Problem #26: Lighthouse
How far is the horizon from the top of a 125.7meterhigh lighthouse? (The earth can be considered spherical with a circumference of 40,000 km.)
Solution #26: Lighthouse
Let S be the top of the lighthouse. Let A be a point on the horizon.
Let O be the center of the earth. Triangle OAS is a rightangle triangle since SA is tangent to the earth, which we consider to be a perfect sphere.
We thus have OS^{2} = OA^{2} + AS^{2}
but OA = 40,000,000/(2pi) meters
and OS = OA + 125.7 = OA + 40pi/OS^{2} ~= OA*OA + 2 OA = 40pi
Consequently,
AS = sqrt(OS^{2}  OA^{2}) ~= sqrt(2*40pi^{.}40,000,000/2pi) = 40,000 m The horizon is therefore approximately 40 km from the top of the lighthouse.
Problem #27: The Demonstration
If we set out by ranks of 10, we will be one short. We will also be one short if we set out by ranks of 9, 8, 7, 6, 5, 4, 3, and even 2. Yet there are fewer than 5000 participants. How many are we?
Solution #27: The Demonstration
Let x be the number of demonstrators.
(x + 1) must be a multiple of 2, 3, 4, ..., 9.
(x + 1) is therefore a multiple of the smallest common multiple of these numbers, that is, a multiple of
2 x 2 x 3 x 3 x 5 x 7 = 2520.
Let x + 1 = k(2520), k being a whole number.
x = k(2520)  1. Since there are fewer than 5000 demonstrators, k = 1. Hence x = 2519, the number of marchers.
Problem #28: Year of Birth
Take away from your year of birth the sum of the four numerals that make it up. You will end of with a number divisible by 9. Why is this?
Solution #28: Year of Birth
Let T, h, t and u be the four digits that make up your year of birth (Thousands, hundreds, tens, and units).
Your hear of birth can be written:
1000T + 100h + 10t + u.
If you take away
T + h + t + u
you are left with
999T + 99h + 9t
which is divisible by 9.
Problem #29: Spanish Arithmetic
4 + 4 + 4 + 4 + 4 = 20
In Spanish, this can be written:
CUATRO
+ CUATRO
+ CUATRO
+ CUATRO
+ CUATRO

VEINTE
How are the 10 numerals represented by the 10 letters A, C, E, I, N, O, R, T, U and V for this sum to come out right?
Solution #29: Spanish Arithmetic
170469
+ 170469
+ 170469
+ 170469
+ 170469

852345
(not yet completed)
Problem #30: A Story of 1's and 2's
Write, side by side, the numeral 1 an even number of times. Subtract from the number thus formed the number obtained by writing, side by side, a series of 2s half the length of the first number. You will always get a perfect square. For instance,
1111  22 = 1089 = 33^{2}
Can you say why this is?
Solution #30: A Story of 1's and 2's
11...1  22...2 = 11...1 11...1  2(11...1)
    
2n times n times n times n times n times
= 11...1 00...0  11...1
  
n times n times n times
= 11...1 x (100...0  1)
 
n times n times
= 11...1 x 99...9
 
n times n times
= 11...1 x 9 x 11...1
 
n times n times
= 3^{2} x 11...1^{2}

n times
= 33...3^{2}

n times
Example: 11  2 = 9 = 3^{2}.
Problem #31: How Many Children Were You?
How many children were you?
If you ask me such a question, I will reply that my mother dreamed of having at least 19 children but her dream did not come true, that my sisters were three times as more numerous than my first cousins, and that I had half as many brothers as I had sisters.
Solution #31: How Many Children Were You?
The number of my sisters must be a multiple of 3 and a multiple of 2. It must therefore be a multiple of 6.
Total number of children: [(multiple of 6) x (1 + 1/2)] + 1.
Since this number must be less than 19, the only possible solution for the multiple of 6 is 6 exactly. I therefore have 6 sisters and 3 brothers. We are 10 children in all.
Problem #32: Brother and Sister
"Sister, you have as many brothers as you have sisters." "Brother, you have twice as many sisters as you have brothers." Can you deduce from this conversation how many children there are in the family?
Solution #32: Brother and Sister
Let x be the number of boys and y the number of girls.
What the boy says to the girl can be written as x = y  1.
The answer that the sister gives her brother can be written as:
y = 2(x  1) or y = 2x  2
Substituting x = y  1 gives us y = 4 and hence x = 3.
There are therefore seven children in the family.
Problem #33: Ursula and the Cats
If you ask old Ursula how many cats she has at home, she answers sadly: "Fourfifths of my cats plus fourfifths of a cat." How many cats does this add up to?
Solution #33: Ursula and the Cats
Let n be the number of cats.
We have n = 4/5 n + 4/5. Hence n = 4.
Ursula lives with four cats.
Problem #34: New Math
My son learned how to count in a base different from 10, so that, for instance, instead of writing 136, he writes 253. In what base does he count?
Solution #34: New Math
Let b be the unknown base. When my son writes 253 in base b, this can be interpreted as
2b^{2} + 5b + 3 = 136
or
2b^{2} + 5b  133 = 0
or
(b  7)(2b + 19) = 0
Since b is a whole, positive number, the only possible solution is 7; the unknown base is 7.
Problem #35: Chinese Numbers
Think of a number between 1 and 26. Look at the following table of six squares, one square at a time.
++++
 1 4 7  2 5 8  3 4 5 
 10 13 16  11 14 17  12 13 14 
 19 22 25  20 23 26  21 22 23 
++++
 6 7 8  9 10 11  18 19 20 
 15 16 17  12 13 14  21 22 23 
 24 25 26  15 16 17  24 25 26 
++++
Each time the number you have picked belongs to one of the squares, write down the number in the top lefthand corner. Add together all these numbers.
For example, 16 is in the first, fourth, and fifth squares. If the first numbers of each of these squares are added together, we get 1 + 6 + 9 = 16, which is the original number we picked. Can you explain this?
Solution #35: Chinese Numbers
Any number n equal to or smaller than 26 must certainly be smaller than 3^{3} = 27. It can therefore be broken down (in base 3) in the following manner:
n = (0 or 1 or 2) x 3^{0} + (0 or 1 or 2) x 3^{1} + (0 or 1 or 2) x 3^{2}
In other words
n = (0 or 1 or 2) + (0 or 3 or 6) + (0 or 9 or 18)
Thus all number that have a "1" in their makeup will be found in the first square, all number that have a "2" in the second square, a "3" in the third square, a "6" in the fourth square, a "9" in the fifth, and an "18" in the sixth.
The number n can therefore be found by adding the numbers at the top lefthand corner of the squares where n appears.
Problem #36: The Smallest Possible
Which is the smallest number which, when divided by 2, 3, 4, 5 and 6 will give 1, 2, 3, 4 and 5 as remainders, respectively?
Solution #36: The Smallest Possible
Let n be this unknown number. Since n divided by 2 leaves a remainder of 1, n + 1 must be divisible by 2.
Since n divided by 3 leaves a remainder of 2, n + 1 must be divisible by 3. Similarly, n + 1 must be divisible by 4, 5, and 6.
The smallest common multiple of 1, 2, 3, 4, 5, and 6 is 60.
Therefore, n + 1 = 60. Hence n = 59.
Problem #37: For Those Under 16
Tell me in which columns your age appears and I will guess your age by adding the first number of the corresponding columns. Is this magic? squares, one square at a time.
+++++
 2  8  4  1 
 3  9  5  3 
 6  10  6  5 
 7  11  7  7 
 10  12  12  9 
 11  13  13  11 
 14  14  14  13 
 15  15  15  15 
+++++
Solution #37: For Those Under 16
In is not a matter of magic but simply the characteristics of a number written in base 2. Any whole number less than 16 can be written as
x_{0}2^{0} + x_{1}2^{1} + x_{2}2^{2} + x_{3}2^{3}
where each x_{n} is either a 0 or a 1.
When x_{n} = 1, the corresponding age is in the column headed 2^{n}.
When x_{n} = 0, the corresponding age is not in the column headed 2^{n}.
Example: If you are 13 years old, your age will appear in the last three columns because
13 = 1^{.}2^{0} + 1^{.}2^{1} + 0^{.}2^{2} + 1^{.}2^{3} = 8 + 4 + 0 + 1
Problem #38: Product
The product of four consecutive whole numbers is 3024. What are these numbers?
Solution #38: Product
3024 ends in neither a 5 nor a 0; therefore, none of the four numbers given can be divisible by 5 or by 10. If all four numbers were greater than 10, their product would exceed 10,000, which is not the case. The four numbers must either be {1,2,3,4} or {6,7,8,9}. In the first case, the product is 24. The answer can only be {6,7,8,9}, as can be easily verified.
Problem #39: Riverside Drive
Would you like to have dinner with me tonight? Don't go to the wrong house. I live in one of the 11 houses along Riverside Drive. When I am at home, facing the river, and I multiply the number of houses on my left by the number of houses on my right, I get a number which is greater by 5 units than the number that my neighbor to my left would get if he did the same thing. Where along the Drive do I live?
Solution #39: Riverside Drive
Let r be the number of houses to the right of my house and l the number of houses to the left of my house when I look toward the river. We have the following simultaneous equations:
r + l = 10
rl  (r+1)(l1) = 5
Hence, r + l = 10 and r  l = 4.
thus r = 7 and l = 3.
My house is therefore the fourth from the left when facing the river.
Problem #40: Summit Meeting
Two delegations are to meet at the top floor of a skyscraper whose elevators can hold up to nine people at a time. The first delegation to arrive makes up a certain number of elevator loads, filling each one except the last, which has space for five more people. The second delegation does the same, not using onethird of the last elevator load.
At the start of the meeting, each member of each delegation shakes hands with each member of the other delegation, and each time, a photo is taken. Knowing that the photographer was using films with nine exposures on each, how many unexposed frames will he have left on his last film?
Solution #40: Summit Meeting
Let d_{1} be the number of the members in the first delegation and d_{2} the number of the members in the second.
d_{1} = (multiple of 9) + 4
d_{2} = (multiple of 9) + 6
Number of photographs:
d_{1} x d_{2} = (multiple of 9) + (4 x 6)
= (multiple of 9) + (18 + 6)
= (multiple of 9) + 6
The photographer therefore exposed six frames on his last film.
There are three frames left.
Problem #41: A Bag Of Marbles
A group of children share marbles from a bag. The first child takes one marble and a tenth of the remainder. The second child takes two marbles and a tenth of the remainder. The third child takes three marbles and a tenth of the remainder. And so on until the last child takes whatever is left. Knowing that all the children end up with the same number of marble, how many children were there and how many marbles did each one get?
Solution #41: A Bag Of Marbles
Let n be the number of children, x each child's share of marbles, and N the total number of marbles. We have N = nx.
First child's share: 1 + (N1)/10 = x.
Last child's share: x.
Penultimate child's share: n  1 + x/9 = (N/x)  1 + x/9 = x.
Consider the expression for the first child's share: N = 10x  9.
Substituting into the equation for the penultimate child's share gives us
(10x  9)/x  1 + x/9 = x
which can be written
8x^{2}  81x + 81 = 0 or (x  9)(8x  9) = 0
The number of marbles taken by each child being a whole number, the only possible solution is x = 9;
hence N = 81 and n = 9.
There were nine children. Each got nine marbles.
Problem #42: The Five Numbers
Can you find five consecutive whole number that are all positive and such that the sum of the squares of the two largest numbers is equal to the sum of the squares of the three smallest?
Solution #42: The Five Numbers
Let n be then number in the middle of the series. We have the relationship
(n + 1)^{2} + (n + 2)^{2} = n^{2} + (n  1)^{2} + (n  2)^{2}
Expanding and simplifying gives us
n(n12) = 0
The only acceptable solution is n = 12. The five consecutive whole numbers are
10, 11, 12, 13 and 14
Problem #43: 100!
How many zeros are there at the end of 100! ?
Solution #43: 100!
The number of zeros at the end of a number is equal to the number of times that number is a multiple of 10. However, 5 and 2 are factors of 10. The number of zeros will therefore be equal to the smaller of the following two numbers: number of times 2 appears as a factor and number of times 5 appears as a factor in the breakdown of the original number into prime factors.
Here, of course, 2 appears as a factor more often than 5. Let us calculate the number of times 5 appears as a factor: 100! has 20 numbers that are multiples of 5. Some of them (4 of them) are even multiples of 25: 25, 50, 75, and 100. In the breakdown of 100! into prime factors, one will find 5 raised to the power 24 (20 + 4 = 24). There are therefore 24 zeros at the end of 100!.
Problem #44: House of Cards
Do you know how to build card houses?
The first floor is easy: (2 cards)
The second floor is easy, too:
(7 cards)
Here's the third:
(15 cards)
And so forth.
How many cards does it take to build a 47story house?
Solution #44: House of Cards
Note first that in all houses of cards, every level has three cards more than the level immediately above it.
The number of cards for a 47story house is therefore
2
+ 2 + 3
+ 2 + 3^{.}2
+ 2 + 3^{.}3
+ ...
+ 2 + 3^{.}(n1)
+ ...
+ 2 + 3^{.}(471)

47^{.}2 + 46^{.}47/2 x 3 = 3337
Problem #45: Lunch With Friends
Ten couples meet for lunch. After a cocktail they go into the dining room in a random order, one by one. How many people must enter the dining room to assure that:
1. There is at least one married couple in the group?
2. There are at least two diners of the same sex?
Solution #45: Lunch With Friends
As soon as 11 people enter the dining room, there will necessarily be one married couple among them.
As soon as three people enter the room, there will necessarily be two diners of the same sex.
Problem #46: Thirtytwo Cards
Alfred, Brian, Christopher and Damon play with a deck of 32 cards. Damon deals them out unequally, then says: "If you want us to have the same number of cards, do exactly as I say. You, Alfred, divide half of your cards between Brian and Christopher. Then, Brian, you do the same with Christopher and Alfred. Finally, Christopher, you follow suit with Alfred and Brian." How did Damon distribute the cards?
Solution #46: Thirtytwo Cards
At the end of the game each of the four players has 8 cards. Christopher, having just shared half his cards between Brian and Alfred, must have had 16 cards and Alfred and Brian 4 each. But Brian had just shared half his cards between Christopher and Alfred: before that shareout, Brian must have had 8, Alfred 2, and Christopher 14. But this was just after Alfred shared half his cards between Brian and Christopher.
Hence at the start Alfred had 4 cards, Brian 7, Christopher 13, and Damon (who was not involved in the three various shareouts) 8.
Problem #47: Checker Pattern
How many turquoise squares will be required to build the twentieth figure in this pattern?
Solution #47: Checker Pattern
761.
One visual representation for the pattern is n^{2} + (n1)^{2} (outer squares + inner squares)
Problem #48: Odd or Even
Does 124_{FIVE} represent an odd number? How can you determine whether a number is odd by looking at its basefive representation?
Solution #48: Odd or Even
Yes. The number 124_{FIVE} is represented by the basefive pieces that follow. Each piece has an odd number of units. The case of 124_{FIVE} has an odd number of pieces, 7, so 124_{FIVE} is odd. In general, if the sum of the digits in base five is odd, the number is odd.
Problem #49: Five Times Five
The digits in this addition problem have been replaced by letters. Replace each letter with a different digit to obtain a correct sum that is as large as possible. What is the value of "FIVE"?
FIVE
FIVE
FIVE
FIVE
+ FIVE

ISLE
Solution #49: Five Times Five
1970.

F must equal 1 because anything higher would make the sum greater than four digits.

(not yet completed)
Problem #50: Largest Product
Use each digit 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 exactly once to form two fivedigit numbers that when multiplied produce the largest quantity.
Solution #50: Largest Product
96,420 and 87,531 with a product of 8,439,739,020.
A reasonable first guess is two numbers that begin 97... and 86..., however these do not actually give the largest product.
Problem #51: Multiple Choice
Given that one and only one answer is correct, which of the following is true?

All of the below

None of the below

One of the above

All of the above

None of the above

None of the above
Solution #51: Multiple Choice
E. Consider the choices. Contradictions between answers, such as between C and D, eliminate A as a possibility. B is false because if it were correct, then C would also be true. It follows that C is false because both A and B are false. Similarly, it follows that D is false. Aha! E is true! Of course, that result makes F false.
Problem #52: A Square Number
For what bases does the number 121_{BASE} represent a square number?
Solution #52: A Square Number
For any base!
In base b, 121 = 1^{.}b^{2} + 2^{.}b + 1 = (b + 1)^{2}.
Problem #53: Reverse Multiplication
Find all fivedigit numbers that are reversed when multiplied by 4.
Solution #53: Reverse Multiplication
21,978 is the only such number. Suppose that ABCDE x 4 = EDCBA. Since 4A < 10 and A is even, A = 2. Therefore, E = 8. Note that E x 4 must end in 2. It follows that B <= 2 because 23 x 4 > 89. In fact, B = 1 because the twodigit number B2 must be a multiple of 4.
Thus far we have found that 21CD8 x 4 = 8DC12. Observe that D >= 4 and that D8 x 4 ends in 12. Therefore, D = 7. The only possible value of C is 9.
Problem #54: Odd Product
The product of three consecutive odd numbers is 357,627. What is the smallest of the three?
Solution #54: Odd Product
69. The cube root of 357,627 is slightly less than 71. It seems that 69 x 71 x 73 is a possible product. The result can be verified. Note that 9 x 1 x 3 = 27, thus making the last digit a 7, which is a good visual check.
Alternatively, solve a cubic equation by setting up a product of successive odd numbers, say, a2, a, and a+2. This setup produces a more workable equation than does 2n1, 2n+1, and 2n+3.
Problem #55: Carnival Dice
A carnival game offers you the opportunity to bet $1 on a number from 1 through 6 on a single roll of the two dice. If your number comes up on one die, you win $2 and keep th $1 you be. If it comes on both dice, you win $5 and keep the $1 you bet. Only if the number does not appear on either die do you lose your $1 bet. Does this game favor you or the carnival?
Solution #55: Carnival Dice
The game is fair and does not favor either you or the carnival. The chart shows the eleven winning and twentyfive losing combinations for betting on the number 1. One outcome will win $5 for you, ten will win $2, and twentyfive will lose $1. Since these outcome are all equally likely, your expected winnings are:
($2 x 10) + ($5 x 1)  ($1 x 25) = 0
1 2 3 4 5 6
1 1,1 1,2 1,3 1,4 1,5 1,6
2 2,1 2,2 2,3 2,4 2,5 2,6 L
3 3,1 3,2 3,3 3,4 3,5 3,6 O
4 4,1 4,2 4,3 4,4 4,5 4,6 S
5 5,1 5,2 5,3 5,4 5,5 5,6 E
6 6,1 6,2 6,3 6,4 6,5 3,6 
WIN LOSE+
Problem #56: Sum Cubes
23 = 2^{3} + 2^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3} + 1^{3}
Only one other integer requires nine cubes of positive integers to represent its value as a sum of cubes. This other integer is between 200 and 300. Can you find it?
Solution #56: Sum Cubes
239. The sum can be expressed as:
23 = 4^{3} + 4^{3} + 3^{3} + 3^{3} + 3^{3} + 3^{3}. + 1^{3} + 1^{3} + 1^{3}
