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

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Director
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A group of n students can be divided into equal groups of 4  [#permalink]

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New post 03 Nov 2007, 05:32
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B
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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

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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Re: A group of n students can be divided into equal groups of 4  [#permalink]

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New post 13 Sep 2012, 08:15
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8
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.

Answer: B.

Hope it helps.
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Director
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New post 03 Nov 2007, 05:45
1
1
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



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New post 03 Nov 2007, 06:58
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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  [#permalink]

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New post 03 Nov 2007, 11:01
1
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]

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New post 03 Nov 2007, 11:11
1
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]

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New post 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]

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New post 11 Dec 2013, 05:52
2
1
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]

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New post 11 Dec 2013, 12:28
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.
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Re: A group of n students can be divided into equal groups of 4  [#permalink]

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New post 16 Oct 2016, 11:07
B is correct. Here's why:

Given the information in the question we can create the following two equations:

n = 4x+1 (5,9,13,17,21,25,29,33)
n = 5y+3 (8,13,18,23,28,33)

The parentheses following the equations represent potential value of n given various values of x (i.e. 1+). We can see from these calculations that 13 and 33 line up with both equations. Therefore, we can add 13 and 33 and we are done :)
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Re: A group of n students can be divided into equal groups of 4  [#permalink]

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New post 03 Nov 2016, 03:03
n=4q+1

n= 5k+3

The smallest numbers that satisfies the equations are 13 and 33...so Answer B
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Re: A group of n students can be divided into equal groups of 4  [#permalink]

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New post 26 Jan 2018, 14:09
Hi All,

You would likely find it easiest to 'brute force' this question (simply write down enough of the possibilities until you either spot the pattern involved or have the exact answer on your pad).

Equal groups of 4 with 1 left over COULD be... 5, 9, 13, 17, 21, 25, 29, 33.....
Equal groups of 5 with 3 left over COULD be... 8, 13, 18, 23, 28, 33....

The two SMALLEST values that fit BOTH groups are 13 and 33. We're asked for the sum of those values...

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Re: A group of n students can be divided into equal groups of 4 &nbs [#permalink] 26 Jan 2018, 14:09
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