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Divisibility and Remainder !

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by raka3399 » Tue Aug 25, 2009 12:11 pm
What is the remainder of 3^4n+3 divided by 5 ???

how best to solve such sums efficiently?
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by grockit_jake » Tue Aug 25, 2009 2:20 pm
For this question, we need to know what n is to find the answer.

With remainders, it helps to look for patterns, though. 3 raised to a positive integer will leave a one's digit in the following pattern:

3^1 = 3
3^2 = 9
3^3 = 7
3^4 = 1
3^5 = 3 ......

Because (4n + 3) will yield an ODD number (assuming n is a positive integer) we know that the one's digit will be either 3 or 7.

When dividing by 5, the remainders will either be 3 or 2 respectively.

Again, this depends on n.
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Re: Divisibility and Remainder !

by raka3399 » Wed Aug 26, 2009 9:37 am
raka3399 wrote:What is the remainder of 3^4n+3 divided by 5 ???

how best to solve such sums efficiently?
Jake , letz assume that n is positive as in 1,2,3,4....

doesn't it make 3^4n+3 have a 7 as its unit digit giving a remainder of 2 when divided by 5 ALWAYS!!!

correct me if am wrong !!

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by grockit_jake » Wed Aug 26, 2009 10:09 am
There is a difference between 3^(4n+3) and 3^4n + 3.

Assuming it's 3^(4n+3):

n=1, 3^(4n+3) = 3^7 = 2187 (remainder of 2)
n=2, 3^(4n+3) = 3^11 = 117147 (remainder of 2)

So yes, you are right. Because of the coefficient of 4, each increase of 1 in n, makes the exponent jump four spots in the 3,9,7,1 pattern, leaving you back at 7.
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by raka3399 » Wed Aug 26, 2009 1:07 pm
grockit_jake wrote:There is a difference between 3^(4n+3) and 3^4n + 3.

Assuming it's 3^(4n+3):

n=1, 3^(4n+3) = 3^7 = 2187 (remainder of 2)
n=2, 3^(4n+3) = 3^11 = 117147 (remainder of 2)

So yes, you are right. Because of the coefficient of 4, each increase of 1 in n, makes the exponent jump four spots in the 3,9,7,1 pattern, leaving you back at 7.
Sorry Jake guess it could be termed as a presentation error ... yes its 3^(4n+3) divided by 5 :)

many thanks again!

you rock !!!
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by Ian Stewart » Thu Aug 27, 2009 8:51 am
We certainly need to know that n is a positive integer here. If that's given, we do not need any further information about n to answer the question, regardless of whether the expression is written as (3^4n) + 3 or as 3^(4n + 3). In the first case, we have

(3^4n) + 3 = (3^4)^n + 3 = 81^n + 3.

Since 81^n will end in 1 regardless of the value of n, 81^n + 3 will end in 4, and the remainder will be 4 when 3^4n + 3 is divided by 5.

In the second case, we have

3^(4n + 3) = [3^(4n)]*[3^3] = [81^n]*[3^3] = [81^n]*[27]

and since we are multiplying one number which ends in 1 with a number which ends in 7, the product must end in 7, so the remainder will be 2 when 3^(4n + 3) is divided by 5.
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