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Director
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A group of n students can be divided into equal groups of 4 [#permalink]
 03 Nov 2007, 05:32
Question Stats:
64% (01:37) correct
35% (01:06) wrong based on 4 sessions
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 I got this so far
n = 4q + 1
n = 5q + 3
4q+1 + 5q+3 = 9q+4
plugging in value for q
q=1
q=2
q=3
q=4
q=5 = 45+4 = 49 ? not sure please help _________________
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Director
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n = 4q + 1
n = 5q + 3
I'll start with the first equation: n = 5+k4 where k = 0,1,2,3, ... etc
also, n = 8+m5 where m = 0,1,2,3,.. etc
for first equation: 5,9,13,17,21,25,29,33,37,41,45
for second equation: 8,13,18,23,28,33,38,43,48,53
The sum of minimum n's = 13 + 33 = 46
B
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VP
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my eq is 4x+1 = 5y+3
so 4x = 5y + 2
if y=2 x=3
ify=6 x=8
is the smallest group 8*4 + 1 =33
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SVP
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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?
33
46
49
53
86
4x+1 = 5y+3...........ie: 4x-5y = 2
x,y must be >1 and y is even ie ( 2,4,6,..etc)
if y = 2 thus x = 3 and thus n = 13
if y = 4 thus x is a fraction ( not possible)
if y = 6 thus x = 8 and n= 33
13+33 = 46..... B
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CEO
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alimad 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?
33
46
49
53
86
I got this so far
n = 4q + 1
n = 5q + 3
4q+1 + 5q+3 = 9q+4
plugging in value for q
q=1
q=2
q=3
q=4
q=5 = 45+4 = 49 ? not sure please help
Man ughhhh haha, I couldnt figure this question out forever. Was wondering why everyone was getting 46. I was like comon its 33.
question is really asking what is the SUM of the two possible values of n.
so ya 13+33=46.
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Re: A group of n students can be divided into equal groups of 4 [#permalink]
13 Sep 2012, 07:58
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.
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Re: A group of n students can be divided into equal groups of 4 [#permalink]
13 Sep 2012, 08:15
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siddharthasingh wrote: 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#p1028654From, 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.
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Re: A group of n students can be divided into equal groups of 4
[#permalink]
13 Sep 2012, 08:15
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