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

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.

There are 24 to 57 members right?

1st condition, we can make a group of 4 members & a final group of 3 members.
4M + 3
(M is nothing but a name given to a group of 4 members, just like Ivy League is a name given to a group of 8 universities)

2nd condition, we can make a group of 5 members & a final group of 4 members.
5N + 4
(N is nothing but a name given to a group of 5 members, just like G7 is a name given to a group of 7 highly developed countries)

So take a number, for ex - 7. If we fit 7 into 1st condition we can make a group of 4 & a final group of 3. But 7 won't fit into 2nd condition, which needs 5 in a group & 4 in the final group.

Thus, we must find a number which fits into both the conditions. Here LCM comes into picture. LCM of 4 & 5 is 20. So 20 fits perfectly into both the conditions.

But there is problems with using 20.

Problem - It gives us two different number count.
As per 1st condition the count becomes 23 (20+3) & as per 2nd condition it becomes 24 (20+4). This shouldn't be the case. The count should always be the same.

The reason this problem arises is because the second halves of both the equations are different.

Thus we must find a common number for 3 & 4 as well.

LCM of 3 & 4 is 12.

So the ideal number has to be between 12 (LCM of 3 & 4) and 20 (LCM of 4 & 5).

Try each number from 12 to 20. You'll get 19 as the ideal one.

But 19 is too small. The number has to be in between 24 and 57.

To find the ideal number use the range 12 to 20 as the base & multiple it with 2 & so on. 24 to 40 & 36 to 60 are the other number ranges.

(Because, multiple of a LCM of 2 numbers is also a multiple of those same numbers; 40 & 60 are also the multiples of both 4 & 5; 12 & 36 are also the multiples of 3 & 4)

Ideal numbers are 19, 39 and 59

But, 59 will be too much due to the condition of number of members has to be between 24 and 57.

39 is perfect.

Divide into groups of 11.

39/11 leaves a remainder of 6.

Thus 6 is the answer.


P. S. - Dear Bunuel, I would request you to please check my explanation and tell me whether it is correct or not, so that others can refer to it as well.

Secondly, I'm sorry for such a long explanation. I personally had a problem understanding the only explanation given by RR88,so I solved explained myself.

Thank you.

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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.