Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 27 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Jul 28 07:00 PM EDT  08:00 PM EDT Strategies and techniques for approaching featured GMAT topics. Sunday, July 28th at 7 PM EDT
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 12 Jul 2007
Posts: 47

When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
15 Sep 2007, 11:19
Question Stats:
76% (01:39) correct 24% (01:46) wrong based on 538 sessions
HideShow timer Statistics
When the positive integer n is divided by 25, the remainder is 13. What is the value of n? (1) n < 100 (2) When n is divided by 20, the remainder is 3. it is c, and i got 63 as n using both statements, but wasn't certain there didn't exist another one.
any quick way to do this.
thx
Official Answer and Stats are available only to registered users. Register/ Login.



VP
Joined: 08 Jun 2005
Posts: 1064

When the positive integer n is divided by 25, the remainder is 13.
n = 25*x+13
statement 1
n < 100
n = 13,38,63,88
insufficient
statement 2
when n is divided by 20, the remainder is 3.
n = 20*y + 3
20*y + 3 = 25*x+13
insufficient
both statements
n = 13,38,63,88
n = 20*y + 3 > 3,23,43,63,83
n = 63
sufficient
the answer is (C)



Director
Joined: 01 May 2007
Posts: 742

DS: GMATPrep Remainder
[#permalink]
Show Tags
03 Feb 2008, 13:07
When the positive integer n is divided by 25, the remainder is 13. What is the value of n?
1. n < 100 2. When n is divided by 20, the remainder is 3.



CEO
Joined: 17 Nov 2007
Posts: 3372
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: DS: GMATPrep Remainder
[#permalink]
Show Tags
03 Feb 2008, 14:22
C1. 38,63,88. insuff. 2. 63,163. insuff. 1&2. 63. suff.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Director
Joined: 01 May 2007
Posts: 742

Re: DS: GMATPrep Remainder
[#permalink]
Show Tags
03 Feb 2008, 14:26
I tried solving this via the remainder formula. Is their anyway to do this without picking #s?
I got as far as:
n = 25k + 13 n = 20k +3
No idea if I was on the right track with this one. I hate remainder problems.



CEO
Joined: 17 Nov 2007
Posts: 3372
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: DS: GMATPrep Remainder
[#permalink]
Show Tags
03 Feb 2008, 14:29
I also hate remainder problems Smart number approach seems to be faster way.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



CEO
Joined: 17 Nov 2007
Posts: 3372
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: DS: GMATPrep Remainder
[#permalink]
Show Tags
03 Feb 2008, 14:57
n = 25k + 13 n = 20m +3 25k + 13 = 20m +3 25k + 10 = 20m 5k + 2 = 4m m=(5k+2)/4  m has to be integer. k has to be even and not divisible by 4 k=2,6,10 ==> 63, 163, 263  our magic integers
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Director
Joined: 01 May 2007
Posts: 742

Re: DS: GMATPrep Remainder
[#permalink]
Show Tags
03 Feb 2008, 17:39
I missing you at this part:
m=(5k+2)/4  m has to be integer. k has to be even and not divisible by 4
Why does k have to be even, and not divisible by 4?



CEO
Joined: 17 Nov 2007
Posts: 3372
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: DS: GMATPrep Remainder
[#permalink]
Show Tags
03 Feb 2008, 23:43
jimmyjamesdonkey wrote: I missing you at this part:
m=(5k+2)/4  m has to be integer. k has to be even and not divisible by 4
Why does k have to be even, and not divisible by 4? 1. m has to be an integer. 2. (5k+2) has to be even and divisible by 4 3. (5k+2) is even when 5k is even. Therefore k is even. 4. if 5k is divisible by 4, (5k+2) will not divisible by 4: 5*4i+2=4*(5i)+2. Therefore, (5k+2) has not to be divisible by 4 => 5k has not to be divisible by 4 5. k is even and indivisible by 4. 6: 2,6,10....
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Director
Joined: 05 Jan 2008
Posts: 613

Re: DS: GMATPrep Remainder
[#permalink]
Show Tags
04 Feb 2008, 03:31
My approach is :
n=25x+13
multiples of 25 are > 25,50,75 thus n would be >38,63,88
Apply 1> I have 3 values > Not sufficient Apply 2> I will have one value >63, there could be more above 100, thus applying both 1&2 you get n=63



Manager
Joined: 16 Feb 2011
Posts: 190

Re: DS GMAT prep Remainder
[#permalink]
Show Tags
04 Sep 2011, 02:47
per ques stem, we can get values > 13,38,63,88,113 etc so not a single value of n is derived..
per stat 1 , we dnt get to knw anything.. per stat 2, we can get values > 3,23,43,63,83,103 etc
Now, there is only one value (63) which satisfies both the ques stenm condition and stat 2 condition..
somebody pls help me understand what am i missing to get the answer...



Retired Moderator
Joined: 20 Dec 2010
Posts: 1733

Re: DS GMAT prep Remainder
[#permalink]
Show Tags
04 Sep 2011, 03:15
DeeptiM wrote: per ques stem, we can get values > 13,38,63,88,113 etc so not a single value of n is derived..
per stat 1 , we dnt get to knw anything.. per stat 2, we can get values > 3,23,43,63,83,103 etc
Now, there is only one value (63) which satisfies both the ques stenm condition and stat 2 condition..
somebody pls help me understand what am i missing to get the answer... St1 & Stem: n can be: 13, 38, 63, 88 Not Sufficient. St2 & Stem: n can be: 63, 163, 263... Not Sufficient. Using both, we have just one integer in common "63". 63 must be the value because it satisfies both statements and the stem. Sufficient. Ans: "C"
_________________



Senior Manager
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 433
Location: India
GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
25 Jan 2013, 23:58
I used the number pluggin approach until I got this no: 63. When I was about to hit B as the answer, I saw the option A .. Highly unlikely that A couldn't have been there for a reason. . thought that there must be numbers greater than 100 which will satisfy B. So Chose C :D
_________________
hope is a good thing, maybe the best of things. And no good thing ever dies.Who says you need a 700 ?Check this out : http://gmatclub.com/forum/whosaysyouneeda149706.html#p1201595My GMAT Journey : http://gmatclub.com/forum/endofmygmatjourney149328.html#p1197992



Math Expert
Joined: 02 Sep 2009
Posts: 56357

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
26 Jan 2013, 04:38
When the positive integer n is divided by 25, the remainder is 13. What is the value of n?Given that \(n=25q+13\), so n could be 13, 38, 63, 88, 113, ... (1) n < 100. n could be 13, 38, 63, or 88. Not sufficient. (2) When n is divided by 20, the remainder is 3 > \(n=20p+3\). n could be 3, 23, 43, 63, 83, 103, ... Not sufficient. (1)+(2) The only value of n which both statements is 63. Sufficient. Answer: C.
_________________



Director
Joined: 29 Nov 2012
Posts: 731

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
06 Oct 2013, 01:55
Bunuel wrote: When the positive integer n is divided by 25, the remainder is 13. What is the value of n?
Given that \(n=25q+13\), so n could be 13, 38, 63, 88, 113, ...
(1) n < 100. n could be 13, 38, 63, or 88. Not sufficient.
(2) When n is divided by 20, the remainder is 3 > \(n=20p+3\). n could be 3, 23, 43, 63, 83, 103, ... Not sufficient.
(1)+(2) The only value of n which both statements is 63. Sufficient.
Answer: C. Out of the highlighted values only 63 satisfies we need to find another value isn't it otherwise we can't claim insufficiency? How do we find the other number quickly... 3,23,43,83 and 103 don't satisfy the 1st equation



Director
Joined: 29 Nov 2012
Posts: 731

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
06 Oct 2013, 03:33
What I meant was we are given n = 25q + 13 => possible values 13,38,63,88,113
statement 2
n=20p + 3 => 3,23,43,63,83,103
For the second statement all the values in red don't satisfy the first equation only 63 does... How do you find another value quickly to claim insufficiency?
don't we have to match the equations here?



Math Expert
Joined: 02 Sep 2009
Posts: 56357

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
06 Oct 2013, 03:47
fozzzy wrote: What I meant was we are given n = 25q + 13 => possible values 13,38,63,88,113
statement 2
n=20p + 3 => 3,23,43,63,83,103
For the second statement all the values in red don't satisfy the first equation only 63 does... How do you find another value quickly to claim insufficiency?
don't we have to match the equations here? Well, there cannot be only one value that satisfies both \(n=25q+13\) (13, 38, 63, 88, 113, ...) and \(n=20p+3\) (3, 23, 43, 63, 83, 103, ...). Next, there is a way to derive general formula for \(n\) (of a type \(n=mx+r\), where \(x\) is a divisor and \(r\) is a remainder) based on above two statements:Divisor \(x\) would be the least common multiple of above two divisors 25 and 20, hence \(x=100\). Remainder \(r\) would be the first common integer in above two patterns, hence \(r=63\). Therefore general formula based on both statements is \(n=100m+63\). Hence n can be 63, 163, 263, ... For more about this concept check: manhattanremainderproblem93752.html#p721341, whenpositiveintegernisdividedby5theremainderis90442.html#p722552, whenthepositiveintegeraisdividedby5and125591.html#p1028654Hope this helps.
_________________



Director
Joined: 29 Nov 2012
Posts: 731

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
06 Oct 2013, 03:58
Really neat trick... I'm gonna save this absolutely brilliant... Thanks Bunuel!



Director
Joined: 29 Jun 2017
Posts: 786

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
10 Jan 2018, 01:32
[quote="combres"]When the positive integer n is divided by 25, the remainder is 13. What is the value of n?
(1) n < 100 (2) When n is divided by 20, the remainder is 3.
[spoiler=My take]it is c, and i got 63 as n using both statements, but wasn't certain there didn't exist another one.
look at choice 2. when facing this kind, do the following way find the first number satisfying two conditions, in this case is 63 , by pick specific numbers. then find the smallest multiple of 25 and 20, which is 100 the formular is the number required=63+100 x, x can be 1, 2. .... this numbeer satisfy both condition, this number can be divided by 25 with redundant 13 and divided by 20 with redundant 3



Intern
Joined: 29 Oct 2016
Posts: 19
Location: India
GPA: 3.84

Re: When the positive integer n is divided by 25, the remainder
[#permalink]
Show Tags
18 Jan 2018, 01:44
When the positive integer n is divided by 25, the remainder is 13.
n = 25*x+13
statement 1
n < 100
n = 13,38,63,88
insufficient
statement 2
when n is divided by 20, the remainder is 3.
n = 20*y + 3
20*y + 3 = 25*x+13
insufficient
both statements
n = 13,38,63,88
n = 20*y + 3 > 3,23,43,63,83
n = 63




Re: When the positive integer n is divided by 25, the remainder
[#permalink]
18 Jan 2018, 01:44



Go to page
1 2
Next
[ 21 posts ]



