What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n

Question

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

kungfury42 · Answer

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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MidhilaMohan · Answer

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

mdsaddamforgmat · Answer

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

IanStewart · Answer

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.

TarunKumar1234 · Answer

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.

desertEagle · Answer

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

zhanbo · Answer

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.

BrushMyQuant · Answer

What is the remainder when  (9^{2n + 3})(5^n)  is divided by 10

Theory : Remainder of a number by 10 is same as the unit's digit of the number

( Watch this Video  to Learn How to find Remainders of Numbers by 10)

Using Above theory Remainder of  (9^{2n + 3})(5^n)  by 10 = Unit's digit of  (9^{2n + 3})(5^n)

Now, Let's find the unit's digit of  (9^{2n + 3})

Now to find the unit's digit of  9^{2n + 3} , we need to find the pattern / cycle of unit's digit of power of 9 and then generalizing it.
Unit's digit of  9^1  = 9
Unit's digit of  9^2  = 1
Unit's digit of  9^3  = 9
Unit's digit of  9^4  = 1
=> Units' digit of Odd Power of 9 = 9
=> Units' digit of Even power of 9 = 1

2n + 3 = Even + Odd = Odd
=> Unit's digit of  9^{2n + 3}  = 9

Unit's digit of  5^n   = 5 (as all positive powers of 5 have units' digit of 5)

=> Unit's digit of  (9^{2n + 3})(5^n)  = Unit's digit of 9 * 5 = 5

So,  Answer will be D 
Hope it helps!

MASTER How to Find Remainders with 2, 3, 5, 9, 10 and Binomial Theorem

https://www.youtube.com/watch?v=h1HTEhmUb1w

What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n

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What is the remainder when (9^(2n+3))(5^n) is divided by 10, where n

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