Problem Solving - Lone Star College System



Problem Solving

Caution, This Induction May Induce Vomiting

1. Observe that [pic], [pic], and

[pic].

Use inductive reasoning to make a conjecture about the value of [pic].

Use your conjecture to determine the value of [pic].

Oh Brother! No, Oh Sister!

2. A boy has twice as many sisters as brothers, and each sister has two more sisters than brothers. How many brothers and sisters are in the family?

{Hint: Let b be the number of boys and g the number of girls. Now write down some equations.}

Exactly How Do You Want Your Million?

3. Find a positive number that you can add to 1,000,000 that will give you a larger value than if you multiplied this number by 1,000,000? Find all such numbers.

{Hint: Let the positive number be x, and solve [pic].}

Interesting Is In The Eye Of The Beholder

4. There is an interesting five-digit number. With a 1 after it, it is three times as large as with a 1 before it. What is the number?

{Hint: If [pic] is the five-digit number, then[pic].}

Twenty-one, But Not Blackjack

5. Find the 21-digit number so that when you write the digit 1 in front and behind, the new number is 99 times the original number.

{Hint: If [pic] is the 21-digit number then [pic],

and [pic],

and [pic]}

Stand On Your Heads And Get It Together

6. The sum of two numbers is 50, and their product is 25. Find the sum of their reciprocals.

{Hint: [pic], [pic], so divide the first equation by the second equation.}

The Last Two Standing

7. What are the final two digits of [pic]?

{Hint: Look for a pattern:

|Power of 7 |Final two digits |

|[pic] |49 |

|[pic] |43 |

|[pic] |01 |

|[pic] |07 |

|[pic] |49 |

|[pic] |[pic] |

}

Don’t Give Up; Don’t Ever Give Up!

8. Given that [pic] and [pic] for all [pic], find[pic]. First find [pic].

{Hint: Look for a pattern:

|n |f(n) |

|11 |11 |

| |[pic] |

|14 | |

| |[pic] |

|17 | |

| |[pic] |

|20 | |

|23 |11 |

|[pic] |[pic] |

}

Mind Your Four’s And Two’s

9. What is the value of [pic] if [pic]?

{Hint: Factor [pic] and use the fact that [pic].}

A Lot Of Weeks, But How Many Days Left Over?

10. What is the remainder when [pic] is divided by 7?

{Hint: Look for a pattern in the remainders:

|Power of 2 |Remainder when divided by 7 |

|[pic] |2 |

|[pic] |4 |

|[pic] |1 |

|[pic] |2 |

|[pic] |4 |

|[pic] |[pic] |

}

Can You Just Tell Me How Old Your Children Are!

11. A student asked his math teacher, “How many children do you have, and how old are they?” “I have 3 girls,” replied the teacher. “The product of their ages is 72, and the sum of their ages is the same as the room number of this classroom.” Knowing that number, the student did some calculations and said, “There are two solutions.” “Yes, that is so,” said the teacher, “but I still hope that the oldest will some day win a math prize at this school.” The student then gave the ages of the three girls. What are the ages?

{Hint:

|Triple factors of 72 |Sum of the factors |

|1,1,72 |74 |

|1,2,36 |39 |

|1,3,24 |28 |

|1,4,18 |23 |

|1,6,12 |19 |

|1,8,9 |18 |

|2,2,18 |22 |

|2,3,12 |17 |

|2,4,9 |15 |

|2,6,6 |14 |

|3,3,8 |14 |

|3,4,6 |13 |

}

Cover All Your Bases, If It’s Within Your Power.

12. Solve for [pic] if [pic].

{Hint: Any number raised to the zero power, except zero itself, equals 1. 1 raised to any power is equal to 1. -1 raised to an even power is equal to 1.}

Who Needs Logarithms?

13. If [pic] and [pic], then find the value of [pic].

{Hint: Substitute the first equation into the second equation, and use an exponent property.}

I Refuse To Join Any Club That Would Have Me As A Member.

14. A club found that it could achieve a membership ratio of 2 Aggies for each Longhorn either by inducting 24 Aggies or by expelling x Longhorns. Find x.

{Hint: Let L be the number of Longhorns and A be the number of Aggies, to get [pic].}

I Cannot Tell A Fib(onacci), My Name Is Lucas.

15. If [pic], [pic], and [pic] for [pic]

a) What is the value of [pic]?

{Hint: [pic],[pic],…Keep going.}

Amazingly, [pic] can be represented as [pic] for [pic]

b) Find the values of [pic] and [pic].

{Hint: [pic], [pic], this should be enough to find values of x and y.}

Seven Heaven or Seven…

16. Find the largest power of 7 that divides 343!. [pic]

{Hint: The multiples of 7 occurring in the expansion of 343! are [pic].

The multiples of [pic] occurring in the expansion of 343! are [pic]

The multiple of [pic] occurring in the expansion of 343! is just 343.

There are no multiples of higher powers of 7 occurring in the expansion of 343!}

A Special Case Of The Chinese Remainder Theorem

17. The positive integer n, when divided by 3, 4, 5, 6, and 7, leaves remainders of 2, 3, 4, 5, and 6, respectively. Find the smallest possible value of n.

{Hint: [pic].

This means that

[pic]. So [pic] is a common multiple of 3, 4, 5, 6, and 7. What’s the least common multiple?}

How Low Can It Go?

18. The grades on six tests all range from 0 to 100 inclusive. If the average for the six tests is 93, what is the lowest possible grade on any one of the tests?

{Hint: [pic]

[pic], so [pic] will be as small as possible when [pic] is as large as possible.}

Just Your Average Joe.

19. If Joe gets 97 on his next math test, his average will be 90. If he gets 73, his average will be 87. How many tests has Joe already taken?

{Hint: Let n be the number of tests he has already taken, and T the total number of points he has already earned on the tests. Then [pic].}

A Whole Lotta Zeroes

20. How many zeroes are at the end of the number [pic]? [pic]

{Hint: Zeroes come from factors of 10. Factors of 10 come from 5’s and 2’s. See the hint for problem #16.}

The Last One Standing

21. Find the ones digit of [pic].

{Hint: Look for a pattern:

|Powers of 13 |One’s-digit |Powers of 17 |One’s digit |

|[pic] |3 |[pic] |7 |

|[pic] |9 |[pic] |9 |

|[pic] |7 |[pic] |3 |

|[pic] |1 |[pic] |1 |

|[pic] |3 |[pic] |7 |

|[pic] |9 |[pic] |9 |

|[pic] |[pic] |[pic] |[pic] |

}

Happy 2009!

22. Find the 2009th digit in the decimal representation of [pic].

{Hint: [pic], so use a pattern.}

Zero, The Something That Stands For Nothing.

23. How many zeroes are at the end of the number [pic]?

{Hint: Zeroes come from factors of 10. Factors of 10 come from 5’s and 2’s.}

Looky Here Son, This Is A Problem, Not A Chicken.

24. Foghorn C sounds every 34 seconds, and foghorn D sounds every 38 seconds. If they sound together at noon, what time will it be when they next sound together?

|Foghorn C |12:00 |12:00:34 |

|1! |1 |1 |

|2! |2 |3 |

|3! |6 |9 |

|4! |4 |3 |

|[pic] |[pic] |[pic] |

}

The Collapse Of Rationalism

27. Find the exact value of [pic].

{Hint: Rationalize the denominators. For example:[pic].}

The Beast With Many Fingers And Toes

28. How many digits does the number [pic] have?

{Hint: Zeroes come from factors of 10. Factors of 10 come from 5’s and 2’s.}

Don’t Get Stumped; Use The Fundamental Theorem Of Arithmetic.

29. Forrest stump heard that there are only two numbers between 2 and 300,000,000,000,000 which are perfect squares, perfect cubes, and perfect fifth powers. He decided to look for them, and so far he has checked out every number up to about 100,000 and is beginning to get discouraged. What are the numbers he is trying to find?

{Hint: Every positive whole number greater than 1 can be written as a product of prime factors. If N is a positive whole number greater than 2, then [pic]. In order for N to be a perfect square, all the positive exponents would have to be multiples of 2; in order for N to be a perfect cube, all the positive exponents would have to be multiples of 3; and in order for N to be a perfect fifth power, all the positive exponents would have to be multiples of 5. So all the positive exponents would have to be common multiples of 2, 3, and 5.}

Sorry, I Can’t Give You Change For A Dollar.

30. What is the largest amount of money in U.S. coins(pennies, nickels, dimes, quarters, but no half-dollars or dollars) you can have and still not have change for a dollar?

{Hint: It’s more than 99 cents. For instance: 3 quarters and 3 dimes is $1.05, but you can’t make change for a dollar.}

Destination Cancellation.

31. Express as a fraction, in lowest terms, the value of the following product of 1,999,999 factors [pic].

{Hint: Look for a pattern:

|[pic] |[pic] |

|[pic] |[pic] |

|[pic] |[pic] |

|[pic] |[pic] |

}

Officer, I Got The License Plate Number, But I Was Lying On My Back.

32. The number on a license plate consists of five digits. When the license plate is turned upside-down, you can still read a number, but the upside-down number is 78,633 greater than the original license number. What is the original license number?

{Hint: The digits that make sense when viewed upside-down are 0, 1, 6, 8, and 9.

| |1st digit |2nd digit |3rd digit |4th digit |5th digit |

|Original plate | | | | | |

| | | | | | |

|Difference of the plates |7 |8 |6 |3 |3 |

}

One Smokin’ Good Problem

33. Mrs. Puffem, a heavy smoker for many years finally decided to stop smoking altogether. “I’ll finish the 27 cigarettes I have left,” she said to herself, “and never smoke another one.” It was Mrs. Puffem’s practice to smoke exactly two-thirds of each complete cigarette(the cigarettes are filterless). It did not take her long to discover that with the aid of some tape, she could stick three butts together to make a new complete cigarette. With 27 cigarettes on hand, how many complete cigarettes can she smoke before she gives up smoking forever, and what portion of a cigarette will remain?

{Hint: With 27 complete cigarettes, she can smoke 27 complete ones and assemble 9 new complete ones…, keep going.}

Just gimme an A.

34. A class of fewer than 45 students took a test. The results were mixed. One-third of the class received a B, one-fourth received a C, one-sixth received a D, one-eight of the class received an F, and the rest of the class received an A. How many students in the class got an A?

{Hint: The number of students in the class must be a multiple of 3, 4, 6, and 8, and must be smaller than 45.}

Working Backwards In Notsuoh

35. A castle in the far away land of Notsuoh was surrounded by four moats. One day the castle was attacked and captured by a fierce tribe from the north. Guards were stationed at each bride over the moats. Johann, from the castle, was allowed to take a number of bags of gold as he went into exile. However, the guard at the first bridge took half of the bags of gold plus one more bag. The guards at the second third and fourth bridges made identical demands, all of which Johann met. When Johann finally crossed all the bridges, he had just one bag of gold left. With how many bags of gold did Johann start?

{Hint: Sometimes working backwards is a good idea. If Johann has 1 bag of gold left, then how many did he have when he approached the fourth guard?

[pic]}

Grazin’ In The Grass Is A Gas, Baby, Can You Dig It?

36. A horse is tethered by a rope to a corner on the outside of a square corral that is 10 feet on each side. The horse can graze at a distance of 18 feet from the corner of the corral where the rope is tied. What is the total grazing area for the horse?

{Hint:

}

Life Is Like A Box Of Chocolate Covered Cherries.

37. Assume that chocolate covered cherries come in boxes of 5, 7, and 10. What is the largest number of chocolate covered cherries that cannot be ordered exactly?

{Hint: If you can get five consecutive amounts of cherries, then you can get all amounts larger. Here’s why: Suppose you can get the amounts [pic], then by the addition of the box of size 5, you can also get [pic], and another addition of the box of size 5 produces [pic]and so on. This would also be true of seven consecutive amounts and ten consecutive amounts, but five consecutive amounts would occur first. So look for amounts smaller than the first five consecutive amounts.}

Easy Come, But Not So Easy Go.

38. A man whose end was approaching summoned his sons and said, “Divide my money as I shall prescribe.” To his eldest son he said, “You are to have $1,000 and a seventh of what is left.” To his second son he said, “Take $2,000 and a seventh of what remains.” To the third son he said, “You are to take $3,000 and a seventh of what is left.” Thus he gave each son $1,000 more than the previous son and a seventh of what remained, and to the last son all that was left. After following their father’s instructions with care, the sons found that they had shared their inheritance equally. How many sons were there, and how large was the estate?

{Hint: Let M be the total value of the estate.

|First Son |Second Son |[pic] |

|[pic] |[pic] |[pic] |

Since each son’s share is the same, solve an equation to determine the value of M, and use it to find the number of sons.}

Round And Round With Sarah And Hillary

39. Sarah and Hillary are racing cars around a track. Sarah can make a complete circuit in 72 seconds, and Hillary completes a circuit in 68 seconds.

a) If they start together at the starting line, how many seconds will it take for Hillary to pass Sarah at the starting line for the first time.

{Hint: Every time Sarah reaches the starting line must be a multiple of 72 seconds, and every time Hillary reaches the starting line must be a multiple of 68 seconds. So they will both be at the starting line at common multiples of 72 and 68.}

b) If they start together, how many laps will Sarah have completed when Hillary has completed one more lap than Sarah?

{Hint: Let n be the number of laps completed by Sarah, then [pic].}

Some divisibility rules for positive integers:

1) A positive integer is divisible by 2 if its one’s digit is even.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as [pic], but [pic], so if the one’s digit, c is even(divisible by 2) then the integer abc will also be divisible by 2.

2) A positive integer is divisible by 3 if the sum of its digits is divisible by 3. This process may be repeated.

Examples: 243 is divisible by 3, but 271 is not.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as [pic], but [pic], so if the sum of the digits, [pic], is divisible by 3 then the integer abc will also be divisible by 3.

3) A positive integer is divisible by 4 if the ten’s and one’s digits form a two-digit integer divisible by 4.

Examples: 724 is divisible by 4, but 726 is not.

Here’s why:

Suppose we have the four-digit integer abcd, then its value can be expressed as

[pic], but

[pic], so if the ten’s and one’s two-digit integer, cd, is divisible by 4 then the integer abcd will also be divisible by 4.

4) A positive integer is divisible by 5 if its one’s digit is either a 5 or a 0.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as [pic], but [pic], so if the one’s digit, c is divisible by 5, then the integer abc will also be divisible by 5, but the only digits divisible by 5 are 0 and 5.

5) A positive integer is divisible by 6 if it’s both divisible by 2 and divisible by 3.

6) A positive integer is divisible by 7 if when you remove the one’s digit from the integer and then subtract twice the one’s digit from the new integer, you get an integer divisible by 7. This process may be repeated.

Examples: 714 is divisible by 7 since 71 – 8 = 63, but 423 is not since 42 – 6 = 36.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as [pic], but

[pic]

, so if the new integer minus twice the one’s digit, [pic], is divisible by 7 then so is the original integer abc.

Or

A positive integer is divisible by 7 if when you remove the one’s digit from the integer and then subtract nine times the one’s digit from the new integer, you get an integer divisible by 7. This process may be repeated.

Examples: 714 is divisible by 7 since 71 – 36 = 35, but 423 is not since 42 – 27 = 15.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as [pic], but

[pic]

, so if the new integer minus nine times the one’s digit, [pic], is divisible by 7 then so is the original integer abc.

Or

A positive integer with more than three digits is divisible by 7 if when you split the digits into groups of three starting from the right and alternately add and subtract these three digit numbers you get a result which is divisible by 7.

Examples: 1412236 is divisible by 7 since 1 – 412 + 236 = -175, but 130747591 is not since 130 – 747 + 591 = -26.

Here’s why:

Suppose we have the five-digit integer abcde, then its value can be expressed as

[pic], but

[pic], so if the integer ab minus the integer cde is divisible by 7 then the integer abcde will also be divisible by 7.

7) A positive integer is divisible by 8 if the hundred’s, ten’s, and one’s digits form a three-digit integer divisible by 8.

Examples: 1240 is divisible by 8, since 240 is, 3238 is not, since 238 is not even divisible by 4.

Here’s why:

Suppose we have the five-digit integer abcde, then its value can be expressed as

[pic], but

[pic], so if the hundred’s, ten’s, and one’s three-digit integer, cde, is divisible by 8 then the integer abcde will also be divisible by 8.

8) A positive integer is divisible by 9 if the sum of its digits is divisible by 9. This process may be repeated.

Examples: 243 is divisible by 9, but 9996 is not.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as [pic], but [pic], so if the sum of the digits, [pic], is divisible by 9 then the integer abc will also be divisible by 9.

9) A positive integer is divisible by 11 if when you remove the one’s digit from the integer and then subtract the one’s digit from the new integer, you get an integer divisible by 11. This process may be repeated.

Examples: 1001 is divisible by 11 since 100 – 1 = 99, but 423 is not since 42 – 3 = 38.

Here’s why:

Suppose we have the three-digit integer abc, then its value can be expressed as [pic], but

[pic]

, so if the new integer minus the one’s digit, [pic], is divisible by 11 then so is the original integer abc.

Or

A positive integer is divisible by 11 if when you subtract the sum of the ten’s digit and every other digit to the left from the sum of the one’s digit and every other digit to the left you get a number divisible by 11. This process may be repeated.

Examples: 9031 is divisible by 11 since [pic], but 423 is not since [pic].

Here’s why:

Suppose we have the five-digit integer abcde, then its value can be expressed as [pic], but

[pic]

, so if the sum of the alternating digits from the one’s digit minus the sum of the alternating digits from the ten’s digit, [pic], is divisible by 11 then so is the original integer abcde.

Or

A positive integer with more than three digits is divisible by 11 if when you split the digits into groups of three starting from the right and alternately add and subtract these three digit numbers you get a result which is divisible by 11.

Examples: 1412290 is divisible by 11 since 1 – 412 + 290 = -121, but 130747591 is not since 130 – 747 + 591 = -26.

Here’s why:

Suppose we have the five-digit integer abcde, then its value can be expressed as

[pic], but

[pic], so if the integer ab minus the integer cde is divisible by 11 then the integer abcde will also be divisible by 11.

Divisibility And Conquer

40. Of the following three numbers determine which are divisible by 2, which are divisible by 3, which are divisible by 4, which are divisible by 6, which are divisible by 8, and which are divisible by 9: [pic], [pic], [pic].

{Hint: Divisibility rules.}

Keep On Dividing And Conquering.

41. Of the following three numbers determine which are divisible by 3, which are divisible by 6, which are divisible by 7, which are divisible by 9, and which are divisible by 11: 1358024680358024679, 864197523864197523, 964197523864197522.

{Hint: Divisibility rules.}

Seven Come Thirteen, Not Eleven.

42. Show that the second divisibility rule for 7 can also be used as a divisibility rule for 13. Modify the explanation to show why it works for 13.

Cogswell Cogs Or Spacely Sprockets?

43. In a machine, a small gear with 45 teeth is engaged with a large gear with 96 teeth. How many more revolutions will the smaller gear have made than the larger gear the first time the two gears are in their starting position?

{Hint: A revolution of the smaller gear is a multiple of 45 teeth, and a revolution of the larger gear is a multiple of 96 teeth. So the gears are again in the starting positions at common multiples of 45 and 96.}

My Tile Cutter Is Broken, So What Now?

| | | | |

| | | | |

| | | | |

| | | | |

| | | | |

| | | | |

44. The figure shows that twenty-four [pic] rectangular tiles can be used to tile a [pic] square without cutting tiles.

a) Is there a smaller sized square that can be tiled without cutting using [pic] tiles? If so, find it.

{Hint: The dimension of the square tile must be a multiple of both dimensions of the rectangular tiles.}

b) What is the smallest square that can be tiled without cutting using [pic] tiles?

{Hint: See the previous hint.}

Solving Without Completely Solving

45. If [pic] and [pic], then find [pic].

{Hint: Square the two equations.}

Odds, Evens, What’s The Difference?

46. What do you get if the sum of the first 8,000,000,000 positive odd integers is subtracted from the sum of the first 8,000,000,000 positive even integers?

{Hint: [pic]}

The Great Luggage Caper

47. Great Aunt Christine is going for her annual holiday to Barbados. She sends her butler John down to the airport with her collection of suitcases, each of which weighs either 18 or 84 pounds, and is informed that the total weight checked-in is 652 pounds. Show that this is impossible without listing and checking all the possible combinations of 18 and 84 pound suitcases.

{Hint: The total weight of the suitcases is of the form [pic], where x and y are nonnegative integers. So the expression must be divisible by the greatest common factor of 18 and 84.}

Gerry Benzel’s Favorite Problem

48. A bottle and a cork together cost $1.10. If the bottle costs $1.00 more than the cork, what does the cork cost?

{Hint: Let x be the cost of the cork and y the cost of the bottle.}

Getting Solutions Without Actually Solving

49. Notice that

[pic]

[pic]

[pic]

[pic].

a) Use inductive reasoning to determine the value of the coefficient of [pic] and the constant term in the expansion of the following product: [pic].

b) Use the previous result to determine the sum of the seventeenth powers of the 17 solutions of the equation [pic].

{Hint: The Fundamental Theorem of Algebra guarantees that the equation [pic] has seventeen solutions(counting duplicates). The seventeen solutions of [pic] are [pic]. So adding the seventeen equations together yields:

[pic]. Now use the previous result.}

Dots And Dashes, But It’s Not Morse Code.

50. The diagram shows a sequence of shapes [pic]. Each shape consists of a number of squares. A dot is placed at each point where there is a corner of one or more squares.

| | | | |

|[pic] |[pic] |[pic] |[pic] |

|Shape |number of rows, n |number of squares, S |number of dots, D |[pic] |

|[pic] |1 |1 |4 |3 |

|[pic] |2 |4 |10 |6 |

|[pic] |3 |9 |18 |9 |

|[pic] |4 |16 |28 |12 |

a) Use inductive reasoning to find a formula for S in terms of n.

b) How many squares are in shape [pic]?

c) Use inductive reasoning to find a formula for D in terms of n.

{Hint: Notice that [pic]in the last column is always a multiple of 3.}

d) How many dots are in shape [pic]?

It Squares; It cubes; It does it all!

51. If [pic] and [pic], then what is the value of [pic]?

{Hint: [pic].}

Two Squares Don’t Get Along – A Difference Of Squares!

52. If [pic], [pic], [pic], [pic], and [pic], then find the value of [pic].

Hint: [pic].

How Touching!

53. Four circles, each of which has a diameter of 2 feet, touch as shown. Find the area of the shaded portion.

{Hint: The area of the shaded portion would be the area of the square minus the area of the four circular sectors.

}

The method of finite differences can be used to produce formulas for lists of numbers. For example, if the list of numbers is 5,7,9,11,13,…, then the list of first finite differences is the list of differences of consecutive numbers: [pic] or more simply, 2,2,2,2,…. Whenever the list of first finite differences is a repetition of the same number, it means that the original list of numbers can be produced by a formula of the form [pic], where n represents the position in the list. Here’s why:

|original |[pic] |[pic] |[pic] |[pic] |[pic] |

|first finite differences |[pic] |[pic] |[pic] |[pic] |[pic] |

In fact, the repeated number in the first finite differences will always be the coefficient of n.

In our example, it must be that [pic] and [pic]. So we get that [pic] and [pic], and a formula for the list of numbers 5,7,9,11,13,… is [pic] Sometimes the list of first differences is not a repetition of the same number. An example of this is the list 3,9,19,33,51,73,…. The list of first finite differences is 6,10,14,18,22,…. As you can see it’s not a repetition of the same number. Now let’s calculate the list of second finite differences: 4,4,4,4,…. Whenever the list of second finite differences is a repetition of the same number it means that the original list of numbers can be produced by a formula of the form [pic], where again n represents the position in the list. Here’s why:

|original |[pic] |[pic] |[pic] |[pic] |[pic] |[pic] |

|first finite differences |[pic] |[pic] |[pic] |[pic] |[pic] |[pic] |

|second finite differences |[pic] |[pic] |[pic] |[pic] |[pic] |[pic] |

In fact, the repeated number in the second finite differences will always be twice the coefficient of [pic]. In our example, it must be that [pic], [pic], and [pic]. Substituting the value of a into the next two equations leads to the system [pic]. Subtracting the first equation from the second equation leads to [pic], so the values are [pic], [pic], and [pic]. So a formula for the list of numbers 3,9,19,33,51,73,… is [pic]. Similar results hold for higher order differences which are repetitions of the same number. You may use the method of finite differences to solve the following two problems.

What’s the Diff?

54. Find a formula for each of the following lists of numbers:

a) 2,5,8,11,…

b) [pic]

c) [pic]

d) 3,7,13,21,31,43

e) [pic]

f) [pic]

g) 1,15,53,127,249,…

h) [pic]

You Want Me To Cut The Pizza Into How Many Slices?

55. A chord is a line segment joining two points on a circle. Here, n is the number of chords.

Associated with each number of chords, n, is the maximum number of regions that the circle is divided into by the n chords. For the first five numbers of chords, the list of the maximum number of regions is [pic]. Use the method of finite differences to find a formula for the maximum number of regions that a circle can be divided into using n chords, and use the formula to predict the maximum number of regions a circle can be divided into using 60 chords.

Now What Was That Last Digit?

56. Peggy is writing the numbers 1 to 9,999. She stops to rest after she has written a total of 1,000 digits. What is the last digit that she wrote?

{Hint:

| |Quantity |Total number of digits |

|1 digit numbers(1-9) |9 |9 |

|2 digit numbers(10-99) |90 |180 |

|3 digit numbers(100-999) |900 |2,700 |

|4 digit numbers(1,000-9,999) |9000 |36,000 |

}

On The Mark, Off The Mark, Or Bull’s Eye?

57. In the following figure, the curves are concentric circles with the indicated radii. Which shaded region has the larger area, the inner circle or the outer ring?

Calculate the area of each region and check your visual estimation ability.

I Hate This Problem To The Nth Degree.

58. Use the following properties of exponents to find the exact value of the given expressions.

[pic], [pic], [pic]

a) [pic] b) [pic]

We use a base 10 number system. A number written with the digits abcd actually represents the number [pic], where the digits a, b, c, and d can be any of the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. If we want to use a base of 2, then abcd would represent the number [pic], where the digits a, b, c, and d can be the numbers 0 or 1.

I’m Thinking Of A Number From 1 To 31.

59. A person is asked to think of a whole number from 1 to 31. The person is shown the following 5 cards and asked to identify the cards that contain the number he is thinking of.

| 1 3 5 7 | 2 3 6 7 | 4 5 6 7 |

| | | |

|9 11 13 15 |10 11 14 15 |12 13 14 15 |

| | | |

|17 19 21 23 |18 19 22 23 |20 21 22 23 |

| | | |

|25 27 29 31 |26 27 30 31 |28 29 30 31 |

| 8 9 10 11 |16 17 18 19 | |

| | | |

|12 13 14 15 |20 21 22 23 | |

| | | |

|24 25 26 27 |24 25 26 27 | |

| | | |

|28 29 30 31 |28 29 30 31 | |

By adding the numbers in the upper left corners of the identified cards, you can always determine the person’s number. Use base 2 numbers to explain how the trick works.

Oh Yeah, Well I’m thinking Of A Number From 1 to 63.

60. See if you can extend the previous trick to work for whole numbers from 1 to 63 using 6 cards instead of 5. Complete the numbers on the 6 cards in order for the trick to work.

| 1 3 5 7 9 11 13 15 | 2 |

| | |

|17 19 21 23 25 27 29 31 | |

| | |

|33 35 37 39 41 43 45 47 | |

| | |

|49 51 53 55 57 59 61 63 | |

| 4 | 8 |

| | |

| | |

| | |

| | |

| | |

|16 |32 |

| | |

| | |

| | |

| | |

| | |

Call it like you see it.

61. Consider the following figure:

a) What fraction of the large square is shaded ?

b) What fraction of the large square is shaded ?

c) What fraction of the large square is shaded ?

Let’s Not Go Overboard.

62. This problem in one form or another has been around for nearly 2,000 years – if not longer. Josephus, a Jewish historian of the first century A. D., mentions it. In all its forms the problem seems to have a violent streak. Here’s our version of it: One-hundred sailors are arranged around the edge of their ship. They hold in order the numbers from 1 to 100. Starting the count with number 1, every other sailor is pushed overboard into the cold North Atlantic waters until there is only one left – the survivor.

a) What number do you want to be holding in order to be the survivor?

b) How many times will the survivor be skipped during this process?

c) Find the number of the last sailor to be pushed overboard.

d) Find the number of the second to last sailor to be pushed overboard.

{Hint: Consider the problem for smaller numbers of sailors and work up: For example with 10 sailors you have

The survivor, 5, is skipped three times. The last sailor to be pushed overboard is 9, and the next to last sailor pushed overboard is 1.}

A Springtime Path

63. Determine the number of different paths for spelling the word APRIL:

| | |

| | |

| | | |

| | | |

| | | |

| | | | |

| | | | |

| | | | |

| | | | |

Put A Sock In It Brad And Angelina.

65. Mr. Smith left on a trip very early one morning. Not wishing to wake Mrs. Smith, Mr. Smith packed in the dark. He had socks that were alike except for color, and his socks came in six different colors. Find the least number of socks he would have had to pack to be guaranteed of getting

a) at least one matching pair of socks. b) at least two matching pairs of socks.

c) at least three matching pairs of socks. d) at least four matching pairs of socks.

{Hint: He could actually pack as many as 6 socks and still not have a matching pair.

|1 |1 |1 |1 |1 |1 |

|Color 1 |

|[pic] |

|[pic] |

|[pic] |

Now subtract these values from largest to smallest if possible from 139 until you are left with a non-negative value less than 12.

[pic], [pic], [pic]. So [pic]. Apply this method to evaluate [pic] and [pic].

Sometimes Reduction Can Get In The Way Of Induction.

92. Observe that

[pic].

Use inductive reasoning to make a conjecture about the value of

[pic].

Use your conjecture to find the value of [pic].

If this process were to continue indefinitely, i.e. [pic], what would be the resulting value?

In College Algebra, you learned something called the Remainder Theorem for evaluating polynomials. It says that if you want to find [pic] where [pic] is a polynomial, then you can divide [pic] by [pic], and the remainder will be [pic]. You also learned synthetic division, which gave you an efficient way to divide [pic] by [pic], and therefore an efficient way to evaluate [pic]. As an example, suppose you want to find [pic], where [pic]. To find the remainder when [pic] is divided by [pic], we’ll use synthetic division:

[pic]

So [pic].

When you are given the base b numeral [pic], its value is [pic] which is what you get when you find [pic] for the polynomial [pic]. So an efficient method of converting the base b numeral [pic] into decimal is to use synthetic division.

[pic]

This approach is called Horner’s Method.

Little Jack Horner Sat In A Corner Converting His Base B Numerals.

93. Convert the following numerals into equivalent decimals using Horner’s Method:

a) [pic] b) [pic] c) [pic]

There is an efficient method for converting base ten(decimal) fractions into base 2. As an example, let’s convert the decimal fraction .8215 into its base 2 equivalent.

| |Whole number part |

|[pic] |1 |

|[pic] |1 |

|[pic] |0 |

|[pic] |1 |

The process stops when the decimal part is zero. So [pic].

Double, Double, Less Toil, And No Trouble.

94. Use the previous algorithm to convert the following base 10 numerals into base 2.

a) 1.6875 b) 2.46875 c) .4

Are You Ready For Prime Time?

95. If you add up all the prime numbers less than one-million, will it be an even number or an odd number? Why?

Cheaper By The Dozen?

96. Find the sum of the divisors of 6480 that are multiples of 12.

What Four?

97. How many four-digit numbers are multiples of 15, 20, and 25?

-----------------------

8

8

18

18

[pic]

[pic]

[pic]

[pic]

[pic]

3

5

4

1

2

3

4

5

6

7

8

9

10

first elimination

9

second elimination

7

1 large and 4 small

1 + 4 = 5

5

9

3

third elimination

1

5

the survivor

1

5

1 large, 4 medium, and 9 small

1 + 4 + 9 = 14

1 extra large, 4 large, 9 medium, and 16 small

1 + 4 + 9 + 16 = 30

Bus

Train

Car

U

6

7

2

x

x

Art

Biology

Chemistry

Drama

U

Fish

Bird

Cat

Dog

U

First-year student

Non-business

Business

Female

Male

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