Download Solutions - Missouri State University

MO-ARML – 2/24/07
Practice Questions
While using calculators to explore solutions during practice is welcome,
justifications for your answers are expected.
Please be mindful that calculators are not allowed during
the Individual and Relay events at the national tournament.
1.
How many ordered pairs (x, y) of positive integers satisfy
1 1
1
+ =
?
x y 2007
Answer:
1 1
1
+ =
x y 2007
xy – 2007x – 2007y = 0
xy – 2007x – 2007y – 20072 = 20072 = (32⋅223)2, where 223 is a prime.
(x – 2007)(y – 2007) = 342232
Since 342232 has (5)(3) = 15 factors, there are 15 ways to factor it as a product
(x – 2007)(y – 2007). (Excel)
2.
How many ordered triples (x, y, z) of positive integers satisfy xyz = 4000?
Answer.
xyz = 4000 = 2553
x = 2a5α, y = 2b5β, z = 2c5γ, 0 ≤ a, b, c, α, β, γ such that a + b + c = 5, α + β + γ = 3
This will involve (1) choosing five times from a, b, c with repetition permissible and
(2) choosing three times from α, β, γ with repetition permissible.
 3 + 5 − 1  7 
 3 + 3 − 1  5 
In (1), there are 
 =   = 21 ways. In (2), there are 
 =   = 10
 5  5
 3   3
ways. Together, there are (21)(10) = 210 ways to factor 4000 as a product of three
 n + k − 1
positive integers x, y, and z. [In general, there are 
 ways to choose k
k


objects out of n objects if repetition is permissible.]
3.
What is the number of different 7-digit numbers that can be made by rearranging the digits
of 3053354?
Answer.
Since 0 can’t be the first digit, the number of possible 7-digits numbers is equal to
(# of possible permutations of 3053354) − (# of possible permutations of 353354)
7!
6!
6!(7 − 1) 720
=
−
=
=
= 360 .
3!2! 3!2!
3!2!
2
4.
A bowl contains 200 colored balls: 52 green, 40 red, 36 blue, 32 yellow, 24 black, and 16
white. If a person is blindfolded, what is the smallest number of balls this person must
pick to guarantee that at least 30 balls of the same color are selected?
Answer.
It is possible to have 29 green, 29 red, 29 blue, 29 yellow, 24 black and 16 white
without having 30 balls of the same color. Therefore, this person must select at least
(29)(4) + 24 + 16 + 1 = 157 balls to guarantee at least 30 of the same color.
5.
A circular arc centered at one vertex of a square of side
length 4 passes through two other vertices. A small
circle is tangent to the large circle and two sides of the
square. What is the radius of the small circle?
Answer.
Let the radius of the small circle be r.
Note that AB = 4 2 = AC + CM + MB =
r=
(
(
)( 2 − 1) = 12 − 8
)( 2 − 1)
4 2−4
4 2−4
=
2 +1
2 +1
A
C
2r + r + 4
r
M
2
B
6.
What is the largest number of pieces into which a circular pizza can be cut with 9 straight
cuts?
Answer.
The maximum number of pieces that can be added by a cut is 1 greater than the
number of lines that the new cut intersects. Hence, the maximum number of pieces
with 9 cuts is 1 + (0 + 1) + (1 + 1) + (2 + 1) + (3 + 1) + ⋅⋅⋅⋅ + (8 + 1) = 1 + 2 + 3 + ⋅⋅⋅⋅ + 9
= 45. (Sketchpad) (Word)
7.
Consider the sum of 3(2007!), 3(2007!) – 1, and 3(2007!) – 2. Add the digit of the sum
and then add the digit of the result. Repeat until the result is a single digit. What is the
final result?
Answer.
3(2007!) + 3(2007!) – 1 + 3(2007!) – 2 = 9(2007!) – 3 . Since 9(2007!) is a multiple of
9, adding the digits of 9(2007!) will produce a multiple of 9. When this process is
repeated until a single digit is left, the result is always 9. When 3 is subtracted from
9(2007!), the final result is always 6.
8.
Suppose Team A typically wins 3 out of every 5 games it plays again Team B. At
season’s end, the teams are tied for the league lead. Team B’s manager can choose
between a one-game playoff or a best-of-three series. Which choice would give Team B
the better opportunity to win the league title?
Answer.
Team B’s probability to win a one-game playoff is 0.4. It has three ways to win a
best-of-three series: WW (p = 0.16), WLW (p = 0.4 × 0.6 × 0.4 = 0.096), LWW (p =
0.096). Team’s probability of winning a best-of-three series is therefore 0.352. The
manager should choose a one-game playoff.
9.
The base of a regular square pyramid is inscribed in the base of a cylinder. The height of
the cylinder is triple the height of the pyramid. Find the ratio of the volume of the
pyramid to the volume of the cylinder.
Answer.
If the radius of the cylinder’s circular base is r, then the side length of the pyramid’s
Vp
 r 
square base is 2 
 = 2r . The ratio is V =
 2
c
h
( 2r )
2
(3h)πr
3
2
=
2
.
9π
10.
Given a, b, and c are integers such that 0 < a < b, and given that the polynomial p(x) =
x(x – a)(x – b) – 89 is divisible by x – c¸ find the value of a + b + c.
Answer.
0 = p(c) = c(c – a)(c – b) – 89. That is, c(c – a)(c – b) = 89, implying c = 1 or 89. (Why
can’t c be –1 or –89?) If c = 89, then c – a = ±1 = c – b, an impossibility because
a < b. Therefore, c = 1, c – b = –89 , c – a = –1. Or equivalently, a = 2, b = 90, and
a + b + c = 93.
11.
Which one of the following numbers can not be expressed as the difference of the squares
of two integers: 314159265, 314159266, 314159267, 314159268, 314159269
Answer.
If k = m2 – n2 for some integers m and n, then k = (m – n)(m + n). Using m – n = 1,
k+1
k− 1
then m + n = k. Solving for m and n, we obtain m =
, which is
,n=
2
2
always possible if k is odd. Therefore, each of 314159265, 314159267, and
314159269 is the difference of two squares. On the other hand, modulo 4, the
squares are congruent to 0 or 1, and k = m2 – n2 must be congruent to 0, 1, or 3.
However, 314159266 ≡ 2 (mod 4), so it can’t the difference of two squares. (How do
we write 314159268 as the difference of two squares?) (Excel)
12.
If i =
Answer.
−1 , find the real numbers a and b such that (1 + i)800 = a + bi.
(
)
(1 + i)800 =  2 cos45o + isin45o 


Therefore, a = 2400, b = 0
800
(Word)
(
)
= 2400 cos(45 ⋅ 800)o + isin(45 ⋅ 800)o = 2400
13.
What is the simplified value k for which the larger root of x2 + 4x + k = 0 is
2+ 3 ⋅ 2+ 2+ 3 ⋅ 2+ 2+ 2+ 3 ⋅ 2− 2+ 2+ 3 ?
Answer.
2+ 3 ⋅ 2+ 2+ 3 ⋅ 2+ 2+ 2+ 3 ⋅ 2− 2+ 2+ 3
(
2−(
=
2+ 3 ⋅ 2+ 2+ 3 ⋅ 4− 2+ 2+ 3
=
2+ 3 ⋅ 2+ 2+ 3 ⋅
=
2+ 3 ⋅ 4− 2+ 3
(
2+ 3
)
)
)
= 2+ 3 ⋅ 2− 3 = 4-3 = 1
If x2 + 4x + k = 0 = (x – 1)(x – r), then r = k and 1 + r = –4. Therefore, k = r = –5.
14.
Suppose the volume of a ball is equal to
π
times the volume of a cube. What is the ratio
6
of the surface area of the ball to that of the cube?
Answer.
4
π
Let r be the radius of the ball and s the side of the cube. Then, πr 3 = s 3
3
6
2
4π π
s
⇒ r = . Thus, the ratio of the areas is 4πr
= .
2 =
6s
2
6⋅4 6
The polynomial P(x) = x3 + ax2 + bx + c has the property that the average of its roots, the
products of its roots, and the sum of its coefficients are all equal. If the y-intercept of the
graph y = P(x) is 2, what is b?
Answer.
The assumptions imply that –a/3 = –c = 1 + a + b + c, and c = 2. Therefore, a = 6
and b = –11. [Note: This polynomial doesn’t have integer roots.]
15.
16.
Find the two factors of 248 – 1 that are between 60 and 70.
Answer.
248 – 1 = (224 + 1)(224 – 1) = (224 + 1) (212 + 1)(212 – 1)
= (224 + 1)(212 + 1)(26 + 1)(26 – 1) = (224 + 1) (212 + 1)⋅65⋅63
The two factors between 60 and 70 are 63 and 65.
[Note: 248 – 1 = (3)2(5)(7)(13)(17)(97)(241)(257)(673)]
17.
What is the greatest positive integer n such that 3n is a factor of 2007!?
Answer.
 2007   2007   2007   2007   2007   2007 
 3  +  9  +  27  +  81  +  243  +  729 
 223   223   223   223 
+
+
+
= 669 + 223 + 
 3   9   27   81 
= 669 + 223 + 74 + 24 + 8 + 2 = 1000. (Word)
18.
Let xn denote the remainder when an integer x is divided by n. For positive integer x,
5
6
compute the sum of all elements in the solution set of x5 ( x5 ) − x 6 − ( x5 ) + x ( x5 ) = 0 .
Answer.
( ) (
0 = x 5 ( x 5 ) − x 6 − ( x5 ) + x x 5 = x 5 − x5
5
6
) (( x )
5
5
−x
)
⇒ x 5 = x5 or ( x5 ) = x .
5
Since x5 = 0, 1, 2, 3, 4, there is only one solution to the equation x 5 = x5 , namely,
x = 1. There are, however, four solutions to the equation ( x5 ) = x , namely, x = 15,
5
25 , 35, 45. The sum of these solutions is 1 + 32 + 243 + 1024 = 1300.