remainder when \((25^{99}*4^{99})^{99}\) is divided by 11?
\((25^{99}*4^{99})^{99} = ((25*4)^{99})^{99} =(100^{99})^{99} =100^{(99*99)}\) Now, when we divide 100/99, 1000/999, 10000/9999.... we always get 1 as remainder, and we can see than 99, 999, 9999 .... is always divisible by 11
Therefore answer is A



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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19 Jul 2019, 23:43
The product 25^99 * 4^99 will result in 100^99.
Number 1 followed by any number of zeros, when divided by 11 will result in a remainder of 1. Eg: 100/11 remainder will be 1, 1000/11 remainder will be 1 and so on.
Hence correct answer A



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 00:08
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 = 100^198 we are to find (100^198)/11 (100^198)/11 = ((99+1)^198)/11 = (99^198)/11 + ((99^197)(1))/11 + ((99^196)(1^2))/11+...+ ((99)(1^197))/11 + (1^198)/11 Since all the terms in the above expression are divisible by 11 with the exception of the last term, 1/11, we know that the remainder is 1.
The answer therefore is A.



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 00:13
(25^99*4^99)^99
(100^99)^99
100 divided by 11 gives the remainder 1
(\(\frac{100}{11}\)R = 1)(\(\frac{100}{11}\)R1)(\(\frac{100}{11}\)R1) (we will do this operation 99 times)
=1*1*1*........99 times = 1^99 = 1 again we will divide this remainder by 11 and get the remainder 1 so 100^99)/11 remainder = 1 now for power outside bracket (1)^99 = 1 divided by 11 gives the remainder 1
(remainder theorem =\(\frac{A*B*C}{M}\) will be the same as \(\frac{(remainder of A) *(remainder of B)*(remainder of C)}{M}\)
A is the answer



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What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 00:34
To find,
Remainder when (25^99 * 4^99)^99 divided by 11
(25^99 * 4^99)^99 => 100^2801 => (99 + 1)^2801
As per binomial theorem, all the terms of this expansion will have a 99 except the last term which will be 1^2801
Hence this would be come,
=> 99K + 1^2801
This divided by 11 will clearly leave a remainder of 1.
Answer: A



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 00:38
What is the remainder when (25^99∗4^99)^99 is divided by 11?
We can write (25^99∗4^99)^99 as following;
(25^99∗4^99)^99 = (100^99)^99 = ((10^2)^99)^99 = (10^198)^99 = 10^(198*99)
When 10^a (a is one of the whole numbers) is divided by 11, the remainder is as per the following:
Remainder when 'a' is odd = 10 Remainder when 'a' is even = 1
Here, (198*99) is an even power. So, the remainder is '1'.
The remainder when (25^99∗4^99)^99 is divided by 11 is '1'.
ANSWER : A



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 00:41
Simplify exponents: if exponents are the same, multiply base 25*4=(100^99)^99. Now find pattern when 100/11 100/11  remainder 1 (even number of zeros) 1000/11  remainder 10 (odd number of zeros) 10000/11  remainder 1 and so on 100^99)^99 (odd number of zeros) will have remainder of 10, E



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 01:17
(25^99 x 4^99)^99= ((99+1)^99)^99)/11= (((1)^99)^99)/11= 1/11 1 reminder Ans A = 1
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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 03:00
Quote: What is the remainder when \((25^{99}∗4^{99})^{99}\) is divided by 11?
A. 1 B. 3 C. 7 D. 9 E. 10 \((25^{99}∗4^{99})^{99}\) = \((100^{99})^{99}\) 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
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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 03:18
x = (25^99*4^99)^99 > (5^198*2^198)^99 >((5*2)^198)^99 > (10)^(198*99) > (10)^even so, if we take even number as 2, then 10^2 = 100, the remainder when 100 is divided by 11 is 1
Answer is A



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 03:38
\((25^{99}*4^{99})^{99} = (5^{198}*2^{198})^{99} = (10^{198})^{99} = 10^{198*99}\)
We need to know the remainder when this term is divided by 11.
\(10^1\) divided by 11 gives a remainder of 10 \(10^2\) divided by 11 gives a remainder of 1 \(10^3\) divided by 11 gives a remainder of 10 \(10^4\) divided by 11 gives a remainder of 1 \(10^5\) divided by 11 gives a remainder of 10
So, we can generalize that \(10^{n}\) divided by 11 gives a remainder of 10 if n is odd and a remainder of 1 if n is even.
Since, in this case, \(n = 198*99\) which is even, the remainder is 1.
Option (A) is correct,



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 04:04
Solution: Expanding the above equation: \(\frac{25^{9801}}{11}\) X \(\frac{4^{9801}}{11}\) For such questions, it is very important to recognize patterns.Let us try with\(\frac{25}{11}\), we get a remainder 3, \(\frac{25^2}{11}\), we get a remainder 9, \(\frac{25^3}{11}\) , we get a remainder 5, and so on if we go, we get a pattern of 3...9...5 , Since 9801 is evenly divisble by 3, the remainder of this division shall be 5 as per the pattern we established. Now let us try \(\frac{4}{11}\) , applying the logic as we did above, we get a pattern of 4...5...9...3...1 , if we divide 9801 by 5, we know that remainder is 1, hence according to the pattern, we must get a remainder of 4 If we multiply both the reaminders i.e 5 X 4 we get 20 to adjust the excess remainder , we divide \(\frac{20}{11}\) then leaves a remainder of 9 Hence the answer must be D
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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 04:51
What is the remainder when (25^99∗4^99)^99 is divided by 11?
(25^99∗4^99)^99 = 11*k + a, a = ?
(25^99∗4^99)^99 = (100^99)^99 = 100^9801 = 10^19602
a: 10 / 11 : a = 1 100 / 11 : a = 10 1000/11 : a = 1 10000/11 : a = 10
If power is odd a = 1, if power is even a = 10 19602  even => a = 10
ANSWER E
A. 1 B. 3 C. 7 D. 9 E. 10



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 05:03
What is the remainder when (25^99∗4^99)^99 is divided by 11?
A. 1 B. 3 C. 7 D. 9 E. 10
Here, {(25^99∗4^99)^99} / 11 = {(5^198* 2^198)^99} / 11 = {(10^198)^99} / 11 = {(100^99)^99} / 11 As, 99 can be divisible by 11, the remainder will be 1. So the correct answer choice is (A)



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 05:42
=What is the remainder when ( 25 99 ∗ 4 99 ) 99 (2599∗499)99 is divided by 11?
A. 1 B. 3 C. 7 D. 9 E. 10
Solution :
(\(a^x\))*( \(b^x\))=\((ab)^x\)
(\((25*4)^99\)\()^99\) = \((100)^9801\)
It implies \((10)^19604\)
the property of 11 as divisor is (sum of odd digits sum of even digits)=0 starting from rightmost digit.. this means for 10^even number, (sum of odd digits sum of even digits) =10=1 so 1 is the remainder...
or, we can solve by the following logic also,
its 100 raise to an odd power.When we divide 100 by 11 we get 1 as remainder .So the remainder is 1. Hence A IMO.



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 05:53
What is the remainder when (25^99 x 4^99)^99 is divided by 11?
25^99=5^(2*99) 4^99=2^(2*99)
(25^99 x 4^99)^99=(5^(198)*2^(198))^99=(10^198)^99=((111)^198)^99
By using the Binomial theorem, (ab)^n, Every term above is divisible by 11, except the last term ((1)^198)^199. So, the last term is ((1)^198)^199=1. 1<11 > the remainder will be 1.
The answer choice is A.



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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 06:25
Unit's digit of 25^99 = 5 Unit's digit of 4^99 = 4 (cyclicity of 4 is 2) This reduces question to finding remainder when 20^99 divided by 11. Cyclicity of 2 is 4. And hence remainder will be 8000%11 = 3 Option B. IMO.
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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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20 Jul 2019, 06:26
What is the remainder when (25^99 ∗ 4^99)^99 is divided by 11?
(25^99 ∗ 4^99)^99/11= (100^99) ^99/11
Lets consider 100/11 has 1 as remainder with 11 100^2/11 has 1 as remainder 100^3/11 has 1 as remainder so all powers of 100 has 1 as remainder with 11
so Option A is the answer




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