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Manager  S
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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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1
1 is the answer.

(25^99 x 4^99)^99 could be written as (100^99)^99---------> ((10^2)^99)^99

11 will leave -1 remainder with 10, and this -1 on squaring will become +1.

then finally 11 will leave 1 remainder with (1^99)^99

pretty bad explanation but I think 1 has to be answered.
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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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1
We can write expression as

100^99*99

So 100 mod 11 is always 1

So remainder will always be 1

Hence 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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1
It can be rewritten as: $$100^{99*99}$$ = $$10^{2*99*99}$$ = $$(11-1)^{2*99*99}$$ --> $$(-1)^{2*99*99}$$ --> $$1$$ (because the power is even)
or can be rewritten as : $$100^{99*99}$$ = $$(99 + 1)^{99*99}$$ --> $$(1)^{99*99}$$ --> $$1$$
so 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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Quote:
What is the remainder when (25^99∗4^99)^99 is divided by 11?

$$25 * 4 = 100$$
$$100 = 11* 9 + 1$$ => 1 is remainder

$$25^2 * 4^2 = 10,000$$
$$10,000 = 11*909+1$$ => 1 is remainder

Independent of the degree, the remainder in this calculation is always 1.
Thus, (25^99∗4^99)^99 / 11 will have 1 as a remainder.

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

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

First, we can simplify this expression by multiplying 25*4 and power it to 99, so we will have:

(100^99)^99

Then we have that when we divide 100 by 11, the reminder is 1, so we can replace this reminder in the expression:

(1^99)^99, so the reminder will be also 1.

So (A) is our answer.

Originally posted by Mizar18 on 19 Jul 2019, 08:56.
Last edited by Mizar18 on 19 Jul 2019, 19:41, 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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What is the remainder when (25^99 * 4^99)^99 is divided by 11?
Solution:
(25^99 * 4^99)^99 = ((25*4)^99)^99=100^(99*99)
if 100 is divided by 11, reminder is 1 => 100 = 11*9+1
=> 100^2=10000 = 11*909+1
=> 100^3=1000000 = 11*90909+1
similarly: if 100^n is divided by 11, reminder is 1, so if 100^(99*99) is divided by 11, reminder is 1

A. 1 --> correct
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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(25^99 ∗ 4^99)^99 ==>> (5^(2(99)) * 2^(2(99)))^99 (factoring out the common exponents) ==>> (5 * 2)^(2(99^2)) ==>> 10^(2(99^2)).

Now we need to find the remainder when 10^(2(99^2)) is divided by 11. Lets try to find the remainder with lesser powers of 10 and see if there is a pattern:

10^1=10 divided by 11 - r=10
10^2=100 divided by 11 - r=1
10^3=1,000 divided by 11 - r=10 (990 is 90 times 11)
10^4=10,000 divided by 11 - r=1 (9,999 is 99 times 11)

So what is the pattern we see here? For every odd power of 10, dividing the number by 11 gives a remainder of 10 and dividing any even power of 10 by 11 gives a remainder of 1. Now we know that 10^(2(99^2)) is an even power of 10 (because of 2 is being multiplied with 99^2), we know for certain that the remainder when 10^(2(99^2)) is divided by 11 will be 1.

Answer is thus 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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What is the remainder when $$(25^{99}∗4^{99})^{99}$$ is divided by 11?

This can be rewritten as $$(100^{99})^{99}$$

So, Numerator is a power of 100, and denominator is 11

100/11 gives remainder of 1

10000/11 also gives remainder of 1 ... and so on

all powers of 100 when divided by 11 always give remainder of 1

ANSWER: 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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When 25 is multiplied by 4 it will give 100
No matter what power 100 is raised to when divided by 11,
Remainder will always be 1
Answer is A
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GMAT 1: 570 Q43 V26 GMAT 2: 660 Q48 V34 Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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Given (25^99 * 4^ 99) ^ 99 and we need to find the remainder of this number to 11.
The number can be written as ((25*4)^99)^99 Why? Since power is same, bases can be multiplied
This becomes ((100)^99)^99 => (100)^99^2

So our number is 100000000...
Now here is the beauty of this number. Irrespective of the trailing zeros, this number will always give remainder as 10 when divided by 11.

Say X = 1000000000...
X + 1 = 100000.... 1
Notice that X + 1 is divisible by 11
So X + 1 = 11n
X = 11n -1 => X = 11n - (11-10) => X = 11n' + 10
So remainder 10

Answer 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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IMO-A

Easier Approach :
(25^99 x 4^99)^99= (100^99)^99= 100^9801

Therefore ques is, 100^9801 / 11 = (Rem 100/11)^9801 = 1^9801= 1

Another approach-
Note- Any Number N of form, {N^(p-1)}/p leaves remainder 1 -- Where p=prime no. and N,p are coprime

Now, (25^99 x 4^99)^99= (100^99)^99= 100^9801
Therefore ques is, {100^9801}/11

Comparing with above, N=100 & p=11 .... Clearly, N & p are coprimes
so, 100^(11-1)/11-- remainder 1
i.e. 100^10/11--remainder 1

100^9801/11= [(100^10)^9800 X 100] /11= [(100^10)^9800]/11 x 100/11= Rem (1) x Rem (1)= Rem (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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IMO E

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

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

Given, (25^99*4^99)^99 = (100)^99*99 = (100)^…1 [99*99 = “…1” => some odd integer ending with 1]

Now, when the even power of 10 (for e.g. 100, 10000 and so on…) is divided by 11, the remainder is 1
And when the odd power of 10 (for e.g. 10, 1000 and so on…) is divided by 11, the remainder is 10

From above expression we get the power of equation is odd, as 99*99 = “…1” => some odd integer ending with 1
So, when this expression is divided by 11, the remainder will be 10
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What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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$$(25^{99}$$ x $$4^{99})^{99} =$$ $$(100^{99})^{99} =$$ $$(100)^{99 * 99} = (100)^{9801}$$
Note that $$99*99$$ can easily be calculated as $$(100-1)^2 = 100^2 + 1 -2*100 = 9801$$

Let's find the pattern of the remainder when $$(100)^n$$ is divided by 11 ($$n$$ is positive integer).
$$n=1$$ --> $$\frac{100}{11}$$ has quotient of 9 and remainder of 1;
$$n=2$$ --> $$\frac{10,000}{11}$$ has quotient of 909 and remainder of 1;
$$n=3$$ --> $$\frac{1,000,000}{11}$$ has quotient of 90,909 and remainder of 1, etc.

Thus, we can confidently say that when $$(25^{99}$$ x $$4^{99})^{99} = 100^{9801}$$ is divided by 11, the remainder is 1.

Correct answer is (A) $$-$$ one.
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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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Given (25^99 x 4^99)^99 is divided by 11

Firstly, 25^99 x 4^99 = (25 x 4)^99
= (100)^99
Therefore, (100^99)^99 divided by 11 is as follows:

100/11 gives a remainder of 1
(100^2)/11 gives a remainder of 1
(100^3)/11 gives a remainder of 1

therefore, (100^99)^99 divided by 11 will give a remainder of 1

Hence answer choice A.

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GMAT 1: 640 Q48 V28 Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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1) Let's make prime factorization:
$$25^99=5^198$$
$$4^99=2^198$$

2) Let's use the formula: $$a^n*b^n=(ab)^n$$ to get the value in the brackets:
$$5^198*2^198=10^198$$

3) now let's open the brackets to get
(10^198)^99=10^{198*99}
The only thing we will pick from this huge power is that it is even

4) Now lets give 10 more convenient form
10=11-1, so we have (11-1)^{even positive integer}
11 will always be divisible by 11 whatever the positive power
-1 will transform into 1 as the power is even and 1/11 will always leave remainder 1

IMO
Ans: 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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As per exponent laws (25^99 * 4^99)^99 = (100^99)^99. It does not matter what the power is, if we divide by 11 the remainder will always be 1. Option A in my opinion.
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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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(25^99 *4^99)^99 =(100^99)^99 when divided 11 should leave remainder 1
as 100/11 leaves 1
Hence A
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What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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

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

inside the bracket, same power we can multiply to give (100^99)^99.. which gives so ans as 100 power something..
for 100 the remainder is 1 if divided by 11 .
so (100^99)^99 gives 1 power something

so 1 power anything will give only 1.

so ans is 1

ie A
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Originally posted by ccheryn on 19 Jul 2019, 10:12.
Last edited by ccheryn on 20 Jul 2019, 03:52, 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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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
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GMAT 1: 670 Q49 V32 Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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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) Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?   [#permalink] 19 Jul 2019, 11:12

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