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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 [#permalink]  03 Nov 2007, 04:32
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66% (02:28) correct 34% (01:35) wrong based on 135 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

[Reveal] Spoiler:
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
[Reveal] Spoiler: OA

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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
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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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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Re: Remainder [#permalink]  03 Nov 2007, 10:11
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

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, 06:58
Expert's post
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, 07:15
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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#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.

Hope it helps.
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Re: A group of n students can be divided into equal groups of 4 [#permalink]  11 Dec 2013, 02:12
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Re: A group of n students can be divided into equal groups of 4 [#permalink]  11 Dec 2013, 04:52
4x + 1 = n (1)
5y + 3 = n (2)

Equating (1) and (2)
4x + 1 = 5y + 3
4x = 5y + 2
Put y=1,2,3,4,etc.
Since (5y + 2) need to be a multiple of 4 to satisfy the equation on the left side. The 2 minimum values of y are 2 and 6.

So, n = 5y + 3
n = 5(2) + 3 = 13 and
n = 5(6) + 3 = 33

Adding the 2 minimum values of n
13 + 33 = 46

So, the correct answer is B.
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Re: A group of n students can be divided into equal groups of 4 [#permalink]  11 Dec 2013, 11:28
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From question we get N=
=> 4K+1=5P+3
K=P+(P+2)/4

So for P=2 & 6 we get K an integer i.e. K=13 & 33

Sum=13+33=46.
B is correct.
Re: A group of n students can be divided into equal groups of 4   [#permalink] 11 Dec 2013, 11:28
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