For a positive integer x, what is the remainder when 3^x+1 is divided

Question

For a positive integer x, what is the remainder when  3^x+1  is divided by 10?

1) x=4n+2, n is a positive integer
2) x>4

Anannt · Answer

Ans :A 
St 1: According to cyclicity of 3 the last digit of 3^(4n+2) would be 9 hence last digit would be zero. Sufficient. 
St 2: x can be 5,6,7,8 and so on hence last digit of 3^x would be different for each value of x, hence can not determine. Insufficient.

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rohit8865 · Answer

3^1=3
3^2=9
3^3=27
3^4=81
3^5=243
...
after every power of 4 ,units digit repeats itself..(ie unit digit of 3^1 & 3^5= 3)
thus 3^x+1
we need to know x
(1) x=4n+2
thus x will always have units digit = 9
so 3^x+1 always results in last digit be 9+1= ...0
thus 3^x+1 is divisible by 10
suff

(2) x>4
if x= 5 then 3^5= 24 3 
if x=6 then 3^6 = 72 9

thus unit digit may be 4 or 0 
insuff

Ans A

MathRevolution · Answer

==> If you modify the original condition and the question, for the remainder question when divided by 10, the exponent has the period of 4. In other words, you get \(~3^1=~3, ~3^2=~9, ~3^3=~7, ~3^4=~1\), so the ones digits has the period of 3-->9-->7-->1-->3-->…. Thus, for con 1), from x=4n+2, the exponent always gets a period of 4, and since you get \(3^4^n^+^2+1=(~3^2)+1=(~9)+1=~0\), the remainder is always 0, hence it is sufficient.

Therefore, the answer is A.
Answer: A

Therefore, the answer is A.
Answer: A

SuryaNouliGMAT · Answer

Sorry for pulling you back to an old post, could you please check this : MathRevolution   Buneal

Given = (3^x + 1) / 10

we need to find X,

from st1 --- x = 4n+2, n = +ve integer

let n=even, x = Even(Even) + Even = Even
let n=odd, x = Even(Odd) + Even = Even

from this we can conclude x = Even ---> 2,4,6,8........

substitute in given condition :
2 ---> 10/10 = remainder = 0
4 ---> 82/10 = remainder = 2
6 ---> 730/10 = remainder = 0
8 ---> 6563/10 = remainder = 2

from this, the remainder is repeating with a period of 2.

can i say it as sufficient ??

Please shed some light.

Thanks & Kudos in advance

BrushMyQuant · Answer

x is a positive integer, is the remainder 0 when  3^x + 1  is divided by 10?

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

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

Using Above theory Remainder of  3^{x} + 1  by 10 = Unit's digit of   3^{x} + 1

We know the units' digit of 1, so we need to find the units' digit of 3^x

Now to find the unit's digit of   3^{x}   , we need to find the pattern / cycle of unit's digit of power of 3 and then generalizing it.

Unit's digit of  3^1  = 3
Unit's digit of  3^2  = 9
Unit's digit of  3^3  = 7
Unit's digit of  3^4  = 1
Unit's digit of  3^5  = 3

So, units' digit of power of 3 repeats after every  4^{th}  number. 
 => We need to divided x by 4 and check what is the remainder

STAT 1: x = 4n + 2, where n is a positive integer.

=> x divided by 4 is same as 4n + 2 divided by 4
=> 4n when divided by 4 will give 0 remainder and 2 when divided by 4 will give 2 remainder
=> Total remainder = 2
=> Units' digit of 3^x will be same as units' digit of 3^2 = 9
=> Units' digit of 3^x + 1 = 9 + 1(=10) = 0
=> Remainder of 3^x by 10 = 1
=>   SUFFICIENT

STAT 2: x > 4  
 
=> Now depending of value of x units' digit of 3^x can have different values
=>   NOT SUFFICIENT

So,  Answer will be A 
Hope it helps!

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

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

For a positive integer x, what is the remainder when 3^x+1 is divided

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