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Jump Like a Flea

A flea can jump 350 times its body length. If humans could jump like fleas, how far could you jump?

Solution
Answers will vary, depending on the height of the person. The average height of an 11–14-year-old male is 60 inches. The average height of a female of the same age is 62 inches, according to the National Institute of Health. A student who is 5 feet tall could jump about 1750 feet, or 583 yards. For varying answers, multiply the height by 350 and approximate that distance with a known landmark.

From Menu of Problems, April 2001




  
Product of Primes

2006 is the product of three primes, 2 × 17 × 59. Find the first year that occurs after 2006, which is the product of three consecutive primes.

Solution
2431, which is the product of 11 × 13 × 17.

From Menu of Problems, April 2006




  
Adding Digits

If you add the digits in a number, how many numbers between 0 and 1000 will have a sum of 15?

Solution
73. Only two ways are possible to add two digits to get a sum of 15: 6 + 9 and 7 + 8. Each combination results in six numbers: 69, 96, 609, 690, 906, and 960. Eight ways exist to combine three distinct digits for a sum of 15: 1 + 5 + 9, 1 + 6 + 8, 2 + 4 + 9, 2 + 5 + 8, 2 + 6 + 7, 3 + 4 + 8, 3 + 5 + 7, and 4 + 5 + 6. Again, each combination results in six numbers. Four other ways to get 15 are 1 + 7 + 7, 3 + 3 + 9, 3 + 6 + 6, and 4 + 4 + 7. Each combination results in three numbers. Finally, 555 fits the bill. All told, we have 12 + 48 + 12 + 1 = 73.

From April's Menu of Problems, April 1999




  
Twins

The product of the ages of twins and their younger brother is 36. How old are the children?

Solution
1, 6, and 6. The possible ages of the children are 1, 1, 36; 2, 2, 9; 3, 3, 4; and 1, 6, 6. The only choice that has the twins as the oldest children is 1, 6, 6.

From May's Menu of Problems, May 1999




  
5, 7, and 11 Minute Glasses

How can you measure 13 minutes exactly with a 5-minute, 7-minute, and 11-minute hourglass?

Solution
Start the 5- and 11-minute timers at the same time. When the 5-minute runs out, start timing for the 13 minutes. When the 6 minutes remaining on the 11-minute timer are finished, start the 7-minute timer.

From Menu of Problems, April 2002




  
Make the Statement True

Add one straight line segment to make the mathematical sentence 5 + 5 + 5 = 550 a true statement.

Solution
Change one of the plus signs to a number 4 by connecting the left and top endpoints. There might be other possibilities as well.

From April's Menu of Problems, April 2007




  
Smallest Odd Number

What is the smallest odd number you can obtain from the product of four different prime numbers?

Solution
1155. Remember that 1 is not a prime number. Two is the first prime number, but if it is one of the prime numbers used, the product will be even. Therefore, to obtain the smallest value, you would use 3, 5, 7, and 11, the product of which is 1155.

From April's Menu of Problems, April 2007




  
Multiples of 6

Find six consecutive multiples of 6 whose sum is the least common multiple (LCM) of 13 and 18.

Solution
24, 30, 36, 42, 48, and 54. The LCM of 13 and 18 is 13 × 18 = 234, since the numbers are relatively prime. Since we are looking for 6 numbers, we can divide 234 by 6 to get 39. Therefore, we know that 39 is greater than 3 of the numbers and less than 3 of the numbers; 24, 30, and 36 are the multiples of 6 less than 39, and 42, 48, and 54 are the multiples of 6 greater than 39. The sum of 24, 30, 36, 42, 48, and 54 is 234.

From March's Menu of Problems, March 2007




  
Drawing Rectangles

How many different rectangles can be drawn using the hour marks on a clock's face as vertices?

Solution
15.

The rectangles that can be drawn are shown below.

rect

From February's Menu of Problems, February 2004 MTMS.




  
Paper Money

In how many different ways can you receive $20 from your bank if you ask for paper money only?

Solution
Ten, excluding the use of $2 bills.

Students can solve this problem by making an organized list.

Using $2 bills creates another thirty-one choices.

From November's Menu of Problems, November 1999 MTMS.

Contributed by Cynthia Barb, Kent State University--Stark Campus, Canton, Ohio; and Anne Larson Quinn, Edinboro University of Pennsylvania, Edinboro, Pennsylvania.




  
Half of a Half of a Half

What would you get if you divided a half of a half of a half by a half of a fourth of a half?

Solution
2. A half of a half of a half is 1/2×1/2×1/2 = 1/8, and a half of a fourth of a half is 1/2×1/4×1/2 = 1/16; 1/8 divided by 1/16 is 2.

From November's Menu of Problems, November 2006




  
Prime Numbers

Find the sum of the least and greatest two-digit prime numbers whose digits are also prime.

Solution
96.

Consider prime numbers such that each digit is 2, 3, 5, or 7. The smallest two-digit number that can be formed with these digits is 22, but it is not prime. Therefore, 23 is the least value.

The greatest two-digit numbers that can be formed with these digits are 77 and 75, but they are not prime. The number with the greatest value that meets the conditions is 73; therefore, 23 + 73 = 96.

From February's Menu of Problems, February 2005 MTMS.

Courtesy of David Rock at the University of Mississippi.




  
Multiple of 12

Find the greatest ten-digit positive multiple of 12 using each digit once and only once.

Solution
9,876,543,120. For a number to be a multiple of 12, it must be divisible by both 3 and 4. The sum of the ten digits 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 45, which is divisible by 3. To be divisible by 4, the last two digits must be divisible by 4. The largest number obtained using all ten digits once and only once is 9,876,543,210, but 10, the last two digits, is not divisible by 4. The next greatest value is 9,876,543,201, but 01 is not divisible by 4. The next greatest value is 9,876,543,120; since 20 is divisible by 4, the number is therefore both divisible by 3 and 4 and 12.

From March's Menu of Problems, March 2007




  
Positive Primes

What is the last digit (ones) of the product of the positive prime numbers less than 50?

Solution
0. Prime numbers less than 50 end in 1, 3, 7, or 9, with the exception of 2 and 5. Since 2 and 5 will have to be factors and 2×5 equals 10, the product must has a ones digit of 0.

From May's Menu of Problems, May 2007




  
Hands of a Clock

Determine the smaller angle between the hands of a clock at 8:20, 12:20, and 1:30.

Solution
12:20, 110°; 8:20, 130°; 1:30, 135°.

The hour hand moves 360°/12 = 30° each hour, or 10° each 20 minutes and 5° each 10 minutes. The minute hand moves 30° every 5 minutes.

At 12:20, the angle between the hour and minute hands is 4(30°) - 10° = 110°, or 3(30°) + 20°.

At 8:20, the hour hand is 10° past the 8. The angle between the 4 and the 8 at 20 minutes past 8 is 4(30°) - 120°, so the angle between the hour and minute hands at 8:20 is 130°.

At 1:30, the angle between hands is 6(30°) - 45° = 135°, or 4(30°) + 15°.

Students should share other ways that they use to calculate these angles.

From January's Menu of Problems, January 2003 MTMS.




  
4-digit Even Numbers

How many four-digit numbers consist of only even digits?

Solution
500. The even digits are 0, 2, 4, 6, and 8. The first digit cannot be a 0, because the number would not be a four-digit number, so you may choose only 2, 4, 6, or 8 for the first digit. Five digits can be chosen for the remaining columns. Therefore, there are 4 × 5 × 5 × 5, or 500, ways to choose the digits. (Note that if the second digit were also 0, it would not be a three-digit number.)

From October's Menu of Problems, October 2006




  
Squares on a Board

How many squares are on a traditional 8 × 8 checkerboard? (By the way, the answer is not 64.)

Solution
204 squares. You have one 8 × 8 square, four 7 × 7 squares, nine 6 × 6 squares, sixteen 5 × 5 squares, twenty-five 4 × 4 squares, thirty-six 3 × 3 squares, forty-nine 2 × 2 squares, and sixty-four 1 × 1 squares, or 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 = 204. By constructing a table, you can see a pattern. Refer to print version to view table.

From Menu of Problems, September 2006




  
Remainder

What is the remainder when the product (1492)(1776)(1812)(1999) is divided by 5?

Solution
1. When considering the remainder when dividing by 5, we need to be concerned with only the units digit of each number. The product of the units digits is 2 × 6 × 2 × 9, or 216, which leaves a remainder of 1 when divided by 5.

From Menu of Problems, September 1999




  
Prime Factors

Find the difference between the least and greatest prime factors of 33,660.

Solution
15. The prime factorization of 33,660 is 2 × 2 × 3 × 3 × 5 × 11 × 17. The difference between 17 and 2 is 15.

From January's Menu of Problems, January 2007




  
How Many Minutes

How many minutes represent 10 percent of one full week?

Solution
1,008 minutes; 7 days × 24 hours × 60 minutes = 10,080 minutes in a week; 10 percent is 1,008 minutes.

From April's Menu of Problems, April 2007




  
Candy Corn

Each Halloween, 20 million pounds of candy corn are sold. If each piece weighs about 1 gram, how many pieces are bought?

Solution
Roughly 9 billion. Each pound of candy corn is 453 grams, so 20 million x 453 grams ≅ 9 billion pieces of candy corn.

From October's Menu of Problems, October 2002 MTMS.




  
What Number?

If a third is a fourth of it, what is it? Provide your answer in fractional form.

Solution
4/3, or 1 1/3. Translate the statement into a mathematical sentence. If x is it, then a third is a fourth of it would be 1/3 = 1/4(x). Multiplying both sides by 4 yields 4/3 = x. Some students might reason that if you have a fourth of something, then to get all of it you would multiply by 4. In this case, 4 × 1/3 = 4/3.

From August's Palette of Problems, August 2007




  
Divided by 5

What is the remainder when 1999 raised to the 2000 power is divided by 5?

Solution
1. The pattern of units digits for powers of 9 is 9, 1, 9, 1.... The same pattern holds for powers of 1999. The 1's occur when the power is even. Dividing a number that has a units digit of 1 by 5 results in a remainder of 1.

From Menu of Problems, September 2000




  
Between 2/7 and 5/12

Find the fraction between 2/7 and 5/12 that has the smallest denominator.

Solution
1/3. Because 2/7 and 5/12 are both less than 1/2, the next possibility for the smallest denominator is 3, and 1/3 is indeed between these two fractions.

From Menu of Problems, May 2000




  
What Is the Largest Number?

What is the largest number less than 10,000 that is divisible by both 36 and 60?

Solution
9,900.

The least common multiple of 36 and 60 is 180. Dividing 10,000 by 180 gives us 55+ (we do not need the decimal).

To check: 180 × 55 = 9,900.

From March's Menu of Problems, March 2005 MTMS.




  
Del Mar Pirates

The roster for the Del Mar Pirates baseball team has 13 players. Assuming that any player can be placed in any place in the batting order, how many different batting orders of 9 players can Coach Ellingson create?


Solution

259,459,200 different lineups. The coach has 13 choices for the first position followed by 12 for the second, 11 for the third, 10 for the fourth, and so on. Therefore, the solution can be found by multiplying: 13 × 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5.



From February's Menu of Problems, February 1999




  
Bicycle Ride


Solution


From February's Menu of Problems, February 1999




  
Making a Palindrome

What is the smallest positive number that can be added to 50607 to make a palindrome? A palindrome is a number that reads the same backward or forward.


Solution

408. One efficient strategy is to modify 50607 to make a palindrome then find the difference between the palindrome and 50607. Changing the 5 to a 7 creates 70607, which requires adding 20,000 to the original number. Changing the 0's to 1's then changing the last 7 to a 5 yields 51615 with a difference of 1008. An even smaller palindrome is made from 51615 by replacing the 6 with a 0, yielding 51015. The difference between 50607 and 51015 is 408.



From April's Menu of Problems, April 1999




  
Triads

Given these triads—(4, 11, 0), (2, 7, 6), (1, 2, 12), and (5, 8, 2)—fill in the blanks: (3, ____, 5) and (9, 7, ____).


Solution

7, −1. The three numbers in each triad sum to 15. Any other consistent solution is acceptable.



From March's Menu of Problems, March 1999




  
Bill and Marcus

Bill and Marcus wanted to make a rectangular box out of plywood. At the lumber yard, they realized that they had forgotten the dimensions of the box. Bill remembered that one of the sides had an area of 96 square centimeters. Marcus remembered that the other sides had areas of 252 and 168 square centimeters. What are the dimensions of the box?


Solution

8 cm by 12 cm by 21 cm. Look at common factors among the areas, and then guess and check to find possible combinations. An algebraic approach also works. Since the dimensions are a by b by c, then ab = 96, bc = 252, and ac = 168. Multiply the first two equations: ab2c = 96 × 252. Dividing both sides of the equation by ac gives the value of b2 and, therefore, b. Then a and c are easy to find.



From April's Menu of Problems, April 1999




  
2.1 Children

According to 1992 data from the United States, the mean number of children in a family was 2.1. Since no family can actually have 2.1 children, how can this figure be correct? Make up a set of ten families such that the mean number of children is 2.1.


Solution

Many answers are possible. One set of ten families is 1, 1, 1, 1, 1, 1, 1, 1, 1, and 12. Any set of ten whole numbers that sum to 21 will do.



From March's Menu of Problems, March 1999




  
Holding Kings and Queens

Caroline is holding four cards: two kings and two queens. Holt picks two of the cards. What is the probability that Holt will pick a king and a queen?


Solution

2/3. The six equally likely pairs that Holt could pick are K1K2, K1Q1, K1Q2, K2Q1, K2Q2, and Q1Q2. Of these pairs, four consist of a king and a queen.



From March's Menu of Problems, March 1999




  
Ms. Chan and Her Son

The combined ages of Ms. Chan and her son are 35. Ms. Chan is exactly 34 years older than her son. How old is Ms. Chan's son?


Solution

Ms. Chan's son is 6 months old; Ms. Chan's age is 34 years, 6 months.



From March's Menu of Problems, March 1999




  
Area Codes

When telephone area codes were first developed, the first digit could not be 0 or 1 and the second digit had to be 0 or 1. How many area codes are possible, given these constraints?


Solution

160. For the first digit, eight choices are available. For the second digit, only two choices are possible. For the last digit, ten choices are possible. The total number of possible area codes is 8 × 2 × 10 = 160. An organized list can help students see this pattern. Note that in 1995, the constraint on the second digit was removed, yielding 640 additional possible area codes.



From March's Menu of Problems, March 1999




  
Peach Trees

Four sisters inherited a square plot of land with twelve peach trees arranged as shown in the diagram. How can the plot of land be divided into four sections of equal area so that each section contains three trees and each resulting plot has only square corners?


Solution

One solution is shown. Others are possible.



From March's Menu of Problems, March 1999




  
Ms. Moretti's Books

Ms. Moretti likes to keep her four books of poetry on the left end of a shelf. In the middle of the shelf are five books of short stories, and she always keeps her three books about cats at the right end of the shelf. In how many different ways can Ms. Moretti arrange her books?


Solution

17,280 different arrangements. We can solve this problem by breaking it into smaller pieces. For the poetry books, four choices for the first book, three choices for the second, two choices for the third, and one choice for the fourth are possible. So the poetry books can be arranged in 4 × 3 × 2 × 1 = 24 ways. Using the same process, we see that the books of short stories can be arranged in 5 × 4 × 3 × 2 × 1 = 120 ways and that the books about cats can be arranged in 3 × 2 × 1 = 6 ways. For each of the 24 different arrangements of poetry books, 120 arrangements for the short stories and 6 arrangements for the books about cats are possible. Therefore the total is 24 × 120 × 6 = 17,280.



From March's Menu of Problems, March 1999




  
Leaky Faucet

A leaky faucet drips at the rate of 1 drip every 3 seconds, and 165 drips will fill one eight-ounce cup. How many gallons of water will drip in one year?


Solution

4,000 gallons. Set up a series of ratios to find the number of drips in a year: 365 days/year × 24 hours/day × 60 minutes/hour × 60 seconds/minute × 1 drip/3 seconds = 10,512,000 drips in 1 year. Another series of ratios will convert drips to gallons: 10,512,000 drips/year × 1 cup/165 drips × 1 gallon/16 cups = 3,981.8 gallons.



From April's Menu of Problems, April 1999




  
Augie v. The Train

Augie wants to catch a train that leaves at 10:45 A.M. and must ride his bike to the station, which is 6 miles away. The first 2 miles are uphill, the next 2 miles are flat, and the last 2 miles are downhill. Augie can ride at 2 miles per hour (MPH) going uphill, 3 MPH on level ground, and 4 MPH going downhill. He figures his average speed will be 3 MPH, so he plans to leave his house at 8:40 A.M. He thinks he will then have five minutes to lock his bike and get on the train. Will he catch the train? Why or why not?


Solution

No. It takes Augie 1 hour to ride the first section, 40 minutes to ride the middle section, and 30 minutes for the downhill for a total of 130 minutes, or 2 hours and 10 minutes. If Augie leaves at 8:40 A.M., he will not arrive until 10:50 A.M., 5 minutes after the train has left the station.



From April's Menu of Problems, April 1999




  
Shady Region

The radius of circle O is 7 cm. Angle AOC measures 110 degrees. What is the area of the shaded region to the nearest tenth?


Solution

47 square centimeters. The area of the circle is 153.9 square centimeters. The shaded region is 110/360 of the circle. Therefore, the area of the shaded region is 153.9 × 110/360, or 47, square centimeters.



From April's Menu of Problems, April 1999




  
Jelly Beans!

Kara bought a bag of jelly beans. As she was walking home, she noticed a hole in the bottom of the bag and that only a handful of jelly beans were left. She could not remember how many jelly beans she had started with, but she knew that it was fewer than 100. If she had counted by twos, threes, or fours, one would have been left over, but none would have been left over if she had counted by fives. How many jelly beans could Kara have started with?


Solution

25 or 85. We know that the number is odd and a multiple of 5, leaving 5, 15, 25, 35, 45, 55, 65, 75, 85, and 95. Of these numbers, only 5, 25, 45, 65, and 85 leave a remainder of 1 when divided by 4. Checking the numbers to see which one gives a remainder of 1 when divided by 3 leaves 25 and 85.



From April's Menu of Problems, April 1999




  
Two Clocks


Solution


From April's Menu of Problems, April 1999




  
Making Bows


Solution


From April's Menu of Problems, April 1999




  
Eating Cheese

Small pieces of cheese are placed over each of twelve numbers on the face of a circular clock. A mouse eats the cheese over the number 1 and walks clockwise around the clock, eating every other piece of cheese. What number will be under the last uneaten piece of cheese?

Solution
8. In the mouse's first trip around the clock, it eats the cheese on all the odd numbers. When it has eaten the cheese on number 11, the following numbers still have cheese: 12, 2, 4, 6, 8, 10. Remember, it must eat every other piece encountered. Therefore, after 11, it will eat 2, 6, and 10. The mouse is now on 10, and cheese remains on 12, 4, and 8. It will then eat the cheese on 4, leaving 8 and 12. It will skip 8 and eat 12's cheese, leaving 8 as the last number containing cheese.

From February's Menu of Problems, February 2007




  
How Many Birthdays?

A man was born on February 29, 1928. He died on January 29, 2004. How many actual birthdays did he get to celebrate during his life (not counting the day he was born as a birthday)? Hint: Don’t forget about leap years.

Solution
18. February 29 occurs every four years. 2004 is a leap year, so the man celebrated his February 29 birthday in these years: 1932, 1936, 1940, 1944, 1948, 1952, 1956, 1960, 1964, 1968, 1972, 1976, 1980, 1984, 1988, 1992, 1996, and 2000 but not 2004 because he died before his birthday.

From April's Menu of Problems, April 2007




  
How Many Hearts?

In a typical deck of 52 playing cards, the 4 of hearts card has 6 hearts on it, not 4: 4 hearts in the center of the card and 1 heart in both the upper-left corner and the lower-right corner. How many total hearts would you see when looking at the ace of hearts card along with the cards 2 of hearts through the 10 of hearts?

Solution
75 hearts. Use the following table:

The total is 75, or use Gauss’s formula for the sum of the first n numbers n(n + 1)/2. Use the formula when n = 12 to get 12(13)/2 = 78, then subtract the first two terms not in the table, 1 and 2.



From May's Menu of Problems, May 2007




  
Consumer Interview

A certain product is sold as either a liquid or a powder. Consumers were interviewed, and a survey revealed that—

1/5 do not use the product,
1/3 do not use the powder form,
427 use both the liquid and powder form, and
2/7 do not use the liquid form.

Solution
735 consumers. Divide the consumers into four sets:

A: They do not use the products.
B: They do not use the powder form.
C: They use both liquid and powder.
D: They do not use the liquid form.

We know that—

A + B = 1/3 of the total,
A + D = 2/7 of the total,
A = 1/5 is the total, and
C = 427.

Then, B is 1/3-1/5 = 2/15 of the total. D is 2/7-1/5 = 3/35 of the total. A + B + D is 44/105 of the total. C is the remaining 61/105 of the total. The number of consumers interviewed is 427/(61/105) = 735.



From May's Menu of Problems, May 2007




  
Super Ball

A Super Ball rebounds half the height from the height it is dropped. One day, a mathematics teacher drops a Super Ball from the roof of the school. If the roof is 32 feet off the ground, what is the total distance (up and down) that the ball will have traveled when it strikes the ground for the sixth time?

Solution
94 feet. After the initial drop of 32 feet (first strike on ground), the ball travels 32 feet (16 feet up and 16 feet down) between first and second strike, 16 feet (8 feet up and 8 feet down) between the second and third strike, 8 feet (4 feet up and 4 feet down) between the third and fourth strike, 4 feet (2 feet up and 2 feet down) between the fourth and fifth strike, 2 feet (1 foot up and 1 foot down) between the fifth and sixth strike: 32 + 32 + 16 + 8 + 4 + 2 = 94 feet.

From May's Menu of Problems, May 2007




  
Handshakes

There are 26 students in your class (including yourself). Every student shakes hands with exactly one-half of the students in the room. What is the minimum number of handshakes that occurred?

Solution
169. Each student shakes hands with 13 students; 26 × 13 = 338. Because each handshake is counted twice, you must divide 338 by 2 for the answer of 169. Examining a smaller problem helps. If there are four students, A, B, C, and D, then the handshakes are as follows. There are exactly 6 handshake combinations with 4 students: AB, AC, AD, BC, BD, and CD. Since each student only shakes hands with one-half the students in the class (2 students), one possible solution is AB, AD, BC, and CD. Another possible solution is AC, AD, BC, and BD. The resulting number pattern for other class sizes is the sequence of perfect squares as seen in the table where n is a counting number.



From April's Menu of Problems, April 2007




  
Spare Tire

Your uncle buys a new car that comes with five new tires, one for each wheel and one spare tire. He decides to be economical by using the spare tire as much as the other four tires. If he drives 68,000 miles, what will be the wear in miles on each tire?

Solution
54,400 miles. 68,000/5×4 = 54,400. He must use four tires at a time, and he must use five tires equally. He has five tires: tires A, B, C, D, and E. Divide his total miles into 5 driving intervals: 68,000/5 = 13,600.

Each tire is used for only 4 intervals; 4×13,600 = 54,400 miles.



From May's Menu of Problems, May 2007




  
Digging a Hole

If it takes two men two hours to dig a hole 3 meters long, 3 meters wide, and 3 meters deep, how long would it take the same two men to dig a hole 6 meters long, 6 meters wide, and 6 meters deep if they worked at the same rate?

Solution
16 hours. The first hole has a volume of 3 × 3 × 3 = 27 cubic units. The new hole has a volume of 6 × 6 × 6 = 216 cubic units; 216/27 shows that 8 smaller holes would fit into the larger hole. In other words, the new hole is twice as wide, twice as long, and twice as deep as the original, so it would take 2 × 2 × 2 = 8 times as long. Each of the smaller holes took 2 hours, so it will take 16 hours (2 × 8 = 16) to dig.

From April's Menu of Problems, April 2007




  
New Operation

Imagine that a new mathematical operation is being used. Its symbol is #. See the following equations.

1 # 1 = 2
3 # 5 = 34
6 # 9 = 117
10 # 14 = 296

Find the value of 15 # 19, and explain your reasoning.


Solution
586. a # b = a2 + b2. For example, 3 # 5 = 32 + 52 = 9 + 25 = 34. Therefore, 152 + 192 = 225 + 361 = 586.

From April's Menu of Problems, April 2007




  
Mouse Race

Two mice are racing around the edges of a square whose sides are 2 feet in length. They start at the same vertex (corner) and both go in a clockwise direction. One mouse travels at a constant rate of 1 foot per second, and the second mouse travels at a constant rate of 2 feet per second. After 22 seconds, how far apart will the mice be from each other?

Solution
2 feet apart. Label a square with vertices A, B, C, and D in clockwise order. Suppose the mice start at vertex A. After 22 seconds, the first mouse will be on vertex D, while the second mouse will be on vertex C. Therefore, they will be two feet apart.

From April's Menu of Problems, April 2007




  
Serious Cyclists

Serious bicyclists can learn the best way to shift their gears by making a table based on the number of teeth in the chainrings (by the pedals) and the number of teeth on the cassette (rear wheel). A gear is lower than another if one pedal revolution takes the bike a shorter distance than another. The distance a bike travels for a gear combination is calculated by multiplying the drive wheel diameter by 3.14, multiplying that number by the number of chainring teeth divided by the number of teeth on the rear cogs. A typical mountain bike has chainrings of 24, 36, and 46 teeth, and rear cassettes of 13, 15, 17, 20, 23, 26, and 30 teeth, driving a 26-inch wheel. Make a table for this bike. Given the table, the bicyclist then decides on a shifting pattern going from the lowest gear (shortest distance traveled in one revolution) to the highest gear (longest distance), omitting gear ratios that are very close. Find a shifting pattern for the typical mountain bike.


Solution

One possible shifting pattern from easiest to peddle to hardest to peddle, which is also slowest to fastest: (front, back) (24, 30) → (24, 26) → (24, 23) → (36, 30) → (36, 23) → (36, 20) → (36, 17) → (46, 20) → (46, 17) → (46, 15) → (46, 13). In reality, the lowest and highest ratios are usually not used, as going from the biggest chainring to the smallest cassette or vice versa is both not as efficient and hard on the bike. It is also best to minimize the number of times that you have to shift the front and the back at the same time. Optimum shifting patterns are a matter of great debate among bicyclists, and choosing gear ratios and shifting patterns are a big part of the strategy for professional racers.



From Menu of Problems, May 2002




  
100 in Cat Years

Fluffy was born on 24 June 1983. On her first birthday, Fluffy's age in cat terms was equivalent to that of a 15-year-old human. When she turned 2, her age in human years was 24. At the end of her third year, her human age was 28, and when she was 4, her human age was 32. She continued to age 4 years with each birthday. What year will Fluffy turn 100 in cat years?


Solution

On 24 June 2004, Fluffy can blow out 100 candles if she can stay awake long enough.



From Menu of Problems, March 2001




  
Lobster Years

A lobster's age in years is approximately his weight multiplied by 4, plus 3 years. Write this rule of thumb as a linear equation and determine the age of a 5-pound lobster. How much will a 15-year-old lobster weigh?


Solution

A = 4W + 3; 23 years; 3 pounds. The equation can be symbolized directly as the words are read, replacing "age" with A, "is" with =, and "weight" with W; 4(5) + 3 = 23, and solving 15 = 4W + 3 for W, we find that a 15-year-old lobster weighs approximately 3 pounds.



From Menu of Problems, May 2002




  
Counting Calories

According to the American Institute for Cancer Research, Americans eat 148 more calories per day now than they did 20 years ago. If Americans are not changing any other habits, and if it takes 3500 calories to gain a pound, how many pounds might a person gain in a year at this rate?

Solution
Over 15 pounds.

Since 148 cal./day . 365 days = 54,020 cal., 54,020 cal./3500 cal./lb. = 15.43 lbs.

From April's Menu of Problems, April 2003 MTMS.




  
Sharing Candy

Tommy has a 1.69 ounce bag of M&M's. The bag contains 56 M&M's. He decides to share his candy by giving Lucy two-thirds of three-fourths of his M&M's. Lucy decides to share her M&M's with Melissa by giving Melissa two-thirds of three-fourths of her M&M's. If half of Melissa's M&M's are red, how many red M&M's does she have?

Solution
7 red M&M's. To solve, 3/4 of 56 = 3/4 × 56 = 42; 2/3 of 42 = 2/3 × 42 = 28, or 2/3 × 3/4 × 56 = 28. Tommy gives Lucy 28 M&M's. To continue, 3/4 of 28 = 3/4 × 28 = 21; 2/3 of 21 = 2/3 × 21 = 14, or 2/3 × 3/4 × 21 = 14. Lucy gives Melissa 14 M&M's. If half are red, then Melissa has 7 red M&M's. Also, 2/3 × 3/4 = 6/12, or 1/2. Therefore, 2/3 of 3/4 is equivalent to 1/2.

From March's Menu of Problems, March 2007




  
Old Books

A mathematics teacher collects old books. One day, a student asked him how many old math books he has. He replied, “If I divide the books into two unequal whole numbers, then 64 times the difference between the two numbers equals the difference between the squares of the two numbers.” How many old math books does the teacher have?


Solution

64. Let a and b represent the two unequal whole numbers.
Then 64(a – b) = a2 – b2.
Factoring the right side yields 64(a – b) = (a – b)(a + b).
Since a and b are not equal, (a – b) does not equal zero.
Therefore, you can divide both sides by (a – b),
which yields 64 = a + b.
To solve the problem,
you do not have to know the values of a and b,
just the sum of a and b,
which is 64, the total number of books.



From Menu of Problems, August 2008




  
Sticker Jar II

Mrs. Cook has a jar of stickers in her classroom. She hands out stickers as a reward to students who do well on exams. The jar is getting low, and she will need to buy more stickers. Suppose there were only 5 star stickers and 4 smiley face stickers in Mrs. Cook’s jar. If Mrs. Cook wants to have at least 60 smiley face stickers in the jar, how many more of each sticker will she have to buy to keep the current ratio of smiley face stickers to star stickers?


Solution

56 smiley face stickers and 70 star stickers. One solution uses a proportion. An alternative strategy would be to use a table beginning with 5 star stickers and 4 smiley face stickers. Increase the quantity of star stickers by 5 and the smiley face stickers by 4 for each row until there are 60 smiley face stickers. This reveals 75 star stickers in the jar when the smiley face total reaches 60. Therefore, Mrs. Cook would need to buy 60 – 4 = 56 smiley face stickers and 75 – 5 = 70 star stickers.



From Pallette of Problems , January 2009




  
Sticker Jar

Mrs. Cook has a jar of stickers in her classroom. She hands out stickers as a reward to students who do well on exams. The jar is getting low, and she will need to buy more stickers. There are only 5 smiley face stickers and 3 star stickers left in the jar. If Mrs. Cook wants to have at least 50 smiley face stickers in the jar, how many more of each sticker will she have to buy to maintain the current ratio of smiley face stickers to star stickers?


Solution

45 smiley face stickers and 27 star stickers. We know that the ratio of smiley face stickers to star stickers is 5 to 3. By using a proportion, we find

Therefore, if the same ratio is used, there will be 30 star stickers in the jar after extras are purchased. There were three star stickers in the jar; therefore, 30 – 3 = 27. There were 5 smiley face stickers in the jar; therefore, 50 – 5 = 45. Mrs. Cook needs to buy 45 smiley face stickers and 27 star stickers.



From Pallette of Problems , January 2009




  
How Many?

How many two-digit numbers exist such that when the product of its digits is added to the sum of its digits, the result is equal to the original two-digit number?


Solution

Nine two-digit numbers: 19, 29, 39, 49, 59, 69, 79, 89, and 99. Any two-digit number can be represented by 10a + b. The product of the digits is ab, and the sum is a + b. When does ab + (a + b) = 10a + b? Solving, when ab = 9a, or when b = 9.



From Menu of Problems, August 2006




  
Cards in a Hat

Cards numbered 1 to 50 are put in a hat. What is the probability that the first two cards chosen at random have prime numbers on them?


Solution

3/35, or about .086. There are 15 prime numbers in the range of 1 to 50: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Remember that 1 is not a prime number. The probability of selecting a prime on the first pick is 15 out of 50; for the second pick, the probability is 14/49. The probabilities are combined by multiplying them, so 15/50 × 14/49 = 210/2450, or 3/35, or about .086.



From Pallette of Problems , November 2008




  
Classmates' Birthdays

Your mathematics teacher displays a monthly calendar, which highlights all the birthdays of your classmates. Your mathematics class has 25 students. What is the probability that 3 or more students were born in the same month?


Solution

1, or 100 percent. In a class of 24 students, it is possible for exactly 2 students to have been born in each of the 12 months. For 25 or more students, it is certain that there would be at least 1 month containing 3 or more birthdays. Therefore, the probability is 1.



From Pallette of Problems , November 2008




  
Curveballs

During a baseball game, a pitcher uses the following 4 throws: a fastball, a curveball, a changeup, and a slider. Of all his pitches, 4 out of every 10 are fastballs. He throws twice as many curveballs as sliders and more changeups than curveballs. If he uses each type of throw as least once over 10 pitches, how many of each pitch does he use for each 10 pitches thrown?


Solution

4 fastballs, 3 changeups, 2 curveballs, and 1 slider. The problem states that the pitcher throws 4 fastballs for every 10 pitches, which leaves 6 pitches. Since there must be a 2:1 ratio of curveballs to sliders, if 4 curveballs were thrown, there would be 2 sliders, which account for all 6 pitches and no changeups. Considering only 2 curveballs and 1 slider leaves 3 changeups, which supports the condition that there must be more changeups than curveballs.



From Pallette of Problems , December 2008




  
Food Trade

During lunch, students like to trade their food. The following exchange rates are used in the cafeteria: 3 candy bars can be traded for 2 pieces of pizza, and 1 piece of pizza can be traded for 4 lollipops. Using this exchange rate, how many lollipops would you accept for one candy bar? Explain your reasoning for this exchange.


Solution

3. Let c = candy bar, p = pizza, and l = lollipop. We know that 3c = 2p and 1p = 4l. We need to find x, where x × l = 1c. Begin with 1p = 4l. Multiplying both sides by 2 yields:

Substitute 8l for 2p in the first equation 3c = 2p, then 3c = 8l. By dividing both sides by 3 yields 1c = (8/3)c, we find that 1 candy bar is equivalent to 2 2/3 lollipops. Since 2 2/3 lollipops are unlikely, rounding up to the nearest whole number would be 3 lollipops.



From Pallette of Problems , January 2009




  
Squares and Triangles III

What is the figure number of the 20th square? What is its height?


Solution

Figure 25, with a height of 16 units. According to the pattern, the triangles are figures 1, 4, 8, 13, 19, 26, 34, . . . . Notice the differences between the terms. To get each next term in the list of figure numbers, we continue the pattern of differences. The figure numbers (second column) that are skipped in the table represent squares whose height is double the height of the last triangle. Specifically, the first two squares are of height 1 (figures 2 and 3), followed by 3 squares of height 2 (figures 5, 6, and 7), 4 squares of height 4 (figures 9−12), 5 squares of height 8 (figures 14−18), and 6 squares of height 16 (figures 20−25). This gives 2 + 3 + 4 + 5 + 6 = 20 squares, so the 20th square is figure 25, with a height of 16 units.



From Pallette of Problems , December 2008




  
Squares and Triangles II

What is the figure number of the 10th triangle? What is its height?


Solution

Figure 64, with a height of 9 units.



From Pallette of Problems , December 2008




  
Squares and Triangles I

What are the shape and height of figure 40?


Solution

A square of height 64 units. Notice the location of the triangles in the diagram and its continuation: 1, 4, 8, 13, 19, 26, 34, 43, 53, 64. . . . Figure 1 and figure 4 have heights of 1 unit; figure 8 has a height of 2 units; figure 13, 4 units; figure 19, 8 units; figure 26, 16 units; figure 34, 32 units; and figure 43, 64 units. Since Figure 40 is a square, it has a height of 64 units.



From Pallette of Problems , December 2008




  
What's my Number?

To find my number, take 1 fewer than the square of the smallest two-digit prime number and divide this result by the difference between the squares of the fourth and first prime numbers. What is my number? (Note: Do not include 1 as a prime number.)


Solution

8/3 or 2.666. . . . The smallest two-digit prime number is 11. Its square is 121, and 1 fewer is 120. The fourth and first prime numbers are 7 and 2, respectively. The difference between their squares, 49 and 4, respectively, is 45. When 120 is divided by 45, the result is 8/3.



From Menu of Problems, October 2008




  
Starting Salary

A woman got a job at a starting salary of $35,000 a year. If she received an 8 percent raise each year, how much would her salary be at the beginning of the tenth year? Round your answer to the nearest dollar.


Solution

$69,965.00. Students may think that the woman would make 80 percent more money at the end of ten years, but the interest is compounded. Getting an 8 percent raise means that each year's salary is 108 percent of the previous year's. At the beginning of the second year, the woman would make $37,800.00. For the third year, her salary would be $40,824.00. By continuing this pattern, at the beginning of the tenth year, the woman's salary would be $35,000.00 × (1.08)9, or $69,965.16.



From Menu of Problems, January 2000




  
Townhouses

A contractor is asked to build a new set of townhouses in attached clusters of different sizes. He created plans for one-, two-, and three-house clusters, as shown in the diagram below. The builder used computer software to draw the line segments used to represent the houses. How many line segments are needed to draw 4 houses? 10 houses? 47 houses? n houses? Explain your strategy.


Solution

26; 51; 236; 5n + 1. One house takes 6 segments. Each additional house only requires 5 additional segments. Therefore, the sequence for the number of segments is 6, 11, 16, 21, 26, 31, 36, . . . . If n is the number of houses, then the nth term for this sequence is 5n + 1, or 5 times the number of houses plus 1. The 5th house would require 5(5) + 1 = 26 segments; the 10th house would require 5(10) + 1 = 51 segments; and the 47th house would require 5(47) + 1 = 236 segments. Another method is to begin with 6 segments for the first house and notice that each additional house adds 5 segments to the total. Therefore, the total number of segments is 6 + 5(n – 1) = 5n + 1.



From Pallette of Problems , December 2008




  
Minimal Coinage

What is the least number of coins that you need to pay exactly for any amount up to but not including $1.


Solution

Ten coins (not using fifty-cent pieces): four pennies, two nickels, one dime, three quarters; nine coins if a fifty-cent piece replaces two of the quarters.



From Menu of Problems, March 2001




  
Perfect Square Cube

What is the smallest natural number that is both a perfect square and a perfect cube? What is the next smallest number?


Solution

1; 64. The number 1 is both a square and a cube of itself, and 64 is 82 and 43. One approach is to consider the perfect cubes: 1, 8, 27, 64, and so on. After 1, the number 64 is the first cube that is also a perfect square.



From Menu of Problems, March 2003




  
Funky Fractions

The repeating decimal 0.656565…can be represented by the fully reduced fraction a/b, where a and b are positive integers. Find the sum of a and b if you form this fraction:


Solution

164.

Let x = 0.656565…,
and 100x=65.656565...

Then:

100x = 65.6565…
x = 0.6565…

Subtracting the second equation
from the first equation yields

99x = 65.
Solving for x gives x = 65/99. Since
65 and 99 have no common factors
other than 1, the fraction 65/99 is in
simplest form, and 65 + 99 = 164.



From Menu of Problems, September 2008




  
Life of a Lightbulb

A 60-watt lightbulb is used for 95 hours before it burns out. What is the life of the lightbulb in kilowatt-hours?


Solution

5.7 kilowatt-hours; 60 watts × 95 hours reveals the number of watt hours used, which is 5700 watt hours. To convert watts to kilowatts, multiply by 1/1000 since a watt is 1/1000 of a kilowatt. Therefore, 5700 watt hours yields 5700 × 1/1000 = 5.7 kilowatt hours.



From Menu of Problems, August 2008




  
Guessing the Greatest

If n represents an integer, which of the following expressions yields the greatest value?


Solution

If n ≤ 0, then 2 – n yields the greatest value. If n = 1, then 2n and 2/n both yield the greatest (same) value. If n > 2, then 2n yields the greatest value. Making a table helps to determine the greatest value in each case. (See table.)



From Menu of Problems, August 2008




  
Factor Fun

How many different prime factors are in the set of numbers between 7 and 37 that are multiples of 5?


Solution

Four prime factors: 2, 3, 5, and 7.
Let x represent the product of the
multiples of 5 that are between 7 and 37.
Then x = (10)(15)(20)(25)(30)(35)
= (2)(5) • (3)(5) • (4)(5) • (5)(5) • (6)(5) • (7)(5),
so x has these different prime factors: 2, 3, 5, and 7.
(Note that 4 = (2)(2) and 6 = (2)(3),
so these numbers do not provide any additional prime factors.)



From Menu of Problems, September 2008




  
Ratios

The Fibonacci sequence begins with 1, 1, 2, 3, 5, 8,... (starting with two 1's and adding the previous two numbers to get the next number). Find the values of ratios of consecutive terms 1/1, 2/1, 3/2, 5/3, 8/5,..., 144/89 to the nearest thousandth. What did you discover? Here is an interesting property about this ratio: Square it and subtract 1. What do you get?


Solution

The ratios eventually approach the golden ratio of approximately 1.61803.... If you square the golden ratio and subtract 1, you get the golden ratio.



From Menu of Problems, August 2006




  
Cake Time!

Mom baked a cake for the entire family. Dad ate 1/6 of the cake. Brother ate 1/5 of what was left. Sister ate 1/4 of what was left after that. Spot the dog ate 1/3 of what was left after that. Baby sister ate 1/2 of what was left after that. How much of the original cake was left for Mom to eat? Who ate the most cake?


Solution

1/6 was left for Mom. Everybody eats the same amount of cake.

(1/6)(1) = 1/6 eaten; 1 – 1/6 = 5/6 of the cake that is left.
(1/5)(5/6) = 5/30 = 1/6; 5/6 – 1/6 = 4/6
(1/4)(4/6) = 4/24 = 1/6; 4/6 – 1/6 = 3/6
(1/3)(3/6) = 3/18 = 1/6; 3/6 – 1/6 = 2/6
(1/2)(2/6) = 2/12 = 1/6; 2/6 – 1/6 = 1/6



From Menu of Problems, August 2006




  
Free Throws

If the probability of making a free throw in basketball is 40 percent, what is the most likely number of points you will make in a "bonus" free throw? (With a bonus free throw, you shoot one free throw and get one additional free throw if you make the first one.) What if the probability of making a free throw is 80 percent?


Solution

For a 40 percent free-throw shooter, 0 points are most likely. For an 80 percent shooter, 2 points are likely. If the probability of making a free throw is 40 percent, the probability of missing is 60 percent.

P(0 points) = .60
P(1 point) = .40 × .60 = .24
P(2 points) = .40 × .40 = .16
If the probability of making a free throw is 80 percent, the probability of missing is 20 percent.
P(0 points) = .20
P(1 point) = .80 × .20 = .16
P(2 points) = .80 × .80 = .64



From Menu of Problems, August 2006




  
Cubic Yards and Feet

What fraction of 2 cubic yards is 4 cubic feet?


Solution

2/27, or approximately 0.074. Since there are 3 feet in a yard, there are 3 × 3 × 3 = 27 cubic feet in 1 cubic yard. Therefore, there are 54 cubic feet in 2 cubic yards; 4 cubic feet/54 cubic feet = 2/27.



From Menu of Problems, August 2006




  
Profit?

A person buys a new MP3 player for $150 on the Internet, sells it on the Internet for $175, buys it back for $195, and sells it again for $230. After all these transactions, has the person made a profit? If so, how much?


Solution

The person made a profit of $60. The person paid $345 total ($150 + $195) and received $405 ($175 + $230). Therefore, the person made a profit of $405 – $345, or $60.



From Menu of Problems, May 2006




  
Martian Feet in a Mile

A fictitious news story relates how the Mars probe sent from Earth has disturbed a colony of Martians. They have come to Earth and have taken over our planet. The supreme ruler of Mars has decided to change the unit of length used in the United States from 12 inches in a foot to 10 inches in a foot because the supreme ruler's foot is only 10 inches long. The Martians are allowing the United States to keep the mile the exact same length so that no new road signs and mile markers have to be created. Now that the Martians have taken over our planet, how many feet are in one mile?


Solution

6,336 feet. Before the ruler was changed, there were 12 inches in 1 foot and 5,280 feet in 1 mile, which meant that there were 63,360 inches in 1 mile. There are still 63,360 inches in 1 mile because the lengths of the mile and the inch were not changed. Since there are now 10 inches in 1 foot and 63,360 inches in 1 mile, divide 63,360 by 10 to get 6,336 feet in 1 mile.



From Menu of Problems, May 2006




  
The Three Greatest...

Find the three greatest whole numbers less than 200 such that each number has an odd number of factors.


Solution

144; 169; and 196. Only perfect squares have an odd number of factors. Therefore, 12², 13², and 14² would be the numbers sought since 15² = 225, which is greater than 200.



From Menu of Problems, May 2006




  
Finding a Sum

Find the sum of all two-digit multiples of 6 whose digits sum to 6. For example, 24 is a multiple of 6, and the digits 2 and 4 sum to 6.


Solution

126. The two-digit multiples of 6 are 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96. The numbers 24, 42, and 60 have digits that sum to 6; 24 + 42 + 60 = 126.



From Menu of Problems, May 2006




  
Equivalent to 99

Given the digits 987654321 in descending order, place addition symbols (+) between any of the digits to produce an expression that is equivalent to 99. You may use as many addition symbols as you wish, but you may not use any other mathematical symbols or alter the order of the digits. One such example is 9 + 8 + 7 + 6 + 5 + 43 + 21 = 99. Find another possibility.


Solution

9 + 8 + 7 + 65 + 4 + 3 + 2 + 1 = 99. Can you find another solution?



From Menu of Problems, August 2006




  
Positive Integers

Determine the values for A, B, C, and D, which are all positive integers, given the following information:

A × B = 24
A + B = 14
C × D = 48
A × D = 192
B × C = 6


Solution

(A, B, C, D) = (12, 2, 3, 16). Since all values are positive integers, A and B can be 1 and 24, 2 and 12, 3 and 8, or 4 and 6 to satisfy A × B = 24. Since A + B = 14, A and B have to be 2 and 12. Since B × C = 6, B equals 2, and C equals 3. Since C × D = 48 and C = 3, then D equals 16. Finally, since B = 2, A equals 12.



From Menu of Problems, September 2006