“We – A, B and C – each have some children.

(i) A has at least one girl and twice as many boys as girls.

(ii) B has at least one girl and three times as many boys as girls.

(iii) C has at least one girl and three more boys than girls.

(iv) When I tell you the number of children we have – a number less than 25 – you will know how many children I have, but not how many children each of the others has. Altogether we have……..”

Who is the speaker and how many children does the speaker have?

According to (i), A can have 3, 6, 9, 12, 15, etc children.
According to (ii), B can have 4, 8, 12, 16, 20, etc children.
According to (iii), C can have 4, 5, 6, 7, 8, 9, etc children. (the solution incorrectly states that C can have 5, 7, 9, 11 children.)

Plotting out all the possible solutions on a graph shows:

11 X <-(minimum number of kids possible)
12 X
13 X
14 X X
15 X X X
16 X X X
17 X X X X
18 X X X X
19 X X X X X
20 X X X X X X
21 X X X X X X
22 X X X X X X
23 X X X X X X X
24 X X X X X X X

Since the only total that provides two options (we were told we'd know how many one person had but not the other two) is 14, we can deduce that the total number of children is 14. The two solutions for 14 children are:

A=3, B=4, C=7
or
A=6, B=4, C=4

B is the only one we know for sure, so B is the speaker.

Posted by Erik O. on 2004-06-07 21:42:43