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If x is a positive integer, is the remainder 0 when (3x + 1) [#permalink]
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02 Jan 2013, 09:00
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If x is a positive integer, is the remainder 0 when (3^x + 1)/10? (1) x = 3n + 2, where n is a positive integer. (2) x > 4
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Re: If x is a positive integer, is the remainder 0 when (3x + 1) [#permalink]
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02 Jan 2013, 09:34
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curtis0063 wrote: If x is a positive integer, is the remainder 0 when (\(3^x + 1\))/10? (1) x = 3n + 2, where n is a positive integer. (2) x > 4 The remainder will be zero when x is \(4n2\) i.e. 2, 6, 10, 14 etc. Statement 1 tells us that x=5, 8,11, 14 etc Not sufficient statement 2 is not sufficient On combinng also, the information is not sufficient. Hence +1E Do mention the source
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Re: If x is a positive integer, is the remainder 0 when (3x + 1) [#permalink]
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03 Jan 2013, 14:26
The powers of 3 are as follows: 3,9,27,81,243,729.....The pattern of the units digit is 3,9,7,1,3......
The only way the expression would result in a remainder of 0 is if the numerator is a factor of 10. For that to happen x would need to be 2,6,10,14...and so on.
From statement 1: If n is 4 then there would be a remainder of 0. But if n was 3 that would not hold true From statement 2: Clearly not sufficient. 1+2 Together still not sufficient.
Answer E.



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Re: If x is a positive integer, is the remainder 0 when (3x + 1) [#permalink]
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23 Feb 2014, 09:18
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



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Re: If x is a positive integer, is the remainder 0 when (3x + 1) [#permalink]
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18 May 2014, 10:27
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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.



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Re: If x is a positive integer, is the remainder 0 when (3x + 1) [#permalink]
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26 May 2014, 13:55
Marcab wrote: curtis0063 wrote: If x is a positive integer, is the remainder 0 when (\(3^x + 1\))/10? (1) x = 3n + 2, where n is a positive integer. (2) x > 4 The remainder will be zero when x is \(4n2\) i.e. 2, 6, 10, 14 etc. Statement 1 tells us that x=5, 8,11, 14 etc Not sufficient statement 2 is not sufficient On combinng also, the information is not sufficient. Hence +1E Do mention the source How do you get that remainder will be zero when x=4n  2 ? Could you elaborate a little further on this point? Thanks! Cheers J



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Re: If x is a positive integer, is the remainder 0 when (3x + 1) [#permalink]
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27 May 2014, 21:02
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.



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