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

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03 Nov 2007, 04:32

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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 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 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.

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

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13 Sep 2012, 06:58

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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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, ...

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

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11 Dec 2013, 04:52

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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

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

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16 Oct 2016, 10:07

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

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

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