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03 Nov 2007, 05:32
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:09) correct
27%
(02:11)
wrong
based on 953
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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
A. 33
B. 46
C. 49
D. 53
E. 86
13 Sep 2012, 08:15
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siddharthasingh
Isn't there any arithmetic solution to this question. I mean, just Hit n Trial method. Indeed there must be an arithmetic way out. Using this hit and trial method sometimes takes much longer time, henceforth I needed to go with a systematic approach.
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
Given:
\(n=4q+1\), so \(n\) could be: 1, 5, 9, 13, ...
\(n=5p+3\), so \(n\) could be: 3, 8, 13, ...
General formula for \(n\) based on above two statements will be: \(n=20m+13\) (the divisor should be the least common multiple of above two divisors 4 and 5, so 20 and the remainder should be the first common integer in above two patterns, hence 13). For more about this concept see: manhattan-remainder-problem-93752.html#p721341, when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552, when-the-positive-integer-a-is-divided-by-5-and-125591.html#p1028654
From, \(n=20m+13\) we have that the two smallest possible values of \(n\) are 13 (for \(m=0\)) and 33 (for \(m=1\)).
13+33=46.
Answer: B.
Hope it helps.
21 Jan 2012, 14:59
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enigma123
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?
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?
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:
https://gmatclub.com/forum/manhattan-re ... ml#p721341
https://gmatclub.com/forum/when-positiv ... ml#p722552
https://gmatclub.com/forum/when-the-pos ... l#p1028654
Hope it helps.
