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When the positive integer n is divided by 25, the remainder [#permalink]
 15 Sep 2007, 11:19
Question Stats:
73% (02:19) correct
26% (01:02) wrong based on 38 sessions
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 |
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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)
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DS: GMATPrep Remainder [#permalink]
03 Feb 2008, 13:07
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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.
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Re: DS: GMATPrep Remainder [#permalink]
03 Feb 2008, 14:22
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C1. 38,63,88. insuff. 2. 63,163. insuff. 1&2. 63. suff.
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Re: DS: GMATPrep Remainder [#permalink]
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.
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Re: DS: GMATPrep Remainder [#permalink]
03 Feb 2008, 14:29
I also hate remainder problems Smart number approach seems to be faster way.
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Re: DS: GMATPrep Remainder [#permalink]
03 Feb 2008, 14:47
Anyone...give it a try without plugging in #s?
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Re: DS: GMATPrep Remainder [#permalink]
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
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Re: DS: GMATPrep Remainder [#permalink]
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?
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Re: DS: GMATPrep Remainder [#permalink]
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....
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Re: DS: GMATPrep Remainder [#permalink]
04 Feb 2008, 03:31
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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
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Re: DS GMAT prep Remainder [#permalink]
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...
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Re: DS GMAT prep Remainder [#permalink]
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"
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Re: When the positive integer n is divided by 25, the remainder [#permalink]
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
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Re: When the positive integer n is divided by 25, the remainder [#permalink]
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.
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Re: When the positive integer n is divided by 25, the remainder
[#permalink]
26 Jan 2013, 04:38
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