How Many Marbles?

A number of children played at marbles. Each child won a different number of games and the total number of games won by the boys equaled the total number of games won by the two girls. As a momento of the occasion, they each received a present of marbles, equal to the square of the number of games he or she had won. In total, the boys received as many marbles as the girls, and the total number of marbles received was less than 100. How many marbles were given as presents?

If the number of girls is allowed to vary (not set at two as above) and the total number of marbles increases to no more than 1000, how many solutions are there?

Source: Work This One Out, Longley-Cook, #77, 1960.

Solutions were received from Denis Borris, Sorin Ionescu, Sandy Thompson, Joseph DeVincentis, Philippe Fondanaiche, and Dave Bachtel. The solution to the first question:

82 marbles were given as presents. (3 boys(1,2,6) and 2 girls (4,5)) 1*1+2*2+6*6=4*4+5*5=41

I didn't realize how very many solutions there were to the second problem. Joseph DeVincentis found many solutions for exactly 1000 marbles handed out, and Sorin Ionescu and Philippe Fondanaiche sent other solutions. For example,

Sorin Ionescu:
818 marbles were given as presents. (3 boys(3,12,16) and 3 girls (6,7, 18)) 3*3+12*12+16*16=6*6+7*7+18*18=409

Philippe Fondanaiche: 

The number of girls is allowed to vary and N ==>1000
13 children at the maximum with 6 boys and 7 girls (or vice versa):

B: 1,2,4,7,9,10,14 
G:3,5,6,8,12,13
sum of games won =94  sum of marbles = 894

B:1,2,3,8,10,11,14
G:4,5,6,7,12,15
sum of games won:49  sum of marbles:990

B:1,3,4,6,9,12,14
G:2,5,8,10,11,13
sum of games won:49   sum of marbles:966

B:1,2,3,8,9,12,13
G:4,5,6,7,11,15
sum of games won:48   sum of marbles:944
Joseph DeVincentis' solution for total marbles = 1000: 
For the first problem, the solution is that the girls won 5 and 4 games,
and the boys won 6, 2, and 1 games.  The boys and girls together received
82 marbles.

For the second problem, I will only consider the solutions where the largest
number of games won by any child is by, say, a girl.  For any solution, there
is another with the boys and girls swapped.  At the end, I will multiply
my number of solutions by 2 to account for these.

In addition, I will ignore solutions in which one of the children won
no games.  These are also arguably outside the scope of the problem,
since they require one child to receive a present of no marbles.
However, since any such solution also leads to another solution without
the winless child, and any solution without a winless child leads to
two others, one with a winless boy and another with a winless girl.
If these are to be considered, multiply my number of solutions by 3.

Now, the total number of marbles received by the girls is 500, so
no one child could have won more than 22 (22^2 = 484) games.

The following are the possible numbers of games won by either the
boys or the girls, to allow them to receive exactly 500 marbles,
sorted by the sum of the games played:

22 4                  (26)
20 10                 (30)
21 7 3 1              (32)
20 8 6                (34)

As each of these is the only set with its sum, these do not provide
solutions to the problem.

19 9 7 3              (38)
16 12 10              (38)

These two make one solution. (1)

20 7 5 4 3 1          (40)
19 10 5 3 2 1         (40)
17 13 5 4 1           (40)
19 8 7 5 1            (40)
17 11 9 3             (40)

The only one of these which does not contain 1 is the last, and it makes
a solution with the one other set that does not contain 3 or 17. (2)

19 8 7 4 3 1          (42)
17 11 8 5 1           (42)
16 13 7 5 1           (42)
15 12 11 3 1          (42)
16 12 8 6             (42)
15 13 9 5             (42)

Only the last two of these do not contain 1; they make a solution themselves.
The 16 12 8 6 set has something with each other set here, but 15 13 9 5 makes
a solution with the first set listed. (3,4)

18 11 5 4 3 2 1       (44)
17 11 8 4 3 1         (44)
16 13 7 4 3 1         (44)
15 12 9 7 1           (44)
15 14 7 5 2 1         (44)
19 8 6 5 3 2 1        (44)
18 9 7 6 3 1          (44)
17 10 9 5 2 1         (44)
14 13 11 3 2 1        (44)
17 11 7 6 2 1         (44)
15 13 9 4 3           (44)
17 11 7 5 4           (44)

Only the last two of these do not contain 1, and they share 4.  The only
solution from this group is 15 13 9 4 3 and 17 11 7 6 2 1. (5)

14 13 10 5 3 1        (46)
17 10 9 4 3 2 1       (46)
16 11 8 7 3 1         (46)
16 13 6 5 3 2 1       (46)
15 14 7 4 3 2 1       (46)
14 13 9 7 2 1         (46)
17 9 8 7 4 1          (46)
18 9 7 5 4 2 1        (46)
16 11 9 5 4 1         (46)
15 11 10 7 2 1        (46)
15 13 8 5 4 1         (46)
17 10 7 6 5 1         (46)
15 11 9 8 3           (46)
19 7 6 5 4 3 2        (46)

Only the last two of these don't contain 1, and the only solution from the
group is 15 11 9 8 3 and 17 10 7 6 5 1. (6)

15 13 7 6 4 2 1       (48)
16 12 7 5 4 3 1       (48)
16 10 9 7 3 2 1       (48)
17 10 7 6 4 3 1       (48)
14 12 11 5 3 2 1      (48)
13 12 11 7 4 1        (48)
17 9 8 6 5 2 1        (48)
15 12 9 5 4 3         (48)
17 11 6 5 4 3 2       (48)

Only the last two solutions do not have 1, but every solution has at least
one of 3, 4, and 5, so there are no solutions from this group.

13 12 11 6 5 2 1      (50)
16 11 7 6 5 3 2       (50)
17 9 7 6 5 4 2        (50)
14 13 8 6 5 3 1       (50)
14 12 9 7 5 2 1       (50)
15 11 10 5 4 3 2      (50)
14 13 9 5 4 3 2       (50)
14 11 10 7 5 3        (50)
15 11 8 7 5 4         (50)
15 10 9 7 6 3         (50)
17 9 8 6 4 3 2 1      (50)
15 11 8 7 6 2 1       (50)

Here it is no longer reasonable to use "not having 1" as the first basis
of eliminating pairs.  However, only the last three do not have 5.  These
all have 6, though, so there are no solutions within the set.  The first
four sets also fail to make solutions with these three because they
contain 6.  Each of the others fails to make a solution because it shares
either 7, 4, 3, or 2 with each of the last three sets.

15 10 8 7 6 5 1       (52)
16 9 8 7 6 3 2 1      (52)
14 12 9 7 4 3 2 1     (52)
14 10 9 8 7 3 1       (52)
15 12 7 6 5 4 2 1     (52)
13 11 10 7 6 5        (52)
16 9 8 7 5 4 3        (52)
15 10 9 7 5 4 2       (52)
13 12 10 7 5 3 2      (52)
12 11 10 9 7 2 1      (52)
13 11 9 8 7 4         (52)
13 11 10 8 6 3 1      (52)

Only the last set does not contain 7.  Each other set has at least one
number in common with this set.

13 11 10 7 6 4 3      (54)
13 11 10 8 5 4 2 1    (54)
15 11 8 6 5 4 3 2     (54)
15 10 8 7 6 4 3 1     (54)
13 12 10 6 5 4 3 1    (54)
13 12 9 7 6 4 2 1     (54)
13 11 9 8 6 5 2       (54)
13 10 9 8 7 6 1       (54)

Only the last two sets do not contain 4.  Every other set contains at least
one of 9, 8, and 6, so no solutions arise from this group.

13 12 8 7 6 5 3 2     (56)
15 9 8 7 6 5 4 2      (56)
14 10 9 7 6 5 3 2     (56)
12 11 10 9 5 4 3 2    (56)
12 11 10 8 6 5 3 1    (56)
12 11 9 8 7 5 4       (56)
12 11 9 8 7 6 2 1     (56)

Every set but the last contains 5; every other set shares a number with
that set.

13 10 9 8 6 5 4 3     (58)
12 10 9 8 7 6 5 1     (58)
12 11 10 7 6 5 4 3    (58)

All these sets contain 5.

12 11 9 8 6 5 4 3 2   (60)
14 10 8 7 6 5 4 3 2 1 (60)
12 10 9 8 7 6 4 3 1   (60)

All these sets contain 3.


So, the solutions are:

19 9 7 3      - 16 12 10
19 8 7 5 1    - 17 11 9 3
16 12 8 6     - 15 13 9 5
19 8 7 4 3 1  - 15 13 9 5
17 11 7 6 2 1 - 15 13 9 4 3
17 10 7 6 5 1 - 15 11 9 8 3

This gives six solutions, or 12 if we count the same solutions with the
boys and girls switched, or 36 if we also allow a child to win 0 games.
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