As promised:
The Question: How much was left of the small snowball, when half of the volume of the large snowball had melted?
Solution to: Melting Snowballs
Let r1(t) be the radius of the smallest ball (ball 1) at time t.
Let r2(t) de the radius of the largest ball (ball 2) at time t.
Let r0 = r1(0).
Then holds r2(0) = 2 × r0. The surface Ai(t) of ball i at time t is equal to
4 × pi × (ri(t))2 and the volume Vi(t) of ball i at time t is equal to
4/3 × pi × (ri(t))3.
Then holds:
d Vi(t) / dt = - k × Ai(t)
so
d [4/3 × pi × (ri(t))3] / dt = - k × [4 × pi × (ri(t))2]
for a certain melting factor k independent of i. This gives
4 × pi × (ri(t))2 × [d ri(t) / dt] = - k × 4 × pi × (ri(t))2
so
[d ri(t) / dt] = - k
and
ri(t) = ri(0) - k × t.
Suppose that at time th half of the volume of ball 2 has melted, then
4/3 × pi × (r2(th))3 = 0.5 × 4/3 × pi × (r2(0))3
so
(r2(th))3 = 0.5 × (r2(0))3
and
(2 × r0 - k × th)3 = 4 × (r0)3.
Then holds:
k × th = 2 × r0 - 4(1/3) × r0.
At that time th holds for the small ball (ball 1):
r1(th) = r0 - k × th
= r0 - (2 × r0 - 4(1/3) × r0)
= 4(1/3) × r0 - r0
= (4(1/3) - 1) × r0.
The volume of ball 1 is at that time:
V1(t) = 4/3 × pi × (r1(t))3
= 4/3 × pi × ((4(1/3) - 1) × r0)3
= (4/3 × pi × r03) × (4(1/3) - 1)3
= (4(1/3) - 1)3 × V1(0)
So the volume of ball 1 at that moment is only (4(1/3) - 1)3 × 100% of the original volume. This is approximately 20.27%.
Mark on triangles.
"144, 192, 240"
Looks easy, anybody could do it! Well, I think not!
Let the sides of the triangle be a, b, and c, with c=240 being the hypotenuse.
The triangle has the minimal circumference when a=1 and b=sqrt(c2-12) (approximately 240.0). The circumference in that case is approximately 480.0.
The triangle has the maximal circumference when a and b are equal: a=b=sqrt(1/2×c2) (approximately 169.7). The circumference in that case is approximately 579.4.
The only two squares of whole numbers that lie in the interval [480.0, 579.4] are 529 and 576.
Now we know that a+b=529 or a+b=576. In addition, a2+b2=c2, so a2+b2=57600.
Suppose that a+b=529. Then b=529-a, and when we fill that in a2+b2=57600, we get a2+(529-a)2=57600, so a2-289×a+12960.5=0. This equation has no solutions if a must be integer.
Suppose that a+b=576. Then b=576-a, and when we fill that in a2+b2=57600, we get a2+(576-a)2=57600, so a2-336×a+27648=0. This equation has solutions a=192 (b=144) and a=144 (b=192).
Therefore, the sides of the triangle are a=144, b=192, and c=240.
Mark on farming:
"36 days "
How did he know that? Here is a clue.
The Question: For how long can the three animals graze on the field together?
Some assumptions:
The horse, the goat, and the goose eat grass with a constant speed (amount per day): v1 for the horse, v2 for the goat, v3 for the goose.
The grass grows with a constant amount per day (k).
The amount of grass at the beginning is h.
There is given:
When the horse and the goat graze on the field together, there is no grass left after 45 days. Therefore, h-45×(v1+v2-k) = 0, so v1+v2-k = h/45 = 4×h/180.
When the horse and the goose graze on the field together, there is no grass left after 60 days. Therefore, h-60×(v1+v3-k) = 0, so v1+v3-k = h/60 = 3×h/180.
When the horse grazes on the field alone, there is no grass left after 90 days. Therefore, h-90×(v1-k) = 0, so v1-k = h/90 = 2×h/180.
When the goat and the goose graze on the field together, there is also no grass left after 90 days. Therefore, h-90×(v2+v3-k) = 0, so v2+v3-k = h/90 = 2×h/180.
From this follows:
v1 = 3 × h/180,
v2 = 2 × h/180,
v3 = 1 × h/180,
k = 1 × h/180.
Then holds for the time t that the three animals can graze together: h-t×(v1+v2+v3-k) = 0, so t = h/(v1+v2+v3-k) = h/(3×h/180+2×h/180+1×h/180-1×h/180) = 36. The three animals can graze together for 36 days.
Merlin, who has now been nominated to join the Hall of Fame
(aka ?'The Labyrinth of Zzqeue) for his contribution;
MIDAS:
"I get coins divided equally at least through room 930 (Excel started complaining after that)."
"Solution:
Let number of rooms = x
Let number of coins = y
y=x^3 (number of rooms times boxes per room times coins per box)
The number of coins after one box removed for the barber: x^3-x.
If x^3-x modulo 6=0 then coins are divided equally among the six sons."
A fair division of Midas' coins is indeed possible. Let the number of rooms be N. This means that per room there are N boxes with N coins each. In total there are N×N×N = N3 coins. One box with N coins goes to the barber. For the six brothers, N3 - N coins remain. We can write this as: N(N2 - l), or: N(N - 1)(N + l). This last expression is divisible by 6 in all cases, since a number is divisible by 6 when it is both divisible by 3 and even. This is indeed the case here: whatever N may be, the expression N(N - 1)(N + l) always contains three successive numbers. One of those is always divisible by 3, and at least one of the others is even. This even holds when N=1; in that case all the brothers get nothing, which is also a fair division!
483 623 S56 023
6A7 224 368 F8
276 545 T621 45
82E 857 339 Y75
A well known phrase or saying.
Hint: The answer is in the numbers. (not the numbers themselves)
During the final round of New York's All-State Chess Tournament, the eventual top 4 finishers in the tournament played their final game against 4 different opponents. The tournament boards were numbered from 1 to 50 to facilitate location and identification of games. Each of the top 4 finishers played a different opening in this last game. Use the clues below to determine the first and last names of the first through fourth place winners of the tournament, the chess opening each used in his last game, and the number of the chess board on which he played his final game.
1. Mr. Hart played the King's Indian Defence.
2. Steve placed ahead of the one who used the Ruy Lopez opening.
3. The top 4 players were Larry, Mr. Korn, the contestant who opened with Queen's Gambit, and a player who played on an even-numbered board.
4. Mr. Rose finished exactly 2 places ahead of the player on board 31.
5. The number of the 3rd place winner's board is at least 10 higher or at least 10 lower than that of Mr. Baird.
6. The number of the board of the contestant who placed immediately after Bert is exactly 15 higher than the number of the board played by the one who placed immediately before Tom.
7. The lowest board number of the top 4 finishers was exactly half the number of the board played by the one who placed immediately after the one who played the Giuoco Piano opening.
8. The highest board number was exactly 8 higher than the board number of the man who finished exactly two places after Bert.
Something a little different: Sets of Concrete Propositions, proposed as Premises for Sorties. Conclusions to be found.
Example:
1. Babies are illogical;
2. Nobody is despised who can manage a crocodile;
3. Illogical persons are despised.
Conclusion: Babies cannot manage crocodiles
If you are with me so far, try:
1. My saucepans are the only things that I have that are made of tin;
2. I find all your presents very useful;
3. None of my saucepans are of the slightest use.
1. No potatoes of mine, that are new, have been boiled;
2. All of my potatoes in this dish are fit to eat;
3. No unboiled potatoes of mine are fit to eat.
1. There are no Jews in the kitchen;
2. No Gentiles say "shpoonj";
3. My servants are all in the kitchen.
Knights (who always tell the truth), Knaves (who always lie) and Normals (who sometimes lie and sometimes tell the truth). On this island, a man and a woman may only marry if they are both Normal, or one of them is a Knight and the other one is a Knave. Now you meet Mr. and Mrs. A who tell you the following:
Mr. A: "My wife is not normal"
Mrs. A: "My husband is not normal"
What types of persons are Mr. and Mrs. A
(With a little clue)
