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A group of n students can be divided into equal groups of 4
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A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? A. 33 B. 46 C. 49 D. 53 E. 86 OA is a and this is how I arrived at.
Let say n = 4x+1 and n = 5y+3 > From the question Stem
n=4x+1 n = 5y+3 5 8 9 13 13 18 17 23 25 33 29 38 33 43
I get these above values by putting the same values for x and y. Is my concept correct?
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Originally posted by enigma123 on 21 Jan 2012, 13:37.
Last edited by mau5 on 18 Nov 2013, 21:35, edited 3 times in total.
Edited the OA



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Re: Group of Students
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21 Jan 2012, 13:59
enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86 Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one. A group of n students can be divided into equal groups of 4 with 1 student left over > n=4q+1 > n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4) A group of n students can be divided into equal groups of 5 with 3 students left over > n=5p+3 > n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5) Therefor two smallest possible values of n are 13 and 33 > 13+33=46. Answer: B. Else you can derive general formula based on n=4q+1 and n=5p+3. Divisor will be the least common multiple of above two divisors 4 and 5, hence 20. Remainder will be the first common integer in above two patterns, hence 13 > so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) > 13+33=46. Answer: B. For more about this concept see: manhattanremainderproblem93752.html#p721341whenpositiveintegernisdividedby5theremainderis90442.html#p722552whenthepositiveintegeraisdividedby5and125591.html#p1028654Hope it helps.
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Re: A group of n students can be divided into equal groups of 4
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21 Jan 2012, 14:34
Hi Bunuel  can the values of q and p be ZERO? I don't think they can be and therefore n cannot be 1 & 3. Am I wrong?
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Re: A group of n students can be divided into equal groups of 4
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21 Jan 2012, 14:44
enigma123 wrote: Hi Bunuel  can the values of q and p be ZERO? I don't think they can be and therefore n cannot be 1 & 3. Am I wrong? THEORY: Positive integer \(a\) divided by positive integer \(d\) yields a reminder of \(r\) can always be expressed as \(a=qd+r\), where \(q\) is called a quotient and \(r\) is called a remainder, note here that \(0\leq{r}<d\) (remainder is nonnegative integer and always less than divisor).For example we are told that when positive integer n is divided by 25, the remainder is 13 > \(n=25q+13\). Now, the lowest value of \(q\) can be zero and in this case \(n=13\) > 13 divided by 25 yields the remainder of 13. Generally when divisor (25 in our case) is more than dividend (13 in our case) then the reminder equals to the dividend. For example: 3 divided by 24 yields a reminder of 3 > \(3=0*24+3\); or: 5 divided by 6 yields a reminder of 5 > \(5=0*6+5\). Also note that you shouldn't worry about negative numbers in divisibility questions, as every GMAT divisibility question will tell you in advance that any unknowns represent positive integers. OUR ORIGINAL QUESTION: We are told that "a group of n students can be divided into equal groups of 4 with 1 student left over" > n=4q+1. Here q also can be zero, which would mean that there is only 1 student and zero groups of 4. QUESTIONS TO PRACTICE: PS questions on remainders: search.php?search_id=tag&tag_id=199DS questions on remainders: search.php?search_id=tag&tag_id=198THEORY ON REMAINDERS: compilationoftipsandtrickstodealwithremainders86714.htmlHope it helps.
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Re: A group of n students can be divided into equal groups of 4
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21 Jan 2012, 14:53
You are a true genius buddy.
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Re: A group of n students can be divided into equal groups of 4
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22 Jan 2012, 07:16
Stright and simple B just try numbers for 4*n+1=5*k+3 where k and n are integers, find 2 smallest and your solve it for 30 sec



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Re: A group of n students can be divided into equal groups of 4
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22 Jan 2012, 17:48
enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a)33 b)46 c)49 d)53 e) 86
OA is a and this is how I arrived at.
Let say n = 4x+1 and n = 5y+3 > From the question Stem
n=4x+1 n = 5y+3 5 8 9 13 13 18 17 23 25 33 29 38 33 43
I get these above values by putting the same values for x and y. Is my concept correct? I wrote a blog post discussing this concept in detail. I have discussed a couple of questions very similar to this one in the post. You can check it out if you like. http://www.veritasprep.com/blog/2011/05 ... spartii/
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Re: A group of n students can be divided into equal groups of 4
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23 Jan 2012, 09:01
n=4q+1 > 1,5,9,13,17,21,25,29,33 n=5q+3>3,8,13,18,23,28,33 the first two common numbers are 13 and 33, so add those numbers, you get 13+33=46 so, asnwer is B, 46 hope this helps
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Re: A group of n students can be divided into equal groups of 4
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24 Jan 2012, 15:09
Thanks Karishma. A very helpful post.
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Re: Group of Students
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18 Nov 2013, 19:15
Again since the max student left over are 3 so any no with 3 can be common between the two groups ; 3, 13, 23, 33, 43, 53 etc. using in the formula pf n= 4q+1 = 5r+ 3 ; will get 13 and 33 ; addd then =46
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Re: Group of Students
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19 Nov 2013, 09:48
Bunuel wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86 Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one. A group of n students can be divided into equal groups of 4 with 1 student left over > n=4q+1 > n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4) A group of n students can be divided into equal groups of 5 with 3 students left over > n=5p+3 > n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5) Therefor two smallest possible values of n are 13 and 33 > 13+33=46. Answer: B. Else you can derive general formula based on n=4q+1 and n=5p+3. Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.
Remainder will be the first common integer in above two patterns, hence 13 > so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) > 13+33=46.
Answer: B. For more about this concept see: manhattanremainderproblem93752.html#p721341whenpositiveintegernisdividedby5theremainderis90442.html#p722552whenthepositiveintegeraisdividedby5and125591.html#p1028654Hope it helps. Could you explain what you mean by 'divisor'? Where is there a divisor in those two equations....Is there work here that you did in your head, but left out? I'm trying to learn how to do these, so if you could show any work that was done mentally I would appreciate it! Thanks!



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Re: Group of Students
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19 Nov 2013, 14:19
AccipiterQ wrote: Bunuel wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n? a) 33 b) 46 c) 49 d) 53 e) 86 Yes you can do the way you started by listing the possible values of n for both patterns and then picking first two matching numbers from these lists. Since we are dealing with easy and small numbers this approach probably would be the fastest one. A group of n students can be divided into equal groups of 4 with 1 student left over > n=4q+1 > n can be: 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, ... (basically an evenly spaced set with common difference of 4) A group of n students can be divided into equal groups of 5 with 3 students left over > n=5p+3 > n can be: 3, 8, 13, 18, 23, 28, 33, 38, ... (basically an evenly spaced set with common difference of 5) Therefor two smallest possible values of n are 13 and 33 > 13+33=46. Answer: B. Else you can derive general formula based on n=4q+1 and n=5p+3. Divisor will be the least common multiple of above two divisors 4 and 5, hence 20.
Remainder will be the first common integer in above two patterns, hence 13 > so, to satisfy both conditions, n must be of a type n=20m+13: 13, 33, 53, ... (two two smallest possible values of n are for m=0 and for m=1, so 13, and 33 respectively) > 13+33=46.
Answer: B. For more about this concept see: manhattanremainderproblem93752.html#p721341whenpositiveintegernisdividedby5theremainderis90442.html#p722552whenthepositiveintegeraisdividedby5and125591.html#p1028654Hope it helps. Could you explain what you mean by 'divisor'? Where is there a divisor in those two equations....Is there work here that you did in your head, but left out? I'm trying to learn how to do these, so if you could show any work that was done mentally I would appreciate it! Thanks! 20 is the divisor in n=20m+13. Please follow the links in my post.
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Re: A group of n students can be divided into equal groups of 4
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06 Sep 2017, 11:21
start with 8, then 13,18,23,... try every combination



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Re: A group of n students can be divided into equal groups of 4
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06 Sep 2017, 13:31
enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
A. 33 B. 46 C. 49 D. 53 E. 86 assume difference of 1 between quotients (n1)/4(n3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B



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Re: A group of n students can be divided into equal groups of 4
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06 Sep 2017, 14:19
gracie wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
A. 33 B. 46 C. 49 D. 53 E. 86 assume difference of 1 between quotients (n1)/4(n3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B this can be applied to any problem of this kind?



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Re: A group of n students can be divided into equal groups of 4
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06 Sep 2017, 14:24
gracie wrote: enigma123 wrote: A group of n students can be divided into equal groups of 4 with 1 student left over or equal groups of 5 with 3 students left over. What is the sum of the two smallest possible values of n?
A. 33 B. 46 C. 49 D. 53 E. 86 assume difference of 1 between quotients (n1)/4(n3)/5=1 n=13=smallest n 13+4*5=33=second smallest n 13+33=46 B also, how can you assume that the difference is 1, it can be 2 or 3 or 4



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Re: A group of n students can be divided into equal groups of 4
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05 Oct 2018, 15:41
\(\frac{n}{x}= 4 + \frac{1}{x}\)
\(n = 4x + 1\)
n = 5,9,13,17,21,25,29,33
\(\frac{n}{y}= 5 + \frac{3}{y}\)
\(n = 5y + 3\)
n = 8,13,18,23,28,33
13 + 33 = 46
Answer choice B



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A group of n students can be divided into equal groups of 4
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14 Oct 2018, 22:16
Deriving the general formula n=kx+r, based on both statements, where k is the LCM of the 2 divisors and r is the first common integer: n=4q+1> n=1,5,9, 13... n=5p+3> n=3,8, 13... general formula: n=20x+13> n=13, 33... sum=13+33=46Please hit Kudos if you liked this approach




A group of n students can be divided into equal groups of 4
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