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A local club has between 24 and 57 members. The members of the club ca

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A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 13 Dec 2016, 10:53
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A local club has between 24 and 57 members. The members of the club can be separated into groups of which all but the final group, which will have 3 members, will have 4 members. The members can also be separated into groups so that all groups but final group, which will have 3 members, will have 5 members. If the members are separated into as many pairs as possible, how many members will be in the final group?

A. 1
B. 2
C. 3
D. 5
E. 6

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Re: A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 10 Nov 2018, 02:24
Bunuel wrote:
A local club has between 24 and 57 members. The members of the club can be separated into groups of which all but the final group, which will have 3 members, will have 4 members. The members can also be separated into groups so that all groups but final group, which will have 3 members, will have 5 members. If the members are separated into as many pairs as possible, how many members will be in the final group?

A. 1
B. 2
C. 3
D. 5
E. 6


Responding to a PM..

since we are looking for number of people left when pairs are made..
so we have 2 in each group except the last which can have 1 or 2..
now 1 or 2 will depend on the total number - If total is ODD, only one will be in the last group an dif number is even, 2 will be left over..

How do we check TOTAL..
the line - The members of the club can be separated into groups of which all but the final group, which will have 3 members, will have 4 members. - is enough to get to an answer, nothing else is required.
Now, if we put all in groups of 4, the last one has 3. Thus we have ODD number of people and we will have 1 in the last group.

A
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Re: A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 13 Dec 2016, 13:07
1
4
\(N = 4x + 3\)
\(N = 5y +3\)

Hence: \(N = LCM (4,5)*n + 3 = 20n + 3\)

Our AP: \(3, 23, 43, 63\)

\(24 < N < 57\) -----> \(N=43\)

Now we need to find remainder upon division by 2 (division into pairs)

43 is odd and our remainder is 1.

Answer A

Another fast approach: N=20n + 3 = even + odd = odd will always produce odd number. Hence remainder upon division by 2 will be 1.
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A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 17 Dec 2016, 00:12
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1
We have to find the number of members in the final group when the groups are formed by 2 members max(pair).
So final group will either have 2 members or 1 member (leaving us with choice A and choice B).

For the first condition, assume that there is x number of the members in each group except final one. Total members= x*4+3
For the second condition, assume that there is y number of the members in each group except final one. Total members= y*5+3
x*4+3=y*5+3
4x=5y

the possibe integer values that can satisfy above equation are x=5, y=4 or x=10, y=8 or x=15, y=12

We are given that the number of the members should be between 24 and 57

the only value that satisfy above condition is x=10, y=8

total number of the members 10*4+3=43=> odd number, the number of members in the last group will certainly be 1

The correct answer is A.
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Re: A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 09 Feb 2017, 12:06
The members of the club can be separated into groups of which all but the final group, which will have 3 members, will have 4 members -

This means the number will be a multiple of 4+3 so the number could be 24+3, 28+3, 32+3, 36+3,40+3 etc. Right here is enough to prove that when this number is divided by 2 (pairs) we will get a remaining group of 1 but just to be sure we go to second statement. Here we can say that the number of members is a multiple of 5 and then a+3 so 25+3, 30+3, 35+3 or 40+3

43 is common, 43/2 will give a remainder of 1.

This could be a great DS question too.
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A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 09 Feb 2017, 14:34
Bunuel wrote:
A local club has between 24 and 57 members. The members of the club can be separated into groups of which all but the final group, which will have 3 members, will have 4 members. The members can also be separated into groups so that all groups but final group, which will have 3 members, will have 5 members. If the members are separated into as many pairs as possible, how many members will be in the final group?

A. 1
B. 2
C. 3
D. 5
E. 6


lowest possible membership=4*5+3=23
23/5=Q4+R3
23/4=Q5+R3
23/11=Q2+R1
next lowest membership=23+4*5=43
43/21=Q2+R1
1
A
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Re: A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 05 Nov 2018, 00:53
Since 2 leaves a remainder of 0 or 1 , ans is A
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Re: A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 08 Apr 2019, 08:49
pandeyashwin wrote:
Since 2 leaves a remainder of 0 or 1 , ans is A


Came to say this. Easiest 2sec approach.
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A local club has between 24 and 57 members. The members of the club ca  [#permalink]

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New post 08 Apr 2019, 09:59
I am not sure whether this is the correct way to look at the problem but it does get the answer:

It says maximum groups possible: minimum number of people required in a group i.e 2
Therefore any odd number will leave 1 as a remainder. No need to look at the pairing mentioned in the question.

Kindly if anyone can verify about this approach, it would be really helpful.
Thanks.
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A local club has between 24 and 57 members. The members of the club ca   [#permalink] 08 Apr 2019, 09:59
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