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

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

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New post 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?  [#permalink]

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New post 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. 9

E. 10

E is the answer. :heart
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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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New post 19 Jul 2019, 23:39
1
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?  [#permalink]

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New post 19 Jul 2019, 23:43
1
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?  [#permalink]

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New post 20 Jul 2019, 00:08
1
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?  [#permalink]

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New post 20 Jul 2019, 00:13
1
(25^99*4^99)^99

(100^99)^99

100 divided by 11 gives the remainder 1

(\(\frac{100}{11}\)--R = 1)--(\(\frac{100}{11}\)--R-1)--(\(\frac{100}{11}\)--R-1) (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?  [#permalink]

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New post 20 Jul 2019, 00:34
1
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?  [#permalink]

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New post 20 Jul 2019, 00:38
1
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?  [#permalink]

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New post 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?  [#permalink]

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New post 20 Jul 2019, 01:17
1
(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?  [#permalink]

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New post 20 Jul 2019, 03:00
1
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?  [#permalink]

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New post 20 Jul 2019, 03:18
1
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?  [#permalink]

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New post 20 Jul 2019, 03:38
1
\((25^{99}*4^{99})^{99}
= (5^{198}*2^{198})^{99}
= (10^{198})^{99}
= 10^{198*99}\)

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?  [#permalink]

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New post 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?  [#permalink]

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New post 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?  [#permalink]

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New post 20 Jul 2019, 05:03
1
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?  [#permalink]

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New post 20 Jul 2019, 05:42
1
=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) =1-0=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?  [#permalink]

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New post 20 Jul 2019, 05:53
1
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=((11-1)^198)^99

By using the Binomial theorem, (a-b)^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?  [#permalink]

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New post 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?  [#permalink]

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New post 20 Jul 2019, 06:26
1
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
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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?   [#permalink] 20 Jul 2019, 06:26

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