7

The months with 28 or 30 days have the same number of even and odd days. The seven months with 31 days have one odd day more. So the answer is 7 days.

No

For example, they might both speak on every fourth turn and stay both silent in between (silent for three turns).

60

Sum the weights and divide by 3, or sum the smallest weight (the two lightest statues) and the largest weight (the two larger statues). One could also have summed the two intermediate weights, which must be XL + S and L + M (and it is a coincidence that those two weights are the same, any of the two could be larger). Or one could have summed the second weight (XL + M) and the penultimate weight (L + S). In any case, the answer is 60.

3

The answer is 3, as the car can reach all further corners. By symmetry, it does not matter at which corner we start. With coordinates, let (0,0) be the starting corner, and let (4,3) be the opposite corner. Some paths going to the further 3 corners are:
(0, 0), (1, 0), (1, 1), (2, 1), (2, 2), (3, 2), (3, 3), (4, 3)
(0, 0), (1, 0), (1, 1), (2, 1), (2, 2), (3, 2), (3, 1), (4, 1), (4, 0)
(0, 0), (1, 0), (1, 1), (0, 1), (0, 2), (1, 2), (1, 3), (0, 3).

Forcibly neither true nor false

The assertion cannot be true because, if true, the child would be a liar (as part of the family) telling the truth. The assertion can be false, for example if only half of the family members are liars.

1

It is optimal to start the row with an occupied seat. Adding two chairs does not make a difference if and only if the last seat is also occupied, which means that the remainder after division by 3 equals 1.

7

If you draw 1 green ball and 2 yellow balls and 3 red balls, then you lose (this is the most unfortunate situation). Once you draw any additional ball, you are guaranteed to win. Hence N=7.

27

Call X the missing side length. We must have 8/12=12/18=18/X hence X=27.

3

If Alice and Bob work for X hours, the fraction of the complete task they get done is (1/4)X+(1/12)X. The task gets completed when this fraction equals 1, that is, if X=3.

25%

Say (without loss of generality) that we have 100 pupils in total. Then 10 pupils watched movies A and B, which corresponds to 25% of the pupils who watched movie B.

3

If the amount of nuggets is a multiple of 3 we can buy several 3-bags. If it is a multiple of 3 plus 2 (at least 5), we can buy a bag of 5 and possibly several 3-bags, if it is a multiple of 3 plus 1 (at least 7), then we can buy a 7-bag and possibly several 3-bags. So the only amounts of nuggets that we cannot buy are 1,2, and 4.

36

The amount of sandwich in each bag is in particular a rational number of the form X/6. From the information about 14 sandwiches, we know that 14/10≤ X/6 ≤ 14/9. We deduce that 8< X< 10, hence X=9. So each bag contains 9/6=3/2 sandwiches, so for 24 bags one needs 24· 3/2=36 sandwiches.

Charles

Since Bob and Charles say contradictory things, one of them is a liar. So Alice cannot be the liar. Hence Charles is the liar.

8

Bob needs 3 white roses, 2 red roses, and 3 yellow roses.

9

The most unlucky case is having two cards of each type, namely 8 cards. With 9 cards one is sure to have three cards of the same type.

32

We look for a number from 0 to 40 which is a multiple of 8 and which leaves remainder 2 after division by 5. The only such number is 32.

12

You only have to choose whom to give the banana to (4 choices) and whom the orange (3 choices left). There are 12 possibilities.

1600

You can build a 60cm×60cm frame with a 40cm× 40cm hole (1600 square centimeters). Or you can make a 50cm× 70cm frame with a 30cm× 50cm hole (1500 square centimeters). Or you can make a 40cm×80cm frame with a 20cm× 60cm hole (1200 square centimeters).

(1,3)

(1,2) and (2,1) will both leave 48 problems unsolved. (1,3) will leave 30 problems unsolved. (3,1) will leave 60 problems unsolved. (2,3) and (3,2) solve less problems than (1,3) and (3,1) respectively.

360

The sum of the exterior angles of any regular polygon is 360 degrees.

8

You can take 4 9AUR coins, 2 3AUR coins, 2 1AUR coin. This is optimal because with only 3 or fewer coins of 9AUR, you would need at least 10 coins to pay 44 AUR (3 × 9AUR + 5 × 3AUR + 2 × 1AUR), and with 4 coins of 9 AUR you need at least 8 coins to pay 44 AUR. The given coins allow you to pay any bill in the range; using the 9AUR coins one is left to pay an amount from 1AUR to 8AUR, which can be paid with the given 3AUR or 1AUR coin.

4

Using all sticks, you can build a 2×2 square with area 4, a 3×1 rectangle with area 3, and an L-shaped figure with area 3.

5

Consider the set of results of the first die and the second die, consisting of the 16 pairs made with the numbers from 1 to 4. The numbers 2 and 8 can be obtained each just with one pair. The numbers 3 and 7 can be obtained each with two pairs. The number 4=3+1=2+2 and 6=4+2=3+3 can be obtained each with three pairs. The number 5=4+1=3+2 can be obtained with four pairs.

2

One friend is not enough because from testing only one message you cannot determine the funny word in all possible cases. Two friends are enough: you send the first word only to the first friend, the second word only to the second, the third word to both, and the fourth word to none.

6

More than 5 because the five mentioned koalas do not make up the 100% of the weight. Each intermediate koala weights each at least 12,5% of the total weight and at most 20% of the total weight. Since only 15% of the total weight is missing, there is space for at most one intermediate koala. So there are 6 koalas.

Ben

Consider the number of remaining candies. When you see 0 or 1 candies on the table, you lose. When there are 2,3 or 4 candies, you win (for 4 candies take 3 candies, for 3 candies take 2 or 3 candies, for 2 candies take 2 candies). When there are 5 or 6 candies, you lose (taking either 2 or 3 candies puts the other player in a winning situation). When there are 7,8 or 9 candies you win (Taking either 2 or 3 candies puts the other player in a losing situation.) When there are 10 candies, you lose. (taking either 2 or 3 candies puts the other player in a winning situation). So the second player always wins, i.e. Ben.

6

A1, A2, A3, A4, D2, D3 works, so the answer is at least 6. There are 7 diagonals going bottom left to top right, so the answer is at most 7. However the first and last diagonal are on a same diagonal line in the other direction, so you cannot place bishops on both of them. Thus the answer is at most 6.

15

Let n be the number we are looking for. Let r and y be the number of red and yellow balls, respectively. We are given that
(r(r-1))/(n(n-1))=1/17,
(y(y-1))(n(n-1))=1/15,
hence 7r(r-1)=n(n-1)=5y(y-1). This shows that n(n-1) is divisible by 5 and by 7, thus by 35. The least n≥ 1 for which this is the case is n=15. To see that n=15 is actually suitable, notice that r=6 and y=7 lead to the desired probabilities.

The letter O

Forbidding the letter O leaves 252· 104 allowed car plates, while forbidding the digit 0 only leaves 262· 94. We conclude because 25 · 102>26 · 92.

1

We have (4n)2< 16n< 2+25< (4n+5)2, hence a perfect square of the form 16n2+25 must be among (4n+1)2, (4n+2)2, (4n+3)2 and (4n+4)2. Since 16n2+25 is odd, it is neither (4n+2)2 nor (4n+4)2. Putting 16n2+25=(4n+1)2 gives n=3. Putting 16n2+25=(4n+3)2 yields n=1/23, which is not an integer. Hence only the value n=3 is counted.
Alternative solution: Assume 16n2+25=x2, where x is a positive integer. Then 25=x2-16n2=(x+4n)(x-4n), and since 25=52 is the square of a prime, the only possibility is given by the equations x+4n=25 and x-4n=1, which give n=3 and x=13.

L

Call the weight of a statue with its size, for example let XL be the weight of the largest and heaviest statue. Clearly W1=XL+L and, as the two weights we sum must be distinct, we have W2=XL+M. Similarly, we have W10=S+XS. Thus W2+W10=XL+M+S+XS. Considering that L=W-(W2+W10), we can compute the weight of L.

5

For example, put them on a4, b2, c5, d3, e1. It cannot be more than 5, because there are 5 columns and you can put at most one queen per column.

30

You only have to choose whom to give the banana to (5 choices) and whom the two oranges. Discarding the child who took the banana, there are four children left and you have to select a pair of children for the oranges (12 choices, as there are 4 choices for the first child, 3 choices for the second child). As the ordering does not matter, you divide by 2 and are left with 6 possibilities. Hence there are 30 possibilities in total.

25

Alice runs 1km in 250 seconds, Zoe in 300 seconds, so Alice's speed is 6/5 of Zoe's speed. Hence they will meet after Alice made 6/11 of the total distance and Zoe 5/11 of the total distance. Thus Alice has run 6km, which will take 25 minutes.

4

Dwarves just behind liars are truth-tellers, hence the first dwarf must be a liar. Dwarves just behind truth-tellers are liars, hence liars and truth-tellers alternate, giving a total of 4 liars.

16

You must use the same staircase for going up and down. Then what you can choose is: the staircase (4 possibilities) and the direction of your visit on the lower floor (2 possibilities), and the direction of your visit on the upper floor (2 possibilities). This gives a total of 4×2×2 =16 tours.

1

Let h be the probability of coming up heads and 1-h the one of coming up tails. Getting the same result happens with probability h2+(1-h)2=1+2h2-2h while getting different results happens with probability 2h(1-h)=2h-2h2. The difference is (1+2h2-2h)-(2h-2h2)=1-4h+4h2=(1-2h)2. This is strictly positive, unless h=1/2, which we excluded. So getting the same result is more likely than getting two different results.

4

Pigeonhole principle. With 99 objects of 26 types, there are at least 4 (namely, the ceiling of 99/26) pieces of pasta of the same kind.

6

Call x the result of the first spin, and y the result from the second spin (both x ad y are integers from 1 to 5). When you count all the possible sums the pair (x,y) can add up to, 6 occurs five times, while all other numbers occur less often.