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Question 5 (nov.24th)

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Source: — Data Sufficiency |
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by cramya » Mon Nov 24, 2008 12:50 pm
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

A)
Last edited by cramya on Mon Nov 24, 2008 1:19 pm, edited 1 time in total.
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by niraj_a » Mon Nov 24, 2008 12:57 pm
@cramya

question. i tried 5 different possible sample values in Statement I and II both and i constantly got the answer NO for both statements.

for e.g. in statement 1, if n = 1, 2, 3, 4, 5 then entering the resulting x value into 3x+1 never yields a value divisible by 10.

in statement II, i used the same values 5, 6, and 7.

so couldn't I have chosen D as the answer? please let me know my mistake.
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by cramya » Mon Nov 24, 2008 1:17 pm
for e.g. in statement 1, if n = 1, 2, 3, 4, 5 then entering the resulting x value into 3x+1 never yields a value divisible by 10.
Since the units digit of 3 cylces as follows: 3,9,7,1,3...

U could pick a number for x where the units digit of the 3*that positive integer will give 9 as units digit and when added to 1 the units digit of the final resulting number will be 0

Eg:x=243

3(243)+1 = 730 divisible by 10 Remainder is 0

whereas in the example I have given the remainders are 6,7 etc... I should have included this too in my solution above. INSUFF cant tell if remainder is 0 or not (edited my post to include this also)

Niraj, let me know if u still have questions
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by niraj_a » Mon Nov 24, 2008 1:23 pm
gotcha now, smart thinking, i'll keep a watch for units digits in remainder problems from now on.
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by vittalgmat » Mon Nov 24, 2008 7:13 pm
Hi,
Here is how I solved:

The q stem asks whether the 3x +1 is divisible by 10 or not.
So a remainder of 0 implies YES; anything else implies a NO.

stmt1: substitute x = 4n+2 in 3x +1. => 12n +7.
Try plugging in numbers for n.
n 12n+7 Is remainder 0, when divided by 10?
----------------------------------------------------------------------
0 7 NO
1 19 NO
==> there is no multiple of 12 such that the unit's digit is a 3.
If such a number exists, then 12n+7 will end in a 0.
Since, there is no such number, the ans is NO for all n.
So stmt 1 is sufficient. Another way to look at this:
the units digit of 12 is 2; any number multiplied by 2 will be even.
So the units digit of the product 12n can never end in an odd number 3.


stmt2: x >4
Try plugging in numbers:

x 3x +1 Is remainder 0 , when divided by 10
-----------------------------------------------------------------------
5 16 NO
6 20 YES

So not suficient.

So answer is A

Hope this helps.
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Re: Question 5 (nov.24th)

by nitin86 » Wed Nov 26, 2008 9:27 am
bacali wrote:Q11:

If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4


OA: A
My approach,

Stmt 1 => x = 4n + 2
Hence, x must be an even number because 4n = even, 2 = even, and sum of two even number is even.

On subsituting this value of X in 3x + 1,

3x + 1 => 3 * (4n + 2) + 1
Here, 3 * (4n+2) = even as product of even and odd is always even
1 = odd,

So, 3 * (4n + 2) + 1 has to be odd as sum of even and odd is always odd.

So, if 3x + 1 is odd, it cannot be divisible by 10.

Hence, SUFFICIENT
-------------------------
Stmt 2, is straightforward, as for differenct values of x , where x > 4, x may/may not be divisible by 10.

Hence INSUFFICIENT.

Hope this approach helps
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by gmataug08 » Thu Nov 27, 2008 10:10 pm
from stmt 1

no matter what n is, X will be even.

as X is even 3x+1 will be odd, as odd numbers can never be divided evenly by any even number (vice versa too applies), hence X is not 10 divisible.

hence stmt 1 is enough.

stmt 2 : is just there to satisfy DS question format :) , no suggestive information.
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Re: Question 5 (nov.24th)

by maihuna » Wed Sep 09, 2009 10:46 am
bacali wrote:Q11:

If x is a positive integer, is the remainder 0 when 3x + 1 is divided by 10?
(1) x = 4n + 2, where n is a positive integer.
(2) x > 4


OA: A
I think the Q was 3^x + 1 and not 3x+1
Charged up again to beat the beast :)
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by hariharakarthi » Sat Sep 12, 2009 8:18 pm
(3^x) + 1

x= 4n+2; power of 3's unit digits are 3,9,7,1. In this when n is a positive integer, then unit digit of 3^x is can be 3,9,7 and 1. Hence, we can not definitively say about the remainder.

X>4; reasoning is same as above.

Hence ans is E
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by m-sand » Sun Jan 17, 2010 8:53 am
hariharakarthi wrote:(3^x) + 1

x= 4n+2; power of 3's unit digits are 3,9,7,1. In this when n is a positive integer, then unit digit of 3^x is can be 3,9,7 and 1. Hence, we can not definitively say about the remainder.

X>4; reasoning is same as above.

Hence ans is E
Ans will be A. As, x=4n+2, unit digit of 3^x+1 will be 0. Therefore , remainder will be 0.
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by ajay7640 » Wed Jan 20, 2010 6:55 pm
Yes. 3^x + 1 would always be divisible by 10 as long as x=4n + 2

Note any fourth power or 3 of multiple of 4 ...so 4n th power of 3 will have 1 as units digit
3 raied to 2 is always 9
so Units digit is always 0 for 3^x + 1 as long as x= 4n + 2.

3^1 3
3^2 9
3^3 27
3^4 81 ---------
3^5 243
3^6 729
3^7 2187
3^8 6561 --------- and so on ...see the pattern of 3, 9, 7 , 1 in the units digit.

So A is sufficient and is the answer.

If x > 4 then units digit can be anything. SO can not conclusively say that divisible by 10.
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by prsnt11 » Sun May 01, 2011 12:15 am
For the remainder to be 0, the units digit of x needs to be 3, which will also mean that x needs to be ODD. Only then 3x + 1 will be divisible by 10.
(1): x = 2( 2n + 1). Hence x must be even. Hence x CANNOT be 3 or have unit's digit as 3. So the answer is definitely NO.
(2): x > 4 => x can be even or odd. Hence x MAY have a unit's digit of 3. So the answer is MAYBE.
Hence A is the correct answer.
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