9

2 + 4 + 7 + 9 = 22
5 + 5 + 6 + 6 = 22
11 + 2 + 3 + 6 = 22
1 + 4 + 8 + 9 = 22

8

2 + 8 – 9 = 1
3 + 2 – 4 = 1
3 + 6 – 8 = 1

This puzzle does not have a unique solution (there are multiple).  E and X can be swapped in the highlighted squares to produce another solution.

25 fruits remain

He gives 3 fruits at each of the first 10 stops;
2 fruits at each of the next 15 stops;
1 fruit at each of the last 5 stops.

90 – 3*10 – 2*15 – 5 = 25

The are multiple answers.  Here are some.

POSTMARK, MARKDOWN
FAIRPLAY, PLAYPEN
CAKEWALK, WALKWAY
BASEBALL, BALLGAME
BACKDOOR, DOORWAY
BROKEDOWN, DOWNTOWN

While I believe these are likely the expected words, FAIRPLAY and BROKEDOWN are not actually English words.  Limiting to actual words, there appear to be no solutions for those two.

Use a hexagon arrangement:

It takes 334 Joules to melt one gram of ice at 0C into water at 0C, so melting 20 grams of ice takes 20 x 334 = 6,680 Joules.
It takes 4.186 Joules to raise the temperature of a gram of water by 1C, so 4.186 x 100 x 20 = 8,372 Joules of heat energy available in the water.  Since 8,372 > 6,680, all the ice will melt, leaving 1,692 Joules.

The final temperature of the water at equilibrium is therefore 1692 / (120 * 4.186) = 3.368C.

All this of course assumes an ideal insulator, as well as assuming that the ice was 0C.  The way the question is phrased leads to the assumption that the ice has already started melting, and is therefore at 0C uniformly, but this is not guaranteed.  The interior of the ice could still be below 0C while the exterior is melting at 0C.  If this were the case, additional energy would be used to raise the temperature of the ice to 0C, and thus the equilibrium temperature of the water would be lower.

The boy/girl odds for a single child are 51/49–males are slightly more likely than girls.  Oddly, after having a boy, the odds are 50/50 for the second child to be a boy, even though the odds for a third boy after the first two are boys jumps up to 52.3%.

So the correct answer is also the intuitive answer: 50%

Thanksgiving and Towel Day.

So obviously this can be solved by brute force–trying different values for each letter.  The challenge is how to deduce the answer with as little trial and error as possible.  We can narrow it down with some observations.

Both I’s in LIMITS line up under an I, so let’s start there.
Because S can’t be zero and S + I results in I as the last digit, we know T + T >= 9 (to generate carry) and that S = 9.
Since S = 9, that must mean that H = 0 so that H + I = I with no carry.
Since S = 9, L + N = 9.
There is no carry from L + N, so L + I = T or 10 + T, but we know there is also no carry from L + I and that I is not zero.
Because T + T = 10 + M and M can’t be zero, then T isn’t 5.
Because WITHIN and LIMITS are both six digits, we know that L = W + 1.

So before any trial and error we have:

S = 9
H = 0
T = {6,7,8} = L + I
M = {2,4,6} = 2*T – 10
W = {1,2,3,4,5,6} = L – 1
L = {2,3,4,5,6,7} = W + 1 = 9 – N
N = {2,3,4,5,6,7} = 9 – L
I = {1,2,3,4,5,6} = T – L

Looking at this, we see that L has a relationship to every other letter, so let’s look at that.  Since W = L – 1 and N = 9 – L, L can’t be 5 since that would require both W and N to be 4.  So now the possibilities are:

S = 9
H = 0
T = {6,7,8} = L + I
M = {2,4,6} = 2*T – 10
W = {1,2,3,5,6} = L – 1
L = {2,3,4,6,7} = W + 1 = 9 – N
N = {2,3,5,6,7} = 9 – L
I = {1,2,3,4,5,6} = T – L

Now we’re ready for trial and error.  Since there is only one way to assign 8 and since eight out of ten digits are used (only two unused), try assigning T = 8, which gives M = 6.  Since 6 is taken, neither L nor I can be 2 (since 8 – 6 = 2) nor 4 (since they need to be different and sum to 8).  Also, since L = 9 – N and N can’t be 6, then L can’t be 3 and L = W + 1 and W can’t be 6, so L can’t be 7, leaving no possible value for L.  So we know T can’t be 8, leaving:

S = 9
H = 0
T = {6,7} = L + I
M = {2,4} = 2*T – 10
W = {1,2,3,5} = L – 1
L = {2,3,4,6} = W + 1 = 9 – N
N = {3,5,6,7} = 9 – L
I = {1,2,3,4,5} = T – L

There are only two ways to assign 7, so try assigning T = 7, which gives M = 4.  Since 4 is taken, neither L nor I can be 3 or 4.  Since N can’t be 7 and N = 9 – L, L also can’t be 2, leaving only L = 6:

S = 9
H = 0
T = 7
M = 4
W = 5
L = 6
N =3
I = 1

This give a unique assignment for each letter, so see if it works:  97166 + 517013 = 614179
It does work, on only the second (of three possible) guesses.