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When positive integer n is divided by 5, the remainder is 1
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When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35. A. 3 B. 4 C. 12 D. 32 E. 35 OPEN DISCUSSION OF THIS QUESTION IS HERE: whenpositiveintegernisdividedby5theremainderis166831.html
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Originally posted by Pedros on 03 Jan 2010, 16:00.
Last edited by Bunuel on 24 Jun 2014, 03:55, edited 3 times in total.
Edited the question and added the OA



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Re: When positive integer n is divided by 5, the remainder is 1
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03 Jan 2010, 20:17
Pedros wrote: When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k+ n is a multiple of 35. A) 3 B) 4 C) 12 D) 32 E) 35 Dont want to try numbers in any remainder problem , please hlep. OA here n is divided by 5 and 7 and remainders are 1 and 3. There is a rule wherein if the difference b/w and remainder is same then the number of obtained from LCM of 2 (here 2) numbers and the constant difference. Here constant difference is 51 = 4 and 73 = 4 so the required number if of the form A(LCM of 5 and 7)  constant difference = 35A  4 So to obtain a multiple of 35 we would need to add 4 to 35A  4. Hence B  4



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Re: When positive integer n is divided by 5, the remainder is 1
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05 Jan 2010, 05:31
kp1811 wrote: Pedros wrote: When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k+ n is a multiple of 35. A) 3 B) 4 C) 12 D) 32 E) 35 Dont want to try numbers in any remainder problem , please hlep. OA here n is divided by 5 and 7 and remainders are 1 and 3. There is a rule wherein if the difference b/w and remainder is same then the number of obtained from LCM of 2 (here 2) numbers and the constant difference. Here constant difference is 51 = 4 and 73 = 4 so the required number if of the form A(LCM of 5 and 7)  constant difference = 35A  4 So to obtain a multiple of 35 we would need to add 4 to 35A  4. Hence B  4 The rule is good to solve such problems but sometime we may just solve GMAT problems simply by observation. In this case we see that number n leaves a remainder 1 from 5 so if we add 4 to n, then the number will be divisible by 5. Similarly, the number n leaves remainder of 3 from 7, again adding 4 to n makes it divisible by 7. So in both the cases adding 4 makes the number n divisible by both 5 and 7 and hence it should also be divisible by LCM of 5,7 i.e.35. So 4 is the answer. It is better to remember the rule but just in case you don't then simply observe. Thanks!



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Re: When positive integer n is divided by 5, the remainder is 1
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09 May 2010, 15:35
nifoui wrote: Here is how I solved this, I'm afraid that my method is completely wrong... so plz tell me what's wrong if it is!
positive integer n is divided by 5, the remainder is 1 > n = 5q + 1 (q being the quotient) positive integer n is divided by 7, the remainder is 3  > n = 7q + 3
so I combined both equations 5q + 1 = 7q + 3 q = 1
and solved for n n = 4
I picked 4 because 4  4 = 0 and I considered that 35 * 0 = 0
to double check I tried n = 4 with the other answer choices and found that it couldn't be a multiple of 35...
Does that work at all?? Positive integer n is divided by 5, the remainder is 1 > \(n=5q+1\), where \(q\) is the quotient > 1, 6, 11, 16, 21, 26, 31, ... Positive integer n is divided by 7, the remainder is 3 > \(n=7p+3\), where \(p\) is the quotient > 3, 10, 17, 24, 31, .... You can not use the same variable for quotients in both formulas, because quotient may not be the same upon division n by two different numbers. For example 31/5, quotient q=6 but 31/7, quotient p=4. There is a way to derive general formula for \(n\) (of a type \(n=mx+r\), where \(x\) is divisor and \(r\) is a remainder) based on above two statements:Divisor \(x\) would be the least common multiple of above two divisors 5 and 7, hence \(x=35\). Remainder \(r\) would be the first common integer in above two patterns, hence \(r=31\). Therefore general formula based on both statements is \(n=35m+31\). Thus the smallest positive integer k such that k+n is a multiple of 35 is 4 > \(n+4=35k+31+4=35(k+1)\). More about deriving general formula for such problems at: manhattanremainderproblem93752.html#p721341Hope it helps.
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Re: When positive integer n is divided by 5, the remainder is 1
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10 May 2010, 16:43



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Re: When positive integer n is divided by 5, the remainder is 1
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07 Nov 2010, 20:24
First, let us say I have a number n which is divisible by 5 and by 7. We all agree that it will be divisible by 35, the LCM of 5 and 7. Now, if I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 1, we can say the number is of the form n = 5a + 1 e.g. 5 + 1, 10 + 1, 15 + 1, 20 + 1, 25 + 1, 30 + 1, 35 + 1 etc and n = 7b + 1 e.g. 7 + 1, 14 + 1, 21 + 1, 28 + 1, 35 + 1 etc So when it is divided by the LCM, 35, it will give 1 as remainder (as is apparent above) Next, if I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 3, we can say the number is of the form n = 5a + 1 and n = 7b + 3 Now, the only thing you should try to understand here is that when n is divided by 5 and if I say the remainder is 1, it is the same as saying the remainder is 4. e.g. When 6 is divided by 5, remainder is 1 because it is 1 more than a multiple of 5. I can also say it is 4 less than the next multiple of 5, can't I? 6 is one more than 5, but 4 less than 10. Therefore, we can say n = 5x  4 and n = 7y  4 (A remainder of 3 when divided by 7 is the same as getting a remainder of 4) Now this question is exactly like the question above. So when you divide n by 35, remainder will be 4 i.e. n will be 4 less than a multiple of 35. So you must add 4 to n to make it a multiple of 35 A trickier version is: If I have a number n which when divided by 5 gives a remainder 1 and when divided by 7 gives a remainder 5, what is the remainder when n is divided by 35? n = 5a + 1 = 5x  4 n = 7b + 5 = 7y 2 Nothing common! Now, I will need to check for the smallest such number. I put b = 1. n = 12. Is it of the form 5a + 1? No. Put b = 2. n = 19. Is it of the form 5a + 1? No. Put b = 3. n = 26. Is it of the form 5a + 1? Yes. When 26 is divided by 5, it gives a remainder of 1. When it is divided by 7, it gives a remainder if 5. Next such number will be 35 + 26. Next will be 35*2 + 26 and so on... The remainder when n is divided by 35 will be 26 (or we can say it will be 9). If we want to find the number that must be added to n to make it divisible by 35, that number will be 9.
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Re: When positive integer n is divided by 5, the remainder is 1
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03 Dec 2010, 21:59
tonebeeze wrote: Hello All,
I am working on the Quant Review 2nd edition today and came across PS #68. This problem gave me some difficulty. The explanation in the back of the book was not very clear. I would appreciate it if you all can walk me through your approach to this problem.
Thanks!
68. When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35?
a. 3 b. 4 c. 12 d. 32 e. 35 This says that n = 5a + 1 and n = 7b + 3. For k + n to be a multiple of 35, then: k + 5a + 1 = 35x and k + 7b + 3 should be a multiple of 35 as well. Now, this means that k + n should be a multiple of both 5 and 7. For k + n to be a multiple of 5, k needs to be either 4, 9, 14 and so on if you consider the first condition, and should be 4, 11, 18 and so on for the second condition. The smallest number as you can see is 4. Hence the answer is B.



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Re: When positive integer n is divided by 5, the remainder is 1
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17 Mar 2011, 09:29
you know guys , I tried to figure out the principle of solving such issues, the formulas, and common approaches but then realized that it is not worth doing. BALLPARKING is much better  it is more reliable, faster and easy. Even if you try all the five variants still you will make it less then 2 minutes.



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Re: When positive integer n is divided by 5, the remainder is 1
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19 Dec 2011, 23:28
Positive integer n is divided by 5; the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k + n is a multiple of 35? a) 3 b) 4 c) 12 d) 32 e) 35 Explanation of Answer 1. Set up algebraic equation for n 5p + 1 = 7q + 3 2. Evaluate 5p = 7q + 2 Since the units digit of 5p is 0 or 5, the units digit of 7q + 2 must also end in 0 or 5. The units digit of 7q must be 8, when the units digit of 5p is 0, or 3, when the units digit of 5p is 5. 7q = 28, 63, 98, 133..., where q = 4, 9, 14, 19 and q = 5m + 4, for some multiple 0, 1, 2... n = 7q + 3 = 31, 66, 101... Multiples of 35 are 35, 70, 105. k = (3531) or (7066) or (105101) k = 4



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Re: When positive integer n is divided by 5, the remainder is 1
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22 Dec 2011, 01:24
Guys, picking number saves a lot of time in this question. I have always been of the opinion that questions that hint at solutions that are small numbers are best handled with numbers. For example. From Question Stem 1... Solutions could be: 1, 6, 11, 16, 21, 26, 31, 36, 41 Now look for numbers that satisfy the condition for divisibility by 7........ n could only be 31 and 36 from the numbers above..... Smallest number is 31... 31 + 4 is 35 So answer is B.....
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Re: When positive integer n is divided by 5, the remainder is 1
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16 Jan 2012, 01:32
carcass wrote: When positive integer n is divided by 5, the remainder is 1. When n is divided by 7, the remainder is 3. What is the smallest positive integer k such that k+n is a multiple of 35?
(A) 3 (B) 4 (C) 12 (D) 32 (E) 35
Someone can help me with this one. I 'm not be able to figure out the problem after the second sentence.
In other words:
n=5a+1 is an integer n=7b+3 idem.......
and then ???
Thanks guys. Instead of just giving you the solution here, let me provide you with the complete theory that will help you solve this question on your own in under a minute. Also, the last link discusses a question very similar to this one. http://www.veritasprep.com/blog/2011/04 ... unraveled/http://www.veritasprep.com/blog/2011/04 ... yapplied/http://www.veritasprep.com/blog/2011/05 ... emainders/
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Re: When positive integer n is divided by 5, the remainder is 1
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16 Jan 2012, 02:54
carcass wrote: Thanks karishma. In those days I'm studying remainder advanced strategy and I am pretty comfortable with this argument but in this problem I was stuck. Either way, I 'll follow your advice. Is much better studying over and over again to foster the foundamental, instead to have the solution. Thanks a lot. Sure, no problem! Get back if there are any doubts in the theory/this question.
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Re: When positive integer n is divided by 5, the remainder is 1
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Re: When positive integer n is divided by 5, the remainder is 1
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Re: When positive integer n is divided by 5, the remainder is 1
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16 Jan 2012, 05:50
carcass wrote: Thanks Bunel.
I would ask only if my previous post have some flaws in the reasoning or in that way I could make something wrong.
Very useful your first post. carcass wrote: Ok the most important thing is that I was on the right path to an efficient solution of the problem. I read all your links Indeed it was very simple. Our target was 35 n=5a+1.....> n=5*6+1=30+1=31+ 4=35 n=7b+1......>n=7*4+3=28+3=31+ 4=35 remainder= 4To make simple. correct me if I'm wrong ..... All is OK with your post above: by trial and error you found first, thus smallest common integer to satisfy both conditions (n=31) and got that in order n+k to be a multiple of 35 k must be 4 (min value). In fact as we are dealing with small and easy numbers then (number picking)+(trial and error)+(simple logic) is probably the easiest way to solve this particular problem. Though knowing general theory might be helpful for harder questions or help to double check easy ones.
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Re: When positive integer n is divided by 5, the remainder is 1
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16 Jan 2012, 09:28
carcass wrote: Ok the most important thing is that I was on the right path to an efficient solution of the problem. I read all your links Indeed it was very simple. Our target was 35 n=5a+1.....> n=5*6+1=30+1=31+ 4=35 n=7b+1......>n=7*4+3=28+3=31+ 4=35 remainder= 4To make simple. correct me if I'm wrong ..... Yes, you are right. You can think of it in another way which will help you arrive at the answer immediately. n=5a+1 or we can say the remainder is 4. n=7b+3 or we can say the remainder is 4. So 4 is the common remainder. When we divide n by 35, the remainder will be 4. This means, n will be 4 less than a multiple of 35. I have explained the same thing in this post above: whenpositiveintegernisdividedby5theremainderis90442.html#p814507whenpositiveintegernisdividedby5theremainderis90442.html#p814507
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Re: When positive integer n is divided by 5, the remainder is 1
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02 Feb 2012, 01:07
The numbers that satisfy the first condition are 6,11,16,21,26,31............ The numbers that satisfy the second condition are 10,17,24,31............ The one common to both and closest to 35 is 31. Therefore if we add 4 to 31 we get 35. The answer is therefore 4. Option (B)
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Re: When positive integer n is divided by 5, the remainder is 1
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08 Feb 2013, 09:47
My approach:
n = 5a + 1 and n = 7b + 3................. [1]
Equating both , we get 5a = 7b +2
Checking for values of b so that RHS becomes a multiple of 5 , the first such value of b is 4 (i.e. n = 7*4 +2 = 30).
So, from [1] we get n= 31
=> smallest value of k is 4 to make (n + k)a multiple of 35.



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Re: When positive integer n is divided by 5, the remainder is 1
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24 Jun 2014, 02:12
When positive integer n is divided by 5, the remainder is 1 => n=5a+1 When positive integer n is divided by 7, the remainder is 3 => n=7b+3
Therefore, n+k = 5a+1+k = 7b+3+k
Since it's divisible by 35, it must be divisible by both 5 and 7,
for 5a+1+k to be divisible by 5, (1+k) should be divisible by and similarly for the next case, (3+k) should be divisible by 7.
Obviously,the least value for which this is true is 4!
Answer: B



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Re: When positive integer n is divided by 5, the remainder is 1
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Re: When positive integer n is divided by 5, the remainder is 1 &nbs
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