Problem #57: Tower of Squares
For this tower of nine squares, determine a line passing through point P that will split the area of the nine squares into two equal parts.
Solution #57: Tower of Squares
PA has the area of 5 squares below it, and PB has the area of 3 squares below it. Thus the area of /\PAB is equal to the area of 2 squares. We have to choose Q so that AQ:QB as 1:3 to get a line with the area of 4.5 squares below it.
Problem #58: Sum to 30
Find different ways to express 30 as the sum of two or more consecutive integers.
Solution #58: Sum to 30
The expressions 4 + 5 + 6 + 7 + 8, 6 + 7 + 8 + 9, 9 + 10 + 11, (-3) + (-2) + (-1) + ... + 6 + 7 + 8, (-5) + (-4) + (-3) + ... + 7 + 8 + 9, and (-8) + (-7) + (-6) + ... + 9 + 10 + 11 are possibilities.
Three of the foregoing six sums consist of only positive integers. Any sum of consecutive positive integers (a+1) + (a+2) + ... + (a+n) is unchanged by placing the sum (-a) + (-a+1) + ... + (a-1) + a in front of it. Therefore, the three "positive only" sums ensure that three more possibilities are feasible.
Consider the factors of 30: 1 x 30, 2 x 15, 3 x 10, and 5 x 6. All odd factors greater than 1 produce consecutive integral sums, as follows. Take 3 x 10, for example. Three consecutive integers with 10 in the middle will add to 30, which produces 9 + 10 + 11.
Observe that 30 = 0.5 x 60. The consecutive integers 0 and 1 average 0.5. Another sum can be formed by taking thirty pairs of integers each totaling 1. The sum is given by (-29) + (-28) + (-27) + ... + 28 + 29 + 30. The result is actually the sum based on putting integers in front of 30, which comes from 30 x 1. Note that 1.5 x 40, 2.5 x 24, and 7.5 x 8 correspond to some of the other foregoing sums.
Problem #59: The King's Fortune Telling Scale
The king has a fortune telling scale. For one penny the scale spits out a slip of paper with a fortune on it along with the weight of the king or whatever is on the scale. The king also has five bags of gold that he has given to five trusted friends for safe keeping. Rumor has it that one of his five trusted caretakers is not to be trusted and has been asking a counterfeiter to make phony gold. All the king could find out about the counterfeit gold was that one of the king's five bags is now filled with counterfeit gold and that the weight of that bag is different from the others. He does not know whether it weighs more or less than each of the others.
The king calls back all of his bags of gold. Using his fortune telling scale, the king wants to find out:
which bag contains counterfeit gold, and
how much it weighs, exactly.
Being a very economical king, he wants to find this out using the least number of pennies possible. His wizard says it can be done with just 3 pennies. No one else can see how. Anyone can do it with 5 pennies. Some say they can do it with 4 pennies. Can you figure out how to do it with 3 pennies?
(Assume that the fortunes the scale spits out offer no clues to which bag contains the counterfeit gold.)
Solution #59: The King's Fortune Telling Scale
Weigh each bag once.
(not yet completed)
Solution provided by Alan Fung.
Let's name the five bags of gold as A, B, C, D and E. One of them is odd and we don't know if the odd bag is lighter or heavier.
The first two weights are like these:
A + B + C = 3X ----------------(weight #1)
A + D = 2Y ----------------(weight #2)
X and Y are average unit weight of the combination. That is, they are roughly the weight of each bag of gold.
If X = Y, it means the weight ratio among A, B, C and D are the same. (You may check it by adjusting one of them as lighter or heavier, X will not equal to Y). Because A, B, C and D are the same, E must be odd. So the final weight will be to weigh E and that will be the answer.
If X < Y, it is because "B or C is lighter" or "A or D is heavier".
If X > Y, it is because "B or C is heavier" or "A or D is lighter". The final weight will be:
C + D + E = 3Z ----------------(weight #3)
The reasons behind picking C, D and E is because now we know E has to be normal, we need to pick E as reference and two bags from the lighter and heavier possibilities. I tried to pick A but it wouldn't work because I shouldn't pick the same thing with the same coefficient that appears in all three equations. By the same token, C and B are interchangeable.
Now to make the solution easier to understand, I will start with the solution and work out the weight comparisons.
If X < Y
If B is lighter, X < Z and Y = Z
If C is lighter, X = Z and Y > Z
If A is heavier, X > Z and Y > Z
If D is heavier, X < Z and Y > Z
If X > Y
If B is heavier, X > Z and Y = Z
If C is heavier, X = Z and Y < Z
If A is lighter, X < Z and Y < Z
If D is lighter, X > Z and Y < Z
Because all results from weights comparisons are unique, we covered all possible solutions.
For the second part of the question, to know how exactly does the odd bag weigh, we can use simple substitution to work it out. For example, if we found B is odd, B = 3X - 2Y.
Problem #60: Knights of the Round Table
King Arthur liked to invite his knights over for parties around the round table. When the king had a gift that he could only give to one knight, he had them play a game that went like this:
First, King Arthur numbered the chairs around the table. At the start, every chair was occupied by a knight. (King Arthur himself did not sit at the table.) Then he stood behind the knight in chair 1 and said, "You're In." Next, he moved to the knight in chair 2 and said, "You're Out," and that knight left his seat and went off to stand at the side of the room to watch the rest of the game. Next he moved to the knight in chair 3 and said, "You're In." Then he said, "You're Out" to the knight in the chair 4, and that knight left his seat and went to the side of the room.
He continued around the table in this matter. When he came back around to the knight in chair 1, he said either "You're In" or "You're Out," depending on what he had said to the last knight. (If the last knight was "In," then the first knight was now "Out, " and vice versa.)
The king kept moving around and around the table, alternately saying, "You're In" or "You're Out" to the knights that remained at the table. (If a chair was now empty, he just skipped it.) He continued until only one knight was left sitting at the table. That knight was the winner.
If you were a knight, which chair number would you try to sit in at King Arthur's table?
The number of knights at the table was always changing, so you should come up with a general rule.
Solution #60: Knights of the Round Table
(not yet completed)
Problem #61: Photographers and Cannibals
Three National Geographic photographers and three cannibals are traveling together through a jungle when they come to a river. The largest boat available can carry only two people at a time. The photographers are safe only if on each side of the river there are equal numbers of photographers and cannibals or there are more photographers than cannibals; otherwise, the photographers become dinner. How can they all get across?
(Tokens such as pennies and nickels may be helpful in solving this problem.)
Solution #61: Photographers and Cannibals
A cannibal and a photographer cross.
The photographer returns.
Two cannibals cross.
One cannibal returns.
Two photographers cross.
One cannibal and one photographer return.
Two photographers cross.
One cannibal returns.
Two cannibals cross.
One cannibal returns.
Two cannibals cross.
Problem #62: Handshakes
Is the number of people in the world who have shaken hands with an odd number of people odd or even?
(Dead people count.)
Solution #62: Handshakes
If you were to ask everyone in the world how many hands he or she has shaken, the total would be an even number because each handshake would have been counted twice -- once each by the two people who shook hands. A group of numbers whose sum is even cannot contain an odd number of odd numbers.
Problem #63: Bookworm
A set of encyclopedias consists of volumes that have 1/8-inch covers and one inch of pages. The set is arranged in order on a shelf from left to right. If a bookworm starts at the first page of Volume I and eats its way through to the last page of Volume II, how far does it travel?
Solution #63: Bookworm
Only 1/4 inch -- the thickness of the two covers. Put a paperclip on the first page of one book, and another on the last page of a second book. Put both books on a shelf together, the first on the left and the second on the right, and you will see why this surprising answer is correct.
Problem #64: Match Game
Using six identical matches, make exactly four equilateral triangles.
Solution #64: Match Game
The solution is a three-dimensional tetrahedron.
Those who have trouble with this one assume they must solve it in only two dimensions. Once you realize that the use of a third dimension is permissible, it's easy.
Problem #65: Where There's a Will
A father wishes to divide a square piece of land among his five sons. One son is his favorite, and he wants to give him one quarter of the land, as shown:
How can he divide the remaining land into four plots of equal size and shape?
Solution #65: Where There's a Will
Problem #66: Rope Trick
Is it possible to pick up a piece of rope, one hand holding each end, and tie a knot in the rope without letting go of either end?
Solution #66: Rope Trick
Problem #67: Alphabet Soup
Put the rest of the alphabet in its proper place:
A EF HI KLMN
B D G J
Solution #67: Alphabet Soup
A EF HI KLMN T VWXYZ
B D G J PQR U
C O S
The top row contains letters that are made up of straight lines; the middle row, letters formed by combinations of straight and curved lines; the third row, letters formed only by curves.
Problem #68: All Wet
A recipe calls for 4 cups of water. You have only a 3-cup and a 5-cup container. How can you measure out four cups of water?
Solution #68: All Wet
Here are two solutions:
Fill the 3-cup container and pour it into the 5-cup container.
Fill the 3-cup container again and from it fill the 5-cup container, leaving 1 cup of water in the 3-cup container.
Empty the 5-cup container and pour the remaining cup from the 3-cup container into it.
Now fill the 3-cup container and add it to the 1 cup that is already in the 5-cup container.
The result is the desired 4 cups.
Fill the 5-cup container.
Fill up the 3-cup container using the water in the 5-cup container, leaving 2 cups in the 5-cup container.
Empty the 3-cup container.
Pour the remaining 2 cups from the 5-cup container into the 3-cup container.
Fill the 5-cup container again.
Pour water from the 5-cup container into the 3-cup container (that already has 2 cups of water in it) until the 3-cup container is full.
Now, 4 cups of water remain in the 5-cup container.
Problem #69: The Telltale Number
Write a ten-digit number so that the first digit tells how many zeros there are in the number, the second how many ones, the third how many twos, and so forth.
Solution #69: The Telltale Number
Problem #70: Something Fishy
A fish weighs ten pounds plus half of its weight. How much does it weigh?
Solution #70: Something Fishy
Here is an algebraic solution:
W = 10 + 1/2 W
W - 1/2 W = 10
1/2 W = 10
W = 20
Problem #71: The Baffling Bicycle
A bicycle climbs a certain hill at 10 miles per hour and returns at 20 miles per hour. What is its average speed for the entire trip?
Solution #71: The Baffling Bicycle
The apparent answer -- 15 miles per hour -- is wrong. The correct answer is 13 1/3 miles per hour because speed is determined by dividing the distance by time. Notice, incidentally, that the answer is the same no matter how long the hill is.
Problem #72: Figure Eight
An eight-digit number contains two 1's, two 2's, two 3's, and two 4's. The 1's are separated by one digit, the 2's by two digits, the 3's by three digits, and the 4's by four digits. What is the number?
Solution #72: Figure Eight
Problem #73: The Balancing Brick
A brick balances evenly with three quarters of a pound and three quarters of a brick. What does the whole brick weigh?
Solution #73: The Balancing Brick
Three pounds. If the brick balances with three quarters of a brick plus three quarters of a pound, then one quarter of a brick must weigh three quarters of a pound. Thus a whole brick weighs four times as much, or three pounds.
Problem #74: The Scrambled Salesman
An egg salesman was asked how many eggs he had sold that day. He replied, "My first customer said, 'I'll buy half your eggs and half an egg more.' My second and third said the same thing. When I had filled all three orders I was sold out and I had not had to break a single egg all day."
Solution #74: The Scrambled Salesman
Seven. He sold four eggs to the first customer, two to the second, and one to the third. The problem is solved most easily if you start with the last customer and work backwards.
Problem #75: Wrong Number
Thirteen per cent of the people in a certain town have unlisted phone numbers. You select three hundred names at random from the phone book. What is the expected number of people who will have unlisted numbers?
Solution #75: Wrong Number
None. An unlisted phone number doesn't appear in the phone book.
Problem #76: Magic Numbers
This is a good way to mystify a friend. Ask someone to pick a number, any number at all, but not to tell you what it is. Have him silently multiply it by five, add five to the product, multiply the resulting number by two, add two to it, and tell you the result. Instantly you are able to tell him the number he started with.
What is the secret and why does it work?
Solution #76: Magic Numbers
The secret: To find the number selected, simply delete the last digit from the number given you. Then subtract one from the remaining number. Reason:
(not yet completed)
Problem #77: Case of the Counterfeit Coins
You have ten stacks of ten silver dollars each. They are identical, except that one stack consists entirely of counterfeit dollars. You know the weight of an authentic dollar, and you also know that a counterfeit dollar weights one gram less. How many weighings are needed to reveal which stack is counterfeit?
Solution #77: Case of the Counterfeit Coins
Only one. Weigh one coin from the first stack, two coins from the second, and so forth. The number of grams by which the total is light will correspond to the number of the counterfeit stack.
Problem #78: Strange Series
The following number is the only one of its kind. Can you figure out what is so special about it?
Solution #78: Strange Series
It is the only one that contains all the numerals in alphabetical order.
Problem #79: Boxed In
Three boxes contain two coins each. One contains two nickels, one contains two dimes, and one contains a dime and a nickel. All three boxes are mislabeled. (Each contains the label of one of the other two boxes.) If you are permitted to take out only one coin at a time, how many must you take out in order to be able to label all three boxes correctly?
Solution #79: Boxed In
Only one. Take it from the box labeled "Dime and Nickel." Since you know all three boxes are mislabeled, the box contains either two nickels or two dimes. Put the correct label on that box after inspecting one of the coins from the box. Then simply switch the two remaining labels.
Problem #80: Weighty Question
Which weighs more, a pound of gold or a pound of lead? (You may think you have heard this one before, but you probably haven't.)
Solution #80: Weighty Question
A pound of lead. Lead is weighed in the standard measure, in which 7,000 grains equal a pound. Gold is measured in Troy weight, in which 5,7600 grains equal a pound.
Problem #81: Early Bird
A chauffeur always arrives at the train station at exactly five o'clock to pick up his boss and drive her home. One day his boss arrives an hour early, starts walking home, and is picked up by the chauffeur on the way out to the train station. They arrive at home twenty minutes earlier than usual. How long did she walk before she met her chauffeur?
Solution #81: Early Bird
For fifty minutes. She saved the chauffeur ten minutes of traveling time each way and thus was picked up at 4:50 pm rather than the usual time.
Problem #82: Take a Letter
What three different digits are represented by X, Y, and Z in this addition problem?
Solution #82: Take a Letter
X=4, Y=9 and Z=5.
In the middle column, adding Y to Z gives a digit of Z. This means Y must be either a 0 or a 9. (It can be a 9 if the addition in the far right column causes carrying a 1 into the middle column.) Since Y is the first digit in the number on the bottom, it cannot be a zero. So, Y=9.
The two X's in the left column plus a possible 1 carried from the middle column must add up to Y, which is 9. Since 9 is odd, a 1 must have been carried over from the middle column, leaving X+X=8. So, X=4.
The value of Z can be determined from the right column: Since Y=9 and X=4, Z must be 5 because it is the only digit that can be added to 9 that puts a 4 in the one's place (9+5=14).
Problem #83: Water and Milk
There are two glasses on the table, one containing water and the other one milk. They both contain exactly the same amount by volume. If you take a teaspoon of water and mix it into the milk and then take a teaspoonful from the milk glass and mix it with the water, both glasses become contaminated. But which is the more contaminated? Does the water now contain more milk than the milk does water or the other way round?
Solution #83: Water and Milk
They are both equally contaminated. The water contains exactly as much milk as the milk contains water. The most elegant proof for this celebrated little puzzle is as follows: It does not matter how many transfers are made between the glasses or whether the contents are stirred. Provided that the volumes in the two glasses are equal, then any water in the water glass must be in the milk, there is nowhere else it can be. The milk that it has replaced must be in the water glass. The water glass therefore contains as much milk as the milk contains water.
Problem #84: Four Sheep
Farmer Giles has four sheep. One day, he notices that they are standing in such a way that they are all the same distance away from each other. That is to say, the distance between any two of the four sheep is the same. How can this be so?
Solution #84: Four Sheep
The sheep are standing on the four corner points of an equal-sided pyramid (a tetrahedron). Or to put it another way, three are on the points of an equilateral triangle and the other is on a mound of earth in the center.
Problem #85: Do Helmets Increase Head Injuries?
At the beginning of the first World War, the uniform of the British soldiers included a brown cloth cap. They were not provided with metal helmets. As the war went on, the army authorities and the War Office became alarmed at the high proportion of men suffering head injuries. They therefore decided to replace the cloth headgear with metal helmets. From then on, all soldiers wore the metal helmets. However, the War Office was amazed to discover that the incidence of head injuries then increased. It can be assumed that the intensity of fighting was the same before and after this change. So why should the recorded number of head injuries per battalion increase when men wore metal helmets rather than cloth caps?
Solution #85: Do Helmets Increase Head Injuries?
The number of recorded head injuries increased, but the number of deaths decreased. Previously, if a soldier had been hit on the head by a piece of shrapnel, it would have pierced his cap and probably killed him. This would have been recorded as a death, not a head injury. After helmets were issued it was more likely that a fragment of shrapnel would cause an injury rather than death. Thus, the incidence of head injuries increased, while the incidence of deaths decreased.
Problem #86: Sons
I. Mrs. Jones has two children. At least one is a boy. What are the chances that both are boys?
II. Mrs. Brown has two children. The younger one is a boy. What are the chances that both are boys?
Solution #86: Sons
For two children, there are only four possible combinations:
A Girl Girl
B Girl Boy
C Boy Girl
D Boy Boy
Each of these combinations is equally likely (i.e. there is a one in four chance that any two-child family will have one of the above combinations).
For Mrs. Jones, the possibilities are narrowed down to A, B or C and of these, only A means that both are both boys. Therefore, the chance that both her children are boys is one in three.
For Mrs. Brown, A and C are the only possibilities. There is thus a one in two chance that both her children are boys.
Problem #87: Dinner for Three
An ancient Arabic puzzle goes like this: A hunter met two shepherds, one of whom had three loaves of bread and the other, five loaves. All the loaves were the same size. The three men agreed to share the eight loaves equally between them. After they had eaten, the hunter gave the shepherds eight bronze coins as payment for his meal. How should the two shepherds fairly divide this money?