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555-605 Level|   Exponents|   Remainders|      
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Bunuel
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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 07 Feb 2022, 02:30
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D
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66% (01:17) correct 34% (01:41) wrong based on 103 sessions

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What is the remainder when \((9^{2n+3})(5^n)\) is divided by 10, where \(n\) is a positive integer?

A. 1
B. 2
C. 4
D. 5
E. 8
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D
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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 07 Feb 2022, 05:02
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Bunuel
What is the remainder when \((9^{2n+3})(5^n)\) is divided by 10, where \(n\) is a positive integer?

A. 1
B. 2
C. 4
D. 5
E. 8

 
Let \(n=1\)
The given translates to \(9^5*5^1\)
Dividing the given by \(10\)

\(= Rem\frac{(9^5)*(5)}{10}\)
Cancelling 5 from Numerator and Denominator
\(= 5*Rem\frac{9^5}{2}\)
\(= 5*Rem\frac{(8+1)^5}{2}\)
\(= 5*Rem\frac{\\
(1^5)}{2}\)
\(= 5*1\)
\(= 5 \)

Hence, our answer is option D.

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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 07 Feb 2022, 05:18
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In here let n=1, substituting n in the expression we get,
9^(2n+3)*5^n = 9^(5)*5

the last digit of 9^5 is 1(refer cyclicity of 9) => 1*5 gives us 5-> hence the last digit of this number would be 5

and hence this number when divided by 10 would give us the remainder 5
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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 07 Feb 2022, 09:30
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IMO:D
As to find divisibility result by 10, we have to find the last digit of the expression
For Example = when “2346559” is divided by 10 it will give 9 as the remainder

Now coming to the question- 9^(2n+3) x (5^n) => 3^2(2n+3) x (5^n) => 3^(4n) x 3^6 x 5^n
Following rule of cyclicity, we get last digit of 3^(4n) = 1, 3^6 = 9 and 5^n = 5
Therefore, last digit multiplication gives 45 i.e. last digit will be 5. hence, the reminder of the expression when it is divided by 10 is 5
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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 07 Feb 2022, 10:01
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To find a remainder by 10, we only need to know our units digit. The product here is clearly a multiple of 5, because it includes 5^n, and every multiple of 5 ends in 0 or 5, so those are the only two possible answers no matter what we're multiplying 5^n by. Now if we either notice 9^(2n + 3) is odd, or that '0' isn't an answer choice at all, we can see the answer must be 5.
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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 07 Feb 2022, 10:05
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What is the remainder when \((9^{2n+3})(5^n)\) is divided by 10, where \(n\) is a positive integer?

\(9^{2n+3}\) can have last digit as 9 and \(5^n\) can have last digit as 5.

So, Last digit of \((9^{2n+3})(5^n)\) will be 5.

I think D. :)
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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 07 Feb 2022, 18:47
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What is the remainder when \((9^{2n+3})(5^n)\) is divided by 10, where \(n\) is a positive integer?

A. 1
B. 2
C. 4
D. 5
E. 8

 
\(Rem(\frac{(9^{2n+3})(5^n)}{10)})\)
=\(Rem(\frac{(9^{2n+3})}{10})*Rem(\frac{(5^n)}{10}))\)
=\(Rem(\frac{(10-1)^{2n+3})}{10})*Rem(\frac{(5^n)}{10}))\)
=-1*5=-5

Rem(-5) = Re(10-5) = 5

Ans D­
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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n 12 Feb 2022, 12:44
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Note that 5^n always ends with 5.
The unit digit of M*5 (Let M be any positive integer) can only be 5 or 0.
If M*5 is divided by 10, the remainder can only be 5 or 0.
5, but not 0, is found in the answer choices, so (D) it is.
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