Ten Problems

The fastest time they could finish is 5:13pm.
First, let’s plug in some numbers just to see what kind of results we get. Since Allen moves small boxes faster and Percy move large boxes faster, then we can have Allen move all of the 16 small boxes (taking 32 minutes) and Percy move all of the 10 large boxes (taking 50 minutes). Overall, they would finish in fifty minutes.

We could transfer two of the large boxes to Allen, making him carry 16 small boxes and 2 large boxes, taking 44 minutes. Percy would carry 8 of the large boxes, taking 40 minutes. So in total, they would take 44 minutes.

If we now transfer one small box from Allen to Percy, making Allen carry 15 small boxes and 2 large boxes (42 minutes) and Percy carry 8 large boxes and 1 small box (43 minutes), they’d take 43 minutes to finish.

Is 43 minutes the shortest time? We are now going to mathematically show that it is the case.

Suppose that it took them 42 minutes to finish the job. Then the total amount of working time for both of them would be 84 minutes. Suppose that Allen moves small boxes and large boxes, which would take minutes. Percy would move small boxes and large boxes, taking minutes. But we know that the total working time is less than 42 minutes, so , or .

Since and , then the possible pairs of and are , which produce the following working times:

x	y	Allen Small	Allen Large	Allen Time	Percy Small	Percy Large	Percy Time
16	0	16	0	32	0	10	50
16	1	16	1	38	0	9	45
16	2	16	2	44	0	8	40
15	0	15	0	30	1	10	53
15	1	15	1	36	1	9	48
14	0	14	0	28	2	10	56

In each of these cases, while the total working time is no more than 84 minutes, it always takes longer than 43 minutes to finish. Thus, we can see that it is impossible to finish in 42 minutes or less, and so the answer is that they take 43 minutes to finish.

Since they started at 4:00pm, talked for 30 minutes, then spent 43 minutes working, the earliest time they could finish is 5:13pm.

7 is the maximum.

Suppose that we removed some Legos from the bag, and the remaining Legos do not satisfy the condition. What is the maximum number of Legos that remain in this situation? In order for the condition to not be satisfies, either we don’t have four Legos of any color, or there are at least four Legos of one color, but don’t have three Legos of any other color.

In the second case, we could only have two Legos of one color for two of the colors, and as many as possible for the third color. The maximum number of Legos of one color is 8 (yellow balls), so the maximum number of Legos in this case is . So if we remove 8 Legos from the bag, we might not satisfy our condition; but if we only remove 7, the condition will be satisfied.

Therefore, the maximum number of balls Igor can take out is 7.

Janet will only have a winning strategy for N=35.
First, we consider a few small values of N.

If N=1, Tony wins by picking up the first coin.

If N=2, Tony must take only 1 coin (since he cannot take 3 or 4) and Janet wins by taking the other coin.

If N=3 or 4, Tony again wins by taking all of the coins.

When N=5, if Tony takes 3 coins, then Janet is left with 2 coins and cannot win (since Tony could not win when starting with 2 coins).

When N=6, if Tony takes 4 coins to start, then Janet is again left with no chance to win.

When N=7, if Tony takes 1, 3, or 4 coins to start, then Janet is left with 6, 4, or 3 coins. But choosing from 6, 4, or 3 coins is a winning position for the first person choosing. Thus, N=7 is a losing position for Tony, since Janet can just use Tony’s winning strategy in step two.

As we can now gradually see, Janet will have a winning strategy if give Tony a winning strategy, since if Tony starts and chooses 1, 3, or 4, Janet will now start with , , and .

In other cases, Tony will have a winning strategy.

From the above knowledge we have, we can construct a list of number of coins and the whether there is a winning strategy for each person.

Tony:  1, 3, 4, 5, 6, 8, 10, 11, 12, 13, 15, 17, 18, 19, 20, 22, 24, 25, 26, 27, 29, 31, 32, 33, 34
Janet: 2, 7, 9, 14, 16, 21, 23, 28, 30, 35

So for N between 31 and 35 inclusive, Janet will only have a winning strategy when N=35.

There are several solutions to this problem, but all of them are along the lines of:

1. Put three cubes in one glass, three cubes in the other, and four cubes in the last glass.

2. Now put one of the glasses contain three cubes into the glass contain four cubes.

Now you have two glasses with three sugar cubes in them and one glass with seven!

!

Since is negative, the greater than sign should have switched to a less than sign when both sides of the inequality are multiplied by .