Remainder R

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Remainder R

by joyseychow » Fri Aug 14, 2009 2:30 am
If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

(1) n+1 is divisible by 3
(2) n>20


Help! I always have difficulty with remainders.
Source: — Data Sufficiency |

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by ssmiles08 » Fri Aug 14, 2009 7:04 am
I have issues with remainders as well. I got A for this problem.

the remainder equation with question stem plugged in:

4 + 7N = 3Q + R

here we know that R can have at most 3 values: 0, 1, 2.

1) n + 1 = 3P + 0

N = 3P - 1

substitute N into the first equation: 4 + 7(3P - 1) = 3Q + R

21P - 7 + 4 = 3Q + R
21P - 3 -3Q = R
3(7P - 1 - Q) = R

Now we know all variables are integers; which means R is an integer as well.

divide both sides by 3, and we find out that R is divisble by 3. The only value R can be is 0. SUFFICIENT.

2) N > 20. doesnt limit R. R can still be 0, 1, 2 depending on N.

IMO (A).
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by vikram_k51 » Fri Aug 14, 2009 9:19 am
If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

(1) n+1 is divisible by 3
(2) n>20


Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.

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by robbie523 » Fri Aug 14, 2009 9:42 am
well, for this question, there is not many short cuts, since 4+7n is not a number can be easily broken down... to see if it fit.

well, stm one, use number like 2, 5, 8, 11, to plug in and check the result, and plug 4 in is sure enough to prove something yes or no

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by ssmiles08 » Fri Aug 14, 2009 10:01 am
vikram_k51 wrote:
Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.

How are you getting a remainder of 2??

anyway i guess you can plug in numbers as well and don't really have to solve the equation completely to find out the answer is A.
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by vikram_k51 » Fri Aug 14, 2009 10:42 am
ssmiles08 wrote:
vikram_k51 wrote:
Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.

How are you getting a remainder of 2??

anyway i guess you can plug in numbers as well and don't really have to solve the equation completely to find out the answer is A.
n=5,8,11,14...all give aremainder of 2 when divided by 3

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by ssmiles08 » Fri Aug 14, 2009 10:59 am
vikram_k51 wrote:
ssmiles08 wrote:
vikram_k51 wrote:
Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.

How are you getting a remainder of 2??

anyway i guess you can plug in numbers as well and don't really have to solve the equation completely to find out the answer is A.
n=5,8,11,14...all give aremainder of 2 when divided by 3
4 + 7N

4 + 5*7 = 39.
4 + 8*7 = 60
4 + 11*7 = 81
4 + 14*7 = 102

39, 60, 81 and 102 are all divisible by 3. which give the remainder of 0.

let me know if you think otherwise...and please show your work.
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by scoobydooby » Fri Aug 14, 2009 12:01 pm
stment 1. n+1 is divisible by 3
=>n+1=3k
=>n=3k-1.
=>n=2, 5, 8,.... (as n>0)


so 4+7n =18, 39, 60... divisible by 3, so leaves no remainder
sufficient


stment 2. n>20
remainder can still be 0, 1, 2 depending on n.

hence, A
Last edited by scoobydooby on Fri Aug 14, 2009 12:49 pm, edited 1 time in total.

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by ssmiles08 » Fri Aug 14, 2009 12:15 pm
scoobydooby wrote:stment 1. n+1 is divisible by 3
=>n+1=3k
=>n=3k+1.
=>n=1, 4, 7, 10.... (as n>0)
if n+1 = 3k

shouldn't n = 3k - 1??? and not 3k + 1
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by scoobydooby » Fri Aug 14, 2009 12:42 pm
oops. thanks. will edit :)

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by tohellandback » Fri Aug 14, 2009 9:57 pm
IMO A
1)meean n is of the form 3k+2

3+1 +7(3k+2)
=3+21k+15
all terms are divisible by 3. remainder 0

2)it depends upon the number n, so not sufficient
Last edited by tohellandback on Sun Aug 16, 2009 8:29 pm, edited 1 time in total.
The powers of two are bloody impolite!!

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Re: Remainder R

by yezz » Sat Aug 15, 2009 1:55 pm
joyseychow wrote:If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

(1) n+1 is divisible by 3
(2) n>20


4+7n = (3+6n)+(n+1)

from 1

n+1 is dev by 3 thus as 3+6n is devis by 3 the whole expression 4+7n is too

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by jjk » Mon Aug 17, 2009 4:14 pm
vikram_k51 wrote:If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

(1) n+1 is divisible by 3
(2) n>20


Will be A.

If n+1 is divisible by 3 then n will always give a rem of 2.
The problem does not ask you the value of r when n is divided by 3. It asks you the value of r when 4 + 7n is divided by 3.

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by rockfeld » Wed May 09, 2012 1:49 am
I think the easiest approach would be:
1. 4+7n=4+4n+3n=4*(1+n)+3n
2. (1) 1+n is a multiple of 3 => if we divide 4*(1+n)+3n by 3 then the remainder would be 1/3. That's it. Suff.
3. (2) not suff info.

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by Stuart@KaplanGMAT » Wed May 09, 2012 4:44 am
joyseychow wrote:If n is a positive integer and r is the remainder when 4+7n divided by 3, what is the value of r?

(1) n+1 is divisible by 3
(2) n>20


Help! I always have difficulty with remainders.
Picking numbers is a very efficient way to solve number property DS questions.

From the question stem, we know that (4+7n)/3 has a remainder of r; we want to know that value. What do we need? Information about n.

1) n+1 is a multiple of 3.

If n=2 (we're allowed to pick 2 since 2+1 is a multiple of 3), then (4+14)/3 = 18/3 = 6rem0
If n=5 (we're allowed to pick 5 since 5+1 is a multiple of 3), then (4+35)/3 = 39/3 = 13rem0

at this point you might already be conviced that you'll always get the same answer, but we could try one more just to be safe:

If n=8 (we're allowed to pick 8 since 8+1 is a multiple of 3), then (4+56) = 60/3 = 20rem0

For all 3 plug-ins we get r=0.. sufficient!

2) n > 20

If n=21, then (4+147)/3 = 151/3 = 50rem1

Note: if you really understand DS, you can actually stop right now. Here's a fundamental rule to remember:

If the answer is (D), then both statements will always give the same answer to the question.

Since we got a remainder of 1, and since statement (1) guaranteed a remainder of 0, we know that (2) cannot be sufficient!

However, if you don't see that logic, then you just pick another number:

If n=22, then (4+154)/3 = 158/3 = 52rem2.

(2) has now given us r=1 AND r=2; more than one possible value, therefore insufficient.

(1) is suff, (2) is infuff: choose A!
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