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Los Primeros MATHCOUNTS 2002–2003
Warm-Up 14
1
Problem 1
A satellite completes a 360◦ -revolution around the planet Earth every 90
minutes. 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 .
2
Problem 1, Continued
Method 2:
ass
P
d
ea
2cd
x
Satellite Path in Red
House Path in Blue
erh
Ov
1st Overhead Pass
The Earth
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.
3
Problem 1, Continued
Let’s find the number of hours T between consecutive passes of the satellite over
360◦
the house. The rotational speed of the satellite is 1.5
hr
360◦
rotational speed of the house is 24 hr
= 240 ◦ /hr and the
= 15 ◦ /hr. We can write the angular
velocity equations for the house and satellite as
Angular Velocity × Time
Satellite:
= Angle Traveled
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
4
= 15 .
Problem 2
What is the sum of the three positive integers a, b and c that satisfy
a+
1
b+ 1c
= 7.5?
1
If we rewrite this equation slightly as
b+
1
c
1
0<
b+
1
c
since a
= 7.5 − a we see that
< 1,
≥ 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
1
b+
1
c
1
1
1
=⇒ b + = 2 =⇒
=2−b
=
2
c
c
so that 1/c must be a positive integer implying c
answer to the question is then 7 + 1 + 1
5
= 9.
= 1 and therefore b = 1. The
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 − 4
+ 1 = 13 + 1 = 14 .
10
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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 2 7 ?
Recall that
10! = 10 × 9 × 8 × · · · × 1.
= 2 × 5, 8 = 2 3 , 6 = 2 × 3,
4 = 22 and 2. Therefore, 10! contains 21+3+1+2+1 = 29 as a factor. It
therefore also contains 2 7 as a factor so that the remainder of dividing 10! by 2 7
is 0 .
Factors of 10! that contain 2 as a factor are 10
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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
=
5×6
2
= 15 such solutions. The total number of possible
combinations is 7 C2 = 7!/(2! 5!) = 7 × 6/2 = 21 so that the desired
probability is 15
21
=
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
There are
2 2 2
2 2 2
2 2 2
1 1
1
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
4
25 squares that could be occupied by the king after four moves.
12
4
4
4
4
4
4
4
4
4
4
4
4
4
4
4
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
Therefore, the probability that Meena wins is 1 −
1
9
8
=
.
9
×
1
3
= 19 .
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
either of these two ways is 23
+
×
2
9
2
3
= 29 . The probability that Meena wins by
8
=
.
9
13