My eight favorite riddles (v. difficult) - Page 4

It's a trick question, it stays constant. The water level is dependent on the amount of water being displaced. The amount of water displaced is proportional to the weight of the object. (Since ice has a density of 7/8 of seawater or something, only 1/8 of an iceberg is visible). Thus, since the total weight (boat + person + brick) remains unchanged, the water level is also unchanged.

Q1 Are there exactly 2 goats in 1/2
Q2 Is there exactly 1 goat in 2/3

If you get a Blue Blue lights (or yellow yellow irrelevant) they can either mean yes or no. If they mean Yes Yes then the car must be 3. If it's a no no then there must be 1 or 0 goats in 1/2 and 2 or 0 goats in 2/3. If 0 in 1/2 there must be 2 in 2/3 so there must be 1 in 1/2. Impossible. If there's 1 goat in 1/2 it must be either goat 1 or 2. Regardless of the goat there then must be 1 in 2/3.

SO THERE CAN'T BE NO NO IT MUST BE YES YES

Now for BY or YB (doesn't matter)

If it's a yes to 1 and a no to 2 there must be 2 in 1/2. So there must be a 2 in 2/3 SO THERE CAN'T BE A YES NO. It must be no yes.

If it's a no to 1 then there must be 0 or 1 goats in 1/2. If it's 0, there's 2 in 2/3 and thus 1 in 1/2. THERE CAN'T BE 0 thusly. If there's 1 it must be either 1/2. If it's 2, then 3 must be the other goat, and we have 2 and 3 being goats thus 2 goats in 2/3 and impossible. If it's a 1, then the other one can't be 2 so the other goat must be 3 and thus car is 2

sorry.. there can be a no no (car in 1, goat in 2, goat in 3 - 1 goat in 1/2, 2 goats in 2/3)

Break up the marbles into 3 groups of 4: group 1, 2, and 3.

1st weighing: Find weight of group 1 compared to group 2. Two cases can occur: either group 1's weight is not equal to group 2's, or group 1's weight and group 2's weight are the same, and therefore the marble is in group 3.

If group 1 = group 2, then their weight is known. You can then take one marble from group 1 (which you know is normal, which I'll call a), and 3 marbles from group 3(b, c, d). 2nd weighing: a+b vs c+d. If they are the same, the unknown marble is the remaining marble from group 3, and you can weigh this against a to find heavier/lighter.

If they are different: the remaining marble is a normal marble, which means that either b, c, or d is the different one. 3rd weighing: c vs d. If c = d, then b was different, and you know if it's heavier depending on the 2nd weighing ab> cd or ab < cd. If c != d, (which means b is normal) then you again know which one was different; if ab > cd, then the lighter one in the 3rd weighing (c or d), is the different one. If ab < cd, then the heavier one in the 3rd weighing is the different one.

NOW FOR THE OTHER HALF (and more) OF THE SOLUTION

Back to the beginning! 1st weighing: group 1 vs group 2, and they were DIFFFERENT. So, group 3 had normal marbles. I'm going to say there are marbles a, b, c, and m in each group. The 2nd weighing will take one marble from each group (1m, 2m, 3m) vs TWO marbles from the first group (1a, 1b) and one marble from the second group (2a). 3m is normal. So, 1m2m3m vs 1a1b2a. If 1m2m3m = 1a1b2a, that means the different marble is in the remaining marbles of groups 1 and 2; 1c, 2b, or 2c.

3rd weighing: 2b vs 2c. if they are same, 1c was the different marble, and the weight heavier/lighter is known from the first weighing (1a1b1c1m vs 2a2b2c2m). If 2b != 2c, then again, heavier/lighter is determined by first weighing (normal group 1a1b1c1m vs diff group 2b2c2m heavier/lighter) and which one of 2b/2c was heavier/lighter.

Okay. 1st weighing, group 1 vs group 2, different weights. 2nd weighing, 1m2m3m vs 1a1b2a, now DIFFERENT weights. Since 3m is normal, that means different marble is among 1m, 2m, 1a, 1b, 2a. 4 different scenarios here.

#1: group 1 > group 2 (1a1b1c1m > 2a2b2c2m), and 1m2m3m > 1a1b2a. 2a, or 1m are the different ones. This is found by crossing out each one that does not fit. 2m can't be the one, because it'll mean 2m is heavier, and in the first weighing, group 1 was heavier. 3m can't, because it's normal. 1a can't, because in 1st weighing, 1a's group was heavier. Same with 1b. So, 2a or 1m. Pick one (2a) and weigh it against normal 3m. If same, different was 1m. If different, well, 2a's the different one then. 2a would be lighter; 1m heavier.

#2: group 1 > group 2 (1a1b1c1m > 2a2b2c2m), 1m2m3m < 1a1b2a. Crossing out choices 3m and 1m (1m is larger in 1st weighing but smaller the next), leaves 1a, 1b, and 2m. 2m, if different, is lighter. But also, 1a or 1b could be the different one, in which case it's heavier. 3rd weighing just compares 1a and 1b; the heavier one is different, if both the same, then 2m must have been lighter.

#3: 1a1b1c1m < 2a2b2c2m, 1m2m3m > 1a1b2a. 2m is heavy, or 1a or 1b is light. Again, compare 1a and 1b in the 3rd weighing to find out the lighter, and so different one. If same, 2m is the heavy, different one.

#4: 1a1b1c1m < 2a2b2c2m, 1m2m3m < 1a1b2a. Similar to #1, either 2a is heavy or 1m is lighter. Compare 2a to known 3m in the 3rd weighing; if 2a is heavier, then it's the different one. If 2a = 3m, then 1m is lighter and different.

Turn one of the switches on for awhile (5 minutes or so should be ok), then turn that switch off and turn another one on. Now go to the other room. Obviously you know which switch controls the one that is lighted up. Now feel the other two bulbs. The one that is warmer than the other bulb is the one that you turned on originally. So now you know all three.

It's really cool, although I knew it in another version. The answer is 10

I guess you may wanna know how to solve it. Well..

Let us assume there is only one man cheating on his wife. That wife saw none of the other men to cheat on their wives, yet she hears the stranger say yes. It's obvious that her husband is the only cheater, so she will kick him out.

Case no2: 2 cheaters. Wife 1 will see wife 2 being cheated and will wait for her to kick her husband out. Wife 2 will do the same. However, when Wife 1 sees wife 2 not doing anything she will know that wife 2 is waiting for wife 1 to kick his husband out. Same with wife 2. So on the second day 2 men will be kicked out.

Case no3: 3 cheaters. Following the same line of thought as in case 2, Wife 3 will wait for wife 1 and 2 to kick their husbands out on the second day. Since that won't happen, since w1 and w2 will be doing the same, she will know that she is being cheated on. On the third day, 3 men are kicked out.

Case no10: 10 men are kicked out on the 10th day.

it stays the same, because of Archimede's Law, like someone else said before. ^^

Obviously no1 stays on. This is the exception.

Now, it's all about the number of dividers (except 1 and the no itself) that the no of the bulb has.
If it has no dividers, as in being prime, it's off, because it's state changes 2 times, once for the first person and again for the nth person, n=bulb no.
If the number has an odd number of dividers, it stays on./ The only question now is how to find all the numbers up to 100 that have an odd number of dividers. Taking each one and checking the divider's number is too time consuming, so, we'll have to go around this somehow.

Now, we can take all the possible forms a number can take when broken down in it's prime factors:

x*y means it can divide by x and y, not good.
x^2 only 1 divider, x, meaning: 4, 9, 25, 49.
x*y*z it can divide by x, y, z, x*y, x*z, y*x, not good.
x^2*y it has 4 dividers: x, y, x*y, x^2, not good.
x^3 has as dividers x, x^2, not good.
x*y*z*w has x,y,z,w,x*y,x*z,x*w,y*z,y*w,z*w, x*y*z, x*y*w, x*z*w, y*z*w, not good.
x^2*y*z has x, y, z, x^2, x*y, x*z, x^2*y, x^2*z, 7 dividers, which is good. This means: 60, 84, 90.
x^3*y has x, y, x^2, x*y, x^3, x^2*y, not good.
x^2*y^2 has x, y, x*y, x^2, y^2, x^2*y, x*y^2 7 dividers, meaning: 36, 100.
x^4 has x, x^2, x^3, meaning: 16, 81.
x^5 has x, x^2, x^3, x^4, not good.
x^4*y has x, y, x*y, x^2, x^2*y, x^3, x^3*y, x^4, 8 dividers ftw.
x^3*y^2 has x, y, x*y, x^2, y^2, x^3, x^2*y, y^2*x, x^3*y, x^2*y^2 10 dividers. Can't go on, because the min value that x^3*y*z can take is 2^3*3*5=120>100
x^6 has x, x^2, x^3, x^4, x^5: 64
x^5*y has 10 dividers, gay...
Can't go on, because the lowest value that x^4*y^3 can take is 16*9>100
From x^7 can't go on, because it's lowest value is 2^7>100.

So, the bulbs that are on: 4, 9, 25, 49, 60, 84, 90, 16, 81, 64.

Well, taht was time consuming aswell, pretty sure you can work around that aswell. I might have messed up aswell at some point.