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

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Difficulty:   25% (medium)

Question Stats: 69% (01:16) correct 31% (01:27) wrong based on 295 sessions

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

A. 1
B. 3
C. 7
D. 9
E. 10 This question was provided by Experts Global for the Game of Timers Competition _________________
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What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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3
3
The given number is very big. It has to be in some pattern.

If we take n = 99 then the number is given in the form:
$$(25^n * 4^n)^n$$

Let us find the value with lower terms of n-
n = 1
(25*4) = 100
Remainder ($$\frac{100}{11}$$) = 1

n = 2
$$(25^2 * 4^2)^2$$ = $$(625 * 16)^2$$ = $$(10000)^2$$
Remainder ($$\frac{10000^2}{11}$$) = 1

So, for any value of n, the remainder of $$(25^n * 4^n)^n$$ divided by 11 will always be 1.

Originally posted by Sayon on 19 Jul 2019, 08:10.
Last edited by Sayon on 20 Jul 2019, 01:51, edited 2 times in total.
##### General Discussion
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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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3
$$(25^{99}∗4^{99})^{99} = (100)^{99*99}$$

100 leaves a remainder of 1 when divided by 11 and so (100)^{99*99} must leave a remainder of 1 when divided by 11
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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
What is the remainder when (25^99∗4^99)^99 is divided by 11?

The product is the question stem can be written as (100^99)^99. Now any power of 100 if divided by 99 the reminder is going to be 1.
For eg 100^1 divided by 99. Remainder is 1
100^2 = 10000 divided by 99. Reminder is 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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1

the statement can be written as 100^(99*99). As 1 followed by any number of zeros and divided by 11, will give a reminder of 1.
so 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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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

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
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Originally posted by Kinshook on 19 Jul 2019, 08:22.
Last edited by Kinshook on 19 Jul 2019, 10:12, edited 2 times 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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1
We know that a^n * b^n = (ab)^6
(25^99 ∗ 4^99)^99 = (100^99)^99 = 100^(99*99)

When 100 is divided by 11, remainder is 1
similarly when 100^(99*99) is divided by 11, remainder is 1^(99*99) = 1

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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1
1. Multiply 25^99 x 4^99=100^99=10^100
2. Now we have to multiply powers and we get 10^9900
3. Now, use the binomial theorem (11-1)^9900.
4. Since the power is even then we have positive remainder 1 and this is the answer.

IMO 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 (2599∗499)99 is divided by 11?

The questions test your ability with regards to divisibility.

We know that units digit of 5 is always 5 and 4's unit digit since 99 is odd is 4 now (5*4) ^11 = 20^99 / 11 = 20/11
which is 9.

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

hence the correct answer is 9. 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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What is the remainder when (25^99∗4^99)^99is divided by 11?

Same exponent so can combine 100^99 is going to end in 0, which that result raised to 99 will end in a 0.

some large base 10 number will end with remainder of 1.

Ex: 100/11 = 9 R 1
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GMAT 1: 640 Q45 V35 GMAT 2: 660 Q48 V33 Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?  [#permalink]

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1
(25^99∗4^99)^99 can be written in the form (100^99)^99 as (25*25**25....99 times)*(4*4*4...99 times) will give 100*100*100..99 times

100/11 will have a remainder of 1. So (1^99)^99/11 will still be 1 (Using remainder theorem for polynomials)

IMO 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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1
(25^99∗4^99)^99
Units digit 5 to any power is 5
Units digit 4 to odd power is 4 & even power is 6.In this case unit digit will be 4
Now,unit digit 5*4=20
So, we're looking for solution for units digit 0^99

Multiple of 10 divided by 11 will always leave remainder as 1.

Hence 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?
$$\frac{(25^{99}∗4^{99})^{99}}{11}$$
=> $$\frac{((25*4)^{99})^{99}}{11}$$
=> $$\frac{((100)^{99})^{99}}{11}$$
=> $$\frac{((1)^{99})^{99}}{11}$$ (Using Remainder Theorem or dividing 100 by 11: Remainder is 1)
=> $$\frac{(1)^{99}}{11}$$
=> $$\frac{1}{11}$$
=> 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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$$(25^{99}∗4^{99})^{99}$$ is divided by 11
simplify it to
$$(100^{99})^{99}$$
now 100 when divided by 11 leaves a remainder 1
$$100 = 9*11 +1$$
thus remainder is $$(1^{99})^{99}/11$$ =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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IMO : A

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

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

SOL:

(a^x)(b^x)=(ab)^x so we get (100^99)99.

Now if the power of 100 is odd then the last multiple of 11 before that number will be 999... or (100-1)

so this means we will always have remainder as 1.

Originally posted by abhishekdadarwal2009 on 19 Jul 2019, 08:36.
Last edited by abhishekdadarwal2009 on 19 Jul 2019, 08:45, 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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1
A

The given expression can be written as (10)^2*99*99. Now, notice that when 10^n is divided by 11, remainder is 10 when is odd and 1 when n is even.

Example: 10/11 - remainder is 10
100/11 - remainder is 1
1000/11 - remainder is 10
10000/11 - remainder is 1

So, in the given expression, n = 2*99*99 - even no. So remainder should be 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 A.

$$(25^{99}∗4^{99})^{99} --> (25∗4)^{99 * 99}$$
25*4= 100.
When 100 is divided by 11 we get a remainder 1.
This is equivalent to having $$\frac{(25∗4)^{99 * 99}}{11}$$ giving us,

$$\frac{(1)^{99 * 99}}{11}$$ --> Which is 1/11 , giving a remainder of 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 the correct answer is option A - 1
Explanation given as attachment -
Attachments IMG_20190719_210529.JPG [ 1017.33 KiB | Viewed 2067 times ]

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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 (25ˆ99∗4ˆ99)ˆ99 is divided by 11?

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

(25ˆ99∗4ˆ99)ˆ99 = [5ˆ(2*99)•2ˆ(2*99)]ˆ99 = [10ˆ198]ˆ99 = [(11-1)ˆ198]ˆ99
now, 11 to any power is divisible by 11, so what we need to find is the remainder of [(-1)ˆ198]ˆ99
which is the same as finding the remainder of -1/11
the remainder of a negative is the same as subtracting 11-1=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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1
Simplify.
100^(99*99)
100/11 remainder is 1
1^(99*99) Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?   [#permalink] 19 Jul 2019, 08:45

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