If x is a positive integer, is the remainder 0 when (3^x + 1)/10?

OK first thing's first. We need to know if the expression 3^x + 1 will be divisible by 10 which means that we need to know if units digit will be zero. Now, 3^x has cycle 3,9,7,1 so only if the units digit is in the second place (9) we will get UD of zero. Let's find out if this can be the case.

First statement, x = 3n + 2. Now we are told that n must be a positive integer. We have the following options 5,8,11,14,17,20 etc....for the exponent. If we divide by 4 and gauge the remainders we will get that remainder can be 3,1,7,9 and then the cycle repeats again. Therefore insufficient.

Second Statement tells us that x>4, well this is insufficient because the cycle repeats itself. Both together, statement 2 wasn't helpful at all so this is going to be a clear E

Hope this helps
Cheers
J

Stmt II

x > 4


x=5 3x+1 is 16 then remainder is 6
x=6 3x+1 is 17 then remainder is 7

x can take on many more values for which the remainder value varies(could be x=243 then remainder is 0)

INSUFFICIENT

Stmt I

x = 4n+2

3X+1 = 3(4N+2) = 12N+7

12n+7 will never be divisible by 10 since the units digit of 12*some positive integer n will never be 3 (only if the units digit is 3 will the resulting number when added to 7 have a units digit of 0 to be divisible by 10)

Units digit for 12* n will cycle as follows 2,4,6,8,0,2,4...etc

Since its a yes or no question we can confidently say NO which makes this statement SUFFICIENT to answer the question Hence E.

Basically , when divisor is 10 the remainder will be 0 when the numerator has a 0 in the units digit.

The numerator here is 3^x + 1 . So this means when the units digit of 3 ^x is 9 then the units digit of the complete numerator will be 0. Now lets see in what circumstances will the units digit of 3^x will be 9.

3^1 - Units digit is 3
3^2 - Units digit is 9
3^3 - Units digit is 7
3^4 - Units digit is 1
3^5 - Units digit is 3

So we see that the cyclisity is 4 and the units digit will be 9 on the second iteration. So cyclisity is 4n and units digit is 9 on 4n - 2. I am guessing this is how Marcab reached the conclusion that remainder will be 0 when X = 4n - 2.

With the equation involving 3^x+1, 3^+1 needs to end with a 0 for the equation to be divisible by 10. This can only happen if 3^x ends with a 9.

If you look at all the power of 3's, the unit digit repeats every power of 4.
3^1=ones digit is 3
3^2=ones digit is 9
3^3=ones digit is 7
3^4=ones digit is 1



1. With x=3n+2, I plugged in random numbers
n=1; x=3(1)+2=5; plugging 5 into the equation given...
(3^5+1)/10 is not divisible by 10, because 3^5 will end with a 3 as the ones digit.

n=8; x=3(8)+2=26; plugging 26 into the equation given...
(3^26+1)/10 is divisible by 10, because 3^26 will end with a 9 as the ones digit.

Not Sufficient

2. x>4
Not sufficient.
You can easily test this out by using x=26 or 5 from above.

From using x=26 or 5 you already know that both statements together are not sufficient.

The answer is E.

The remainder in this division will be zero only when the 3^x term ends in 9. Remember, 3^2, 3^6, 3^10 etc end in 9. So the trend is, 4k+2 where k = 0,1,2,3....

1) x = 3n+2 - 2,5,8,11 etc - clearly, the remainder will be 0 when x = 2 and not 0 when x = 5,8,11 - So not sufficient
2) x > 4 - again, when x = 6 remainder = 0, when x = 5, remainder <> 0 - So not sufficient

Combining 1 & 2: 3n+2 > 4 or n = 1,2,3, etc so, X can take values 5,8,11,14,17,20 - 3^5 + 1 will leave 2 as remainder. 3^14+1 - will leave 0 as remainder (14 = 4*3+1 - fits the trend above). So, given information isn't suitable to answer this question.

Hope this helps.

K

Given x is a positive integer
Note that when x=2, then the expression (3^x+1)/10 will give remainder 0
when x=6 then 3^x=729 so 3^x+1=730 /10 will give remainder 0

St 1 says x=3n+2...possible values of x=2,5,8,11 and 14...
So x=2 R(0),x=3,R not equal to 0..

Not sufficient

St 2 says x>4...x=5 Remainder not zero but if x=6,Remainder zero..

two options..

On combining we have that x>4 and we can have R(0) and \(R \neq{0}\)

Ans E

Similar question to practice: if-x-is-a-positive-integer-is-the-remainder-0-when-3-x-109075.html

The Official answer is NOT CORRECT. The answer is Certainly is A...

DATA #1 ) says that IF N Is a POSITIVE INTEGER (SO n CANNOT BE 0) So if we place 1,2 ,3 ,..... in the place of n in the equation we get following pattern respectly as: 5,8.11,.....

so 3^5 + 1 Is not giving ZERO Remaining after diving by 10. also this pattern is continues if we place 8, 11, 14,..... so The answer is NO... the remaining is not 0 when we divide (3^n+1) By 10 so this data is defenetly sufficient...

DATA#2 ) is obviously not sufficient

so Answer IS A NOT E...

Here the official answer should be modified

That's not correct.

If n = 1, then the remainder when 3^(3*1+2) + 1 is divided by 10 is 4.

If n = 4, then the remainder when 3^(3*4+2) + 1 is divided by 10 is 0.

A is the answer.
S1.
As x=3n+2 , x will be always even and Odd^ Even will always be even as n is not equal to 0. Hence, S1 is sufficient to answer the question.
S2.
x>4 give no information hence B gone.


Hence, the eventual answer is A.
Please correct me if I'm wrong some where.

Dear Bunuel Please Correct me if I'm wrong..

in the equation (3^x +1) / 10 the reminder is 0 ONLY X raises to the power of 2,6, 10,14,... for example (3^2 +1 ) is 10 and 10/10 gives the 0 reminder .

AS you know the cyclist pattern in 3 power repeats every 4 times so if 3 raises to the power of 2,6, 10,... gives the unit digit 9 and 9 plus 1 gives unit digit of 0 which is ALWAYS divisible by 10

so here we ONLY we need to know whether x could equal to 2, 6, 10,... OR NOT

STATEMENT number 1 directly give us the answer to this question as it says X= 3n+2 and says that n is POSITIVE integer so X could equal 5,8,11,... we can see that x CANNOT be 2,6,10,...

So this statement is SUfficient ...

AM I right?


Thanks,

No, you are not.

5, 8, 11, 14, 17, 20, ...
2, 6, 10, 14, 18, 22, ...

I see , Thank you very much.....

We are given that x is a positive integer and need to determine the remainder when 3^x + 1 is divided by 10. Since we know that 1/10 produces a remainder of 1, we really need to determine the remainder when 3^x is divided by 10. We should keep in mind that the remainder of any integer divided by 10 is equivalent to the units digit of that number, so we need to determine the units digit of 3^x.

Statement One Alone:

x = 3n + 2, where n is a positive integer.

We see that we cannot determine the units digit of 3^n. If n = 1, then 3^x = 3^5, which has a units digit of 3, and thus a reminder of 3 when divided by 10. However, if n = 2, then 3^x = 3^8 has a units digit of 1, and thus a reminder of 1 when divided by 10. Statement one alone is not sufficient to answer the question.

Statement Two Alone:

x > 4

Knowing that x is greater than 4 is not sufficient to answer the question.

Statements One and Two Together:

Using the statements together, we still cannot answer the question. We can still use n = 1 and n = 2 to obtain 3^x = 3^5 and 3^x = 3^8, respectively. Since 3^5 and 3^8 have different units digits, the remainder when 3^5 + 1 is divided by 10 and the remainder when 3^8 + 1 is divided by 10 will be different.

Answer: E

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.