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Club \(X\) has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table

Expert replies
by VJesus12 » Sun Apr 25, 2021 2:35 pm

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Club \(X\) has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Answer: E

Source: Official Guide
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Source: — Problem Solving |

VJesus12 wrote:
Sun Apr 25, 2021 2:35 pm
Club \(X\) has more than 10 but fewer than 40 members. Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables, and sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables. If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Answer: E

Source: Official Guide
This is a nice remainder question in disguise.
For this question, we'll use a nice rule that that says:
If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.

Let N = the TOTAL number of members.

Sometimes the members sit at tables with 3 members at one table and 4 members at each of the other tables...
With 4 members at each table, then N is multiple of 4
However, we still have one more table to consider.
Since the last table has 3 members, we know that N is 3 greater than a multiple of 4
In other words, when we divide N by 4, the remainder is 3
By the above rule, some possible values of N are: 11, 15, 19, 23, 27, etc
NOTE: I started at 11, since we're told that 10 < N < 49

Sometimes they sit at tables with 3 members at one table and 5 members at each of the other tables
Using the same logic as above, this question tells us that, when we divide N by 5, the remainder is 3
By the above rule, some possible values of N are: 13, 18, 23, 28, 33, 38

Let's check the two results.
First we learned that N can equal 11, 15, 19, 23, 27, 31, 35, 38
Next we learned that N can equal 13, 18, 23, 28, 33, 38
Once we check the OVERLAP, we can see that N equals 23

If they sit at tables with 6 members at each table except one and fewer than 6 members at that one table, how many members will be at the table that has fewer than 6 members?
If N = 23, then we'll have 3 tables with 6 members and the remaining 5 members will sit at the other table.

Answer: E
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