winners

Solution to Puzzle 1:
Let X be the total no gems. Then
X/3 they got from the ground
X/5 they got from the bed
X/6 was in her hand
X/10 he caught
6 rwas remainining
This leads to the equation X= X/3 +X/5 +X/6+X/10+6
Solving this we get X = 30. i.e there were altogether 30 Gems

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Gayathri. V.
Wisconsin, USA

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Solution to Puzzle 2:

(Umesh Clarifies that the problem was not his invention; but the solution is. I have got quite a few responses that you need a minimum of 8 weighings to order those 5 balls. See the solution Umesh has given.)

Consider the worst scenario in each case.
If there are 2 balls, you need only 1 weighing.
If you have 3 balls, you need 3 weighings.
1. Weigh any two - Let the heavy be H and the light be L.
2. Weigh the other(O) with H. If it is heavier, the order is L, H, O.
3. If O is lighter, you need another weighing. Weigh it with L.
If O is lighter, the order is O, L, H.
If O is heavier, the order is L, O, H.
If you have 4 balls, you need 5 weighings. There are two ways.
(A) 1-3. In 3 weighings, arrange 3 balls (L, M, H), as described above.
4. Compare the other (O) with M, the middle one.
5. If O is heavier, compare it with H.
If O is lighter, compare it with L.
The technique is "compare the middle, and eliminate one side".
(B) 1. Compare any two, let the heavier be H1 and the lighter be L1.
2. Compare the other two, let the heavier be H2 and the lighter be L2.
3. Compare H1 with H2.
4-5. If H1 > H2, the order is (L2, H2, H1), and we know L1 < H1. So you need to compare L1 with H2 and L2 to find the final order.
Similar case if H2 > H1.
The technique is "consider in pairs, and merge them in bulk".

If you have 5 balls, you need 7 weighings. You need to combine the two techniques mentioned above.
1. Compare any two balls - Let them be H1 and L1.
2. Compare any two other balls - Let them be H2 and L2. The fifth ball is O.
3. Compare H1 and H2.
If (H1 > H2), You know (L2, H2, H1), L1 < H1.
If (H1 < H2), You know (L1, H1, H2), L2 < H2.
We have used the "consider in pairs", and merge them in bulk" technique.
Now let us use the other technique.
4-5. Using the technique A above, insert O in the sorted list of 3 in 2 weighings.
(i.e., Compare with the middle one, and depending on the result, compare with the lighter or the heavier).
Upto this, it is similar to the 4-ball scenerio B, except that the 4th ball in the list is the (original) fifth one.
6-7. Now we have a sorted list of four balls, and we need to insert the fifth into it. But the fifth one is either L1 or L2, and you know it is lighter than H1 or H2, respectively. So, actually, you need to consider only the other 3 balls in the list, and you know you can do it in 2 weighings, using our "compare the middle, and eliminate one side" technique.
For example, if the sorted list is (L1, H1, O, H2) and the fifth ball is L2,
Compare it with H1, and depending on the result, compare it with L1 or O.
(If it is heavier than all three, you know the order is (L1, H1, O, L2, H2))