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What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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19 Jul 2019, 08:00
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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
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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, 01:51
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.
Answer A
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.




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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, 08:09
\((25^{99}∗4^{99})^{99} = (100)^{99*99}\)
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?
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19 Jul 2019, 08:12
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.
Hence 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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19 Jul 2019, 08:12
IMO answer is A:
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?
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Updated on: 19 Jul 2019, 10:12
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 \((x+y)^n = x^n + C_1^n x^n1 y + .... C_k^n x^nk y^k +.....y^n\) (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?
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19 Jul 2019, 08:25
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?
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19 Jul 2019, 08:25
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 (111)^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?
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19 Jul 2019, 08:25
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?
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19 Jul 2019, 08:28
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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Re: What is the remainder when (25^99 x 4^99)^99 is divided by 11?
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19 Jul 2019, 08:29
(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?
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19 Jul 2019, 08:35
(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 Posted from my mobile device
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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, 08:36
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
IMO the 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, 08:36
\((25^{99}∗4^{99})^{99}\) 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?
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Updated on: 19 Jul 2019, 08:45
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 (1001)
so this means we will always have remainder as 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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19 Jul 2019, 08:37
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?
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19 Jul 2019, 08:39
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?
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19 Jul 2019, 08:39
IMO the correct answer is option A  1 Explanation given as attachment 
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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, 08:45
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 = [(111)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 111=10 Answer (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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19 Jul 2019, 08:45
Simplify. 100^(99*99) 100/11 remainder is 1 1^(99*99) Therefore, answer=1




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