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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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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 4 members, will have 5 members. If the members are separated into as many groups of 11 as possible, how many members will be in the final group?

A. 7
B. 6
C. 3
D. 2
E. 1


First we can represent the total of members with either \(4i+3\) or \(5j + 4\), while i and j are integers.

Then we have \(4i + 3 = 5j + 4\) and \(4i = 5j + 1\). We may pick a solution here such as j = 3 with i = 4 which would give 16=16. Then observe we can always find the next solution by incrementing j by 4 and i by 5, as that would increase both sides by 20 which maintains the equality.

Hence i = 4 or j = 3 is a solution, we go back and find \(4i+3 = 19.\) We can always increase by 20 for the next solution, so 39 and 59 are also possible options. However only 39 is within range.

39%11 = 6 so the answer is B.

Ans: B
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A local club has between 24 and 57 members. The members of the club ca [#permalink]
Alternative method:

4a+3=X => X-3=4a
5b+4=X => X-4=5b

X-3 is divisible by 4, therefore units digit minus three is even, therefore units digit is odd.
X-4 is divisible by 5, therefore units digit minus four is 0 or 5, therefore untis digit is 4 or 9.

Therefore, units digit is 9.

X can be {29, 39, 49}.
X-3 can be {26, 36, 46}. Only 36 is divisible by 4. Thus, X=39.


Brute-force guessing/iteration is still probably faster, though.
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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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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 4 members, will have 5 members. If the members are separated into as many groups of 11 as possible, how many members will be in the final group?

A. 7
B. 6
C. 3
D. 2
E. 1


You COULD go with a elegant solution...or you could get to the right answer at least as fast just using brute force...

"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 we have 27, 31, 35, 39, 43, 47, 53, or 57 members.

"The members can also be separated into groups so that all groups but final group, which will have 4 members, will have 5 members."
This means we have 24, 29, 34, 39, 44, 49, or 54 members.

Only one number is in both lists: 39.

If we make groups of 11, we will use up 33 and be left with 6.

Answer choice B.

There are no bonus point on the GMAT for mathematical elegance or brilliance...just get the right answer as quickly as possible and with as little risk of a silly mistake as possible.
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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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ThatDudeKnows wrote:
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 4 members, will have 5 members. If the members are separated into as many groups of 11 as possible, how many members will be in the final group?

A. 7
B. 6
C. 3
D. 2
E. 1


You COULD go with a elegant solution...or you could get to the right answer at least as fast just using brute force...

"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 we have 27, 31, 35, 39, 43, 47, 53, or 57 members.

"The members can also be separated into groups so that all groups but final group, which will have 4 members, will have 5 members."
This means we have 24, 29, 34, 39, 44, 49, or 54 members.

Only one number is in both lists: 39.

If we make groups of 11, we will use up 33 and be left with 6.

Answer choice B.

There are no bonus point on the GMAT for mathematical elegance or brilliance...just get the right answer as quickly as possible and with as little risk of a silly mistake as possible.


Indeed. In fact, yours is not only faster but also simpler/less prone to error.

I'm just curious of what the formal/analytic solution would look like, even if it is off-topic.
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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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