Los Primeros MATHCOUNTS 2002–2003 - users.vcnet.comusers.vcnet.com/simonp/mathcounts/wu14slides.pdf · Los Primeros MATHCOUNTS 2002–2003 Warm-Up 14 1. Problem 1 A satellite completes

Los Primeros MATHCOUNTS 2002–2003Warm-Up 14

1

Problem 1

A satellite completes a 360◦-revolution around the planet Earth every 90minutes. The satellite travels in a path parallel to, and in the same direction as,

the path your house is on while Earth completes her 360◦ 24-hour rotation. If a

24-hour interval begins with the first pass of the satellite over your house, how

many total times will the satellite pass over your house before the start of the next

24-hour interval?

Method 1: The satellite requires 1.5 hours to make one revolution about the

earth, so it makes 24/1.5 = 16 revolutions in 24 hours. Of course, the house

makes 1 revolution in 24 hours. The number of times that the satellite passes the

house is therefore 16 − 1 = 15 .

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Problem 1, Continued

Method 2:

�

The Earth

�

x

1st Overhead Pass

2cd Overhead Pass

Satellite Path in Red

House Path in Blue

Both the house and the satellite are revolving about the center of the earth,

though the satellite’s angular velocity is much greater. By the time the satellite has

completed one revolution of the earth, the house will have rotated a small angle

away from its starting position. So the satellite will have to make a total angle of

x + 360◦ between the two alignment conditions, while the house will rotate only

x degrees.

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Problem 1, Continued

Let’s find the number of hours T between consecutive passes of the satellite over

the house. The rotational speed of the satellite is 360◦1.5 hr = 240 ◦/hr and the

rotational speed of the house is 360◦24 hr = 15 ◦/hr. We can write the angular

velocity equations for the house and satellite as

Angular Velocity × Time = Angle Traveled

Satellite: 240T = 360 + x

House: 15T = x

If we subtract 360◦ from both sides of the first equation, we can solve for T :

240T − 360 = 15T =⇒ 240T − 15T = 360 =⇒ T = 1.6 hr.

The number of overhead passes in 24 hours is 24/1.6 = 15 .

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Problem 2

What is the sum of the three positive integers a, b and c that satisfy

a + 1b+ 1

c

= 7.5?

If we rewrite this equation slightly as1

b + 1c

= 7.5 − a we see that

0 <1

b + 1c

< 1,

since a ≥ 1 and b ≥ 1. But this means that

0 < 7.5 − a < 1 =⇒ a = 7.

If we put this value for a into the original equation we obtain

1b + 1

c

=12

=⇒ b +1c

= 2 =⇒ 1c

= 2 − b

so that 1/c must be a positive integer implying c = 1 and therefore b = 1. The

answer to the question is then 7 + 1 + 1 = 9 .

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Problem 3

How many positive integers less than 1000 are multiples of 7 and have a units

digit of 8?

The beginning and ending multiples of 7 are

7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, . . . ,

. . . , 931, 938, 945, 952, 959, 966, 973, 980, 987, 994.

Note that the numbers with a units digit of 8 repeat after every 10 items in the

above list. The first is 28 = 4 × 7 and the last is 938 = 134 × 7. The number

of such items in the list is then

134 − 410

+ 1 = 13 + 1 = 14 .

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Problem 4

The U.S. Mint began producing quarters for the 50 state quarters program at the

beginning of 1999. After making the quarters for a given state for approximately

ten weeks, the Mint begins making quarters honoring a new state. In what year

will the U.S. Mint finish making the quarters honoring the 50th state?

It takes approximately 50 × 10 = 500 weeks to make quarters for all the states.

The number of years is 500/52 ≈ 9.6 After nine years it will be the beginning of

1999 + 9 = 2008. After an additional 0.6 years it will still be 2008 .

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Problem 5

Each edge of a cube is decreased by 40%. What is the percent of decrease in

the volume of the cube? Express your answer to the nearest tenth.

Suppose that the original cube had sides of length 1 unit. Its volume was then

13 = 1 cubic unit. If we reduce each side by 40%, then the new side length is

0.6 and the new volume is 0.63 = 0.216 cubic units. The decrease in volume is

1 − 0.216 = 0.784 or 78.4% .

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Problem 6

What is the remainder when 10! is divided by 27?

Recall that

10! = 10 × 9 × 8 × · · · × 1.

Factors of 10! that contain 2 as a factor are 10 = 2 × 5, 8 = 23, 6 = 2 × 3,

4 = 22 and 2. Therefore, 10! contains 21+3+1+2+1 = 29 as a factor. It

therefore also contains 27 as a factor so that the remainder of dividing 10! by 27

is 0 .

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Problem 7

Two numbers are selected simultaneously and at random from the set

{1, 2, 3, 4, 5, 6, 7}. What is the probability that the positive difference

between the two numbers is 2 or greater?

The possible choices having a positive difference of 2 or greater are listed in the

following table:

(1, 7) (1, 6) (1, 5) (1, 4) (1, 3)

(2, 7) (2, 6) (2, 5) (2, 4)

(3, 7) (3, 6) (3, 5)

(4, 7) (4, 6)

(5, 7)

There are T5 = 5×62 = 15 such solutions. The total number of possible

combinations is 7C2 = 7!/(2! 5!) = 7 × 6/2 = 21 so that the desired

probability is 1521 = 5

7 .

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Problem 8

What is the positive integer n such that n + 2, 4n and 5n − 2 are the side

lengths of a right triangle and 5n − 2 > 4n > n + 2?

Algebra Method: Since 5n − 2 is the greatest length, it must correspond to the

hypotenuse. Therefore, by the Pythagorean Theorem

(5n − 2)2 = (4n)2 + (n + 2)2

25n2 − 20n + 4 = 16n2 + n2 + 4n + 4

25n2 − 20n + 4 = 17n2 + 4n + 4

25n2 − 17n2 − 20n − 4n + 4 − 4 = 0

8n2 − 24n = 0

n(8n − 24) = 0

so either n = 0 (unacceptable) or 8n − 24 = 0 =⇒ n = 3 . Alternatively,

you can use guess and check in the first equation.

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Problem 9

The game of chess is played on an eight by eight grid of squares. In one move,

the king may be moved to any of the squares which adjoin the square it currently

occupies, either along an edge or at a corner. If the king starts in a corner square,

how many different squares could it occupy after exactly four moves?

We illustrate the possible locations after 0, 1, 2, 3, and 4 moves:

0 11 1

222 2

222 2 2

333 3

333 3 3

3333 3 3 3

444 4

444 4 4

4444 4 4 4

44444 4 4 4 4

There are 25 squares that could be occupied by the king after four moves.

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Problem 10

Bob and Meena play a two-person game which is won by the first person to

accumulate 10 points. At each turn Bob gains a point with probability of 13 . If he

doesn’t get a point, then Meena gets a point. Meena is now ahead 9 to 8. What is

the probability that Meena will win? Express your answer as a common fraction.

Method 1: The only way that Bob can win is by winning the next two points. The

probability of this happening is (we assume turns are independent) 13 × 1

3 = 19 .

Therefore, the probability that Meena wins is 1 − 19 =

89

.

Method 2: Meena needs only one point to win and can therefore win in two

different ways. The first way is if she wins the next turn. This probability is 23 . The

second way is if Bob wins the next turn and Meena wins the turn after that. The

probability of this outcome is 13 × 2

3 = 29 . The probability that Meena wins by

either of these two ways is 23 + 2

9 =89

.

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