Question 4: More simple problems:

a) How many different 4-digit numbers can be made from the digits {1, 2, 3} if each digit must appear at least once?

b) A painted rectangular prism with sides 4 cm, 5 cm, and 6 cm is cut up into 1 cm cubes. How many of these cubes are not painted at all ?

c) In a room of 100 people if everyone has to shake hands with everyone in the room once, how many handshakes are made altogether?

Answer

a) 36: Consider a 3-digit number formed from {1, 2, 3}, there are 3×2×1 = 6 ways, the fourth digit can be any of the 3 numbers and it can be at any of the 4 places. Total = 6×3×4 = 72 ways. But since there are two same digits in each number, dividing 72 by 2 gives 36 numbers.

b) 24: Assume there is a layer of 1 cm around the cube. Take the layer off, the new cube has dimensions 2×3×4, i.e. 24 cubes.

c) Two people have 1 handshake, 3 people have 3 handshakes, 4 people have 6 handshakes, . . ., the formula is n(n – 1)/2.

For 100 people, there are 100 × 99 ÷ 2 = 4950 handshakes.

Question 5: Mysterious ages 

Two women A and B are talking.

A: I have three daughters whose ages are found by the following clues. Stop me as soon as you know their ages. Clue 1: The sum of their ages is 13.

B: I need more clues.

A: Clue 2: The product of their age is the same as your age.

B: I need more clues.

C: Clue 3: My oldest daughter is sleeping upstairs.

B: Stop. I know their ages.

Do YOU know her three daughters' ages?

Answer 

From the first clue, their ages can be one of the following sets (1, 1, 11), (1, 2, 10), (1, 3, 9), (1, 4, 8), (1, 5, 7), (1, 6, 6), (2, 2, 9), (2, 3, 8), (2, 4, 7), (2, 5, 6), (3, 3, 7), (3, 4, 6), (3, 5, 5) and (4, 4, 5). Their respective products are 11, 20, 27, 32, 35, 36, 36, 48, 56, 60, 63, 72, 75 and 80, of which one is B's age. Of course you do not know B's age but she must know her own age. The only reason that she needs one more clue is there are two sets of ages that give the same product (= her age). Here, both (1, 6, 6) and (2, 2, 9) give the product of 36. The third clue indicates that A has an oldest daughter. So, their ages are 2, 2, and 9.

Question 6: The Missing Dollar

Three men shared a room in a hotel and were charged $30 for the night. They all agreed to share the cost, so each paid $10. The clerk later found that he had overcharged them by $5 so he asked a waiter to return them $5. The waiter reasoned that it was not easy to divide the $5 by three, so he pocketed $2 and returned only $3.

Now, as each man paid $9 apiece, or a total of $27 for the room. Add that to the $2 the waiter kept gives the total of $29. Where did the missing dollar go?

Answer

The three men paid $30 originally, they were refunded $3 so they paid $27 for the room. The waiter kept $2 from this $27, so correctly we must not add $2 with $27, rather it should be subtracted: $27 less $2 gives $25 (The men paid $27, the manager kept $25, the waiter pocketed $2, everything now is correct!).

Question 7: A piece of cake!

Can you cut a cake into 14 pieces with four straight cuts?

Answer

If you draw your own diagram as you read, it would be easier to understand this: 2 cuts give 4 slices. Make the third cut through the previous two cuts then you have 3 more pieces: Total 7 pieces by far. Make the fourth cut through the other three cuts then you have 4 more pieces: Total 11 pieces at most.

However, if you think of the cake as a cylinder, then make the fourth cut parallel with the top surface of the cake: You double the number of slices you have had after three cuts. Total = 7 × 2 = 14 slices!

Question 8: How many squares in a chessboard?

How many squares are there in the 8 squares by 8 squares chessboard?

Answer

64 1x1 squares, 49 2x2 squares, 36 3x3 squares, 25 4x4 squares, 16 5x5 squares, 9 6x6 squares, 4 7x7 squares and 1 8x8 square. Total = 204 squares.

[Harder question: How many rectangles are there in the 8 squares by 8 squares chessboard?]

Question 9: How many triangles? 

With 6 match sticks how many equilateral triangles of side equal to the length of a match stick can you make at most?

Answer 

4, if you build the frame of a tetrahedron (a triangular pyramid)

Question 10: Locker problem

In a hall there were 50 lockers that belonged to 50 year twelve students. In the year twelve muck-up day, the 50 students decided to go through the hall one after another and played this trick:

The first student shut all the lockers. The second student opened every second locker. The third student changed every third locker (i.e. if it was opened, he/she closed it; and if it was closed he/she opened it). The fourth student changed every fourth locker, and so on. How many lockers remained closed after all 50 students had a go?

Answer 

Consider the 12th locker, for example, it was shut by the 1st student, opened by the 2nd student, shut by the 3rd student, opened by the 4th student, shut by the 6th student, opened by the 12th student. These numbers {1, 2, 3, 4, 6, 12} are the factors of 12. So, the principle is: if a locker has an even number of factors it would be opened at the end, and of course if a locker has an odd number of factors it would be closed at the end. Only {1, 4, 9, 16, 25, 36, 49} have an odd number of factors. Therefore, at the end 7 lockers were shut.

Question 11: Successive savings 

An homeowner makes three successive designs that save, in turn, 20%, 55%, and 25% on the heating costs of the house. Find the overall percentage saved.

Answer 

73%: Since 80%×45%×75% gives 27%; a saving of 73%

Question 12: A clock problem 

How many times, not in 24 hour format, can digital clock displays in half a day are perfect square ? (A display such as 12:22 reads as 1222)

Answer

16: The minimum display is 100 (i.e. 1:00) and the maximum is 1259 (i.e. 12:59). The least perfect square is 10² = 100 and the greatest is 35² = 1225, so there are 35 – 10 + 1 = 26 perfect square numbers.

However, as the last two digits cannot be greater than 60, these 10 numbers must be excluded (13, 14, 17, 19, 22, 24, 26, 28, 31, 33 because 13² = 169, 14² = 196, 17² = 289, 19² = 361, 22² = 484, 24² = 576, 26² = 676, 28² = 784, 31² = 961, 33² = 1089).

(Thanks to Edwin Cooper for his solution)

Question 13: Double vision

A water lily doubles itself in area each day. After 12 days the whole pool is completely covered by the water lily. When was the pool half covered?

Answer

11 days.

Question 14: Numbers matter

The value of a three-digit number is twelve times the sum of its digits. What is the number?

Answer 

Let abc be the number, then 100a + 20b + c = 12(a + b + c), or by simplifying, 88a = 2b + 11c. Only one equation but three unknowns! Impossible? Wait, as a, b, c have to be single digits, we can solve it by trial and errors.

Let b = 0, 88a = 11c, so a = 1, c = 8. The number is 108.

For a ³ 2: LHS ³ 176, while the maximum value for the RHS is 2×9 + 11×9 = 117: There are no other values of a, b, c satisfying the given condition. Thus, the only solution is 108.

Question 15: Crossing the river 

Three couples come to a river. There they find a small boat that can carry only two people at a time. How can all six people cross the river if no wife is allowed to be left with any man at any stage without her husband's presence.

[If you find this one easy, try the next one]

Answer

Question 16: Crossing the river again

Three couples come to a river. There they find a small boat that can carry only two people at a time. How can all six people cross the river if:

Rule 1: The number of women must be equal or more than the number of men at any stage;

Rule 2: Of the six people, only one man and his wife can row the boat.

Answer

Let the men be A, B, C and their wives be a, b, c respectively. Assume A and a can row.

Question 17: Time traveller

What is the greatest number of Mondays which can occur in a 45-day period?

Answer

45 ÷ 7 = 6, remainder = 3. Thus, if Monday is the 1st, 2nd or 3rd day of the 45-day period then there will be 7 Mondays within the period.

Question 18: Jack, Jill and a jug

Jack and Jill each have a 1 litre jug of water. The first day, Jack pours 1 mL of water from his jug into Jill's. The second day Jill pours 3 mL of water from her jug into Jack's. The third day Jack pours 5 mL of water from his jug into Jill's, and so on. What is the volume of water in Jack's jug at the end of the 101st day?

Answer

The volumes of water in Jack's jug are:
Day 1: 999 mL, day 3: 997 mL, day 5: 995 mL, and so on. This is an AP, with a = 999, d = – 2 so on day 101 (n = 51) the amount of water = a + (n – 1)d = 999 + 50 × –2 = 899 mL.

Question 19: A 2KY problem

How many years in the 21st century will have the property that, dividing their year number by each of 2, 3, 5 and 7 always leaves a remainder of 1?

Answer

To divide by each of 2, 3, 5 and 7 with a remainder of 1 the number less 1 must be a multiple of 2 × 3 × 5 × 7, i.e. in the form 210k + 1.
Put k = 9 gives 1891, put k = 10 gives 2101. Thus, for the years from 2000 to 2100 there is none that satisfies the given condition.

Question 20: A triangle problem

A triangle ABC is to be formed having all the sides of integer length. If AB = 37 and AC = m, where m is a fixed integer less than 37, how many different lengths are possible for BC?

Answer

This problem can be viewed geometrically: Draw a line AB of length 37 units. At A draw a circle of radius m units, m < 37, and ask yourself how many circles of integer radius can be drawn at B to intersect the first circle. Obviously the smallest radius = 38 – m, and the largest radius = 36 + m. Thus, the number of circles = (36 + m) – (38 – m) + 1 = 2m – 1.

Question 21: Another triangle problem

If we are given a square we can easily construct a triangle enclosing it as shown (i.e. one side of the square lies on a side of the triangle, the other vertices of the square lie on the other sides of the triangle). Obviously there are many such triangles. However, if we are given a triangle how can you construct such a square?

Answer

Let the given triangle's height and base be h units and a units respectively. Assume such a square is drawn inside the triangle, let the side of the square be x units. Also, let its corner N divide the side AC in the ratio p : q

ABC is similar to AMN, so a/x = (p + q)/p = 1 + q/p. Thus, q/p = a/x – 1.
CAH is similar to CNP, so h/x = (p + q)/q = 1 + p/q. Thus, p/q = h/x – 1.
Multiplying these two equations gives 1 = (a/x – 1)(h/x – 1).
Thus, ah/x² – (a + h)/x = 0, on expanding then simplifying
Thus, x = ah/(a + h).
That is, given a triangle of base a units, height h units the square can be drawn with side MN = NP = ah/(a + h).