Download Ithaca College Math Day Competition April 18, 2007 Solutions Part I

Ithaca College Math Day Competition
April 18, 2007
Solutions
Part I
1. What is the value of i1 + i2 + i3 + · · · + i2007 ? (Recall that i =
√
−1.)
−1
Note that i1 + i2 + i3 + i4 = i − 1 − i + 1 = 0, and in fact, for each k = 0, 1, 2, . . . , 500, we have
i4k+1 + i4k+2 + i4k+3 + i4k+4 = i − 1 − i + 1 = 0. So,
2007
X
m=1
im
= i1 + i2 + i3 + i4 + i5 + i6 + i7 + i8 + · · · + i2001 + i2002 + i2003 + i2004 +i2005 + i2006 + i2007
|
|
{z
} |
{z
}
{z
}
0
=
0
0
0 + 0 + 0 + · · · + 0 +i − 1 − i
|
{z
}
501 zeros
= −1.
2. Write the equation of the line (in form y = mx+b) containing the points of intersection of the parabolas
y = 2x2 + x + 3
and
y = x2 − x + 18.
y = −3x + 33
The intersections are found by solving 2x2 + x + 3 = x2 − x + 18 for x.
2x2 + x + 3 = x2 − x + 18 ⇐⇒ x2 + 2x − 15 = 0 ⇐⇒ (x + 5)(x − 3) = 0 ⇐⇒ x = −5, 3.
Plugging into either parabloa’s equation, we find that the points of intersection are (−5, 48) and (3, 24).
24 − 48
The slope of the line connecting these points is m =
= −3. Since (3, 24) lies on this line, we
3 − (−5)
find 24 = −3 · 3 + b =⇒ b = 33.
3. The centers of adjacent faces of a unit cube are joined to form an octahedron. What is the volume of
the octahedron?
1
6
The octahedron can
√ be seen as two pyramids, with square bases, joined together. The square pyramid
has base length 1/ 2 and height 1/2. The volume of a square pyramid is (1/3)(side length)2 (height).
Thus,
√ 2
1
volume of octahedron = 2
(1/ 2) (1/2) = 1/6.
3
4. For how many real numbers x is
q
144 −
√
x an integer?
13
Let k represent an integer. Then, for 0 ≤ x ≤ 1442 ,
q
√
144 − x = k ⇐⇒ x = (144 − k 2 )2 .
Then there are 13 distinct values of x, corresponding to k = 0, 1, 2, . . . , 12.
5. How many of the three digit numbers that can be formed from all of the digits 3, 5, and 7 (used only
once each) are prime?
0
Any 3-digit number formed using each of 3, 5, and 7 exactly once is always divisible by 3.
6. Two boats on the opposite shores of a river start moving towards each other. When they pass each
other they are 500 meters from one shoreline. They each continue to the opposite shore, immediately
turn around and start back. When they meet again they are 150 meters from the other shoreline. Each
boat maintains a constant speed throughout. How wide is the river?
1350 meters
Let w be the width of the river and let x and y be the distances covered by the boats when they first
pass each other. Thus, w = x + y. When they first meet, we know that one of the boats has traveled
500 meters; so, we may assume that x = 500 meters.
By the time of their second passing, each boat has crossed the river once and turned around to meet.
Thus, the sum of the total distance traveled by the boats at the second meeting is three times the
width of the river. Since the boats travel at a constant rate, and they have gone three times as far
as when they first passed, it follows that one of them has traveled a distance of 3x and the other has
traveled a distance of 3y.
When they passed the second time, 150 yards from the other shore, we see that the boat which traveled
500 meters at the first passing has traveled w +150 meters at the second passing. That is 3x = w +150.
This yields three equations
w = x+y
x = 500
3x = w + 150
yielding solution: w = 3 · 500 − 150 = 1350 meters.
7. Call a number “prime-looking” if it is composite but not divisible by 2, 3, or 5. The three smallest
prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many
prime-looking numbers are there less than 1000? (Recall that a composite number is any integer greater
than 1 and not prime.)
100
Of the numbers less than 1000, there are 499 that are divisible by 2, 333 divisible by 3, and 199 divisible
by 5. There are 166 numbers divisible by 6, 99 divisible by 10 and 66 divisible by 15. Finally, there
are 33 numbers divisible by 30. Thus, by the Inclusion-Exclusion principle, there
499 + 333 + 199 − 166 − 99 − 66 + 33 = 733
numbers that are divisible by at least one of 2, 3 or 5. This leaves 999 − 733 = 266 numbers that are
not divisible by 2, 3, or 5. Of these 266 numbers, 165 of them are primes, not including 2, 3, or 5.
Since 1 in neither prime nor composite, we have 266-165-1 = 100 numbers that are “prime-looking.”
8. For real numbers a and b, define the operation a ? b =
a
. What is the value of
a+b
(2 ? 3) ? 4?
1
11
2
2/5
1
2/5
(2 ? 3) ? 4 =
=
=
.
?4=
5
2/5 + 4
22/5
11
9. If you put a square on each side of a unit square and connect the outside vertices, then you get an
octagon. But, this octagon is not regular. Find the dimensions of the rectangles that you would have
to put on the edges of the unit square so that the resulting figure is a regular octagon.
1
1× √
2
In order that the hexagon be a regular hexagon, it is clear that all side lengths of the hexagon must
be 1, the same as the length of the square’s sides. Letting the other dimension of the rectangle be x,
we have via the Pythagorean Theorem,
1
2x2 = 1 =⇒ x = √ .
2
10. Pascal has six yellow marbles and four blue marbles, each distinct from the others. He arranges his
ten marbles in a row. What is the probability that no blue marble lies next to another?
1
6
Placing the 6 (distinct) yellow marbles out in a row, there are 7 possible spots for a blue marble to
occupy: both ends of the row of yellow marbles and the 5 spots between the yellow marbles. So, there
are 7 × 6 × 5 × 4 ways to pick the 4 slots for the blue marbles. Also, there are 6! ways of ordering the
yellow marbles and there are 10! ways of ordering all 10 marbles. Hence, the probability that no blue
marble lies next to another is
7·6·5·4
1
6! · (7 · 6 · 5 · 4)
=
= .
10!
10 · 9 · 8 · 7
6
11. The odometer in Jane’s car reads 15,951 miles. Law-abiding Jane notices that this is a palindrome
and says, “It will be a long time before that happens again.” But 2 hours later, the odometer shows
a different palindromic number. How fast was the car traveling in those 2 hours, assuming that Jane
traveled at a constant speed during the 2 hours?
55 miles per hour
The next two palindromes are 16061 and 16161. We compute Jane’s speed using each of these mileages.
16061 − 15951
= 55 miles per hour and
2
16161 − 15951
= 105 miles per hour.
2
Since Jane is law-abiding, we find that she traveled at 55 miles per hour during the two hours.
12. The zeros of the polynomial x3 − 24x2 + 191x + C are in arithmetic progression. What is the value of
C?
−504
Since the zeros are in arithmetic progession, we denote the three zeros by r − d, r, and r + d, for some
fixed integer d.
Then,
x3 − 24x2 + 191x + C = (x − (r − d))(x − r)(x − (r + d)),
and expanding, we have
x3 − 24x2 + 191x + C = x3 − (3r)x2 + (r(r − d) + r(r + d) + (r − d)(r + d))x − (r(r − d)(r + d))
implying that 3r = 24, or r = 8. Then,
8(8 − d) + 8(8 + d) + (8 − d)(8 + d) = 191 =⇒ 3 · 82 − d2 = 191 =⇒ d = 1.
Finally, C = −r(r − d)(r + d) = −8 · 7 · 9 = −504.
13. Define a sequence x1 , x2 , . . . by
x1 = 1,
x2 = 2,
and
xn =
xn−1
for each integer n ≥ 3.
xn−2
What is x2007 ?
2
The first few terms of the sequence are
1, 2, 2, 1, 1/2, 1/2, 1, 2, 2, 1, 1/2, 1/2, 1, 2, 2, 1, 1/2, 1/2, . . .
Notice that the sequence repeats after every 6 terms. Since the sequence begins 1 2 2 and since
2007 = 334 · 6 + 3, we find that x2007 = 2.
14. If 9x + 9−x = 14, then what is 81x + 81−x ?
194
Let y = 9x . Then,
9x + 9−x = 14 =⇒ y 1 + y −1 = 14,
and
81x + 2 + 81−x = y 2 + 2 + y −2 = (y + y −1 )2 = 142 = 196.
Hence, 81x + 81−x = 196 − 2 = 194.
15. Which is greater, the probability of getting at least one six with four rolls of a single die or the
probability of getting at least one double 6 in 24 rolls of a pair of dice?
probability of getting at least one six with four rolls of a single die
The probability of NOT rolling a six in 4 rolls of a die is (5/6)4 while the probability of NOT rolling
a double six in 24 rolls of two dice is (35/36)24 . Thus, the probability of getting at least one six with
four rolls of a single die is 1 − (5/6)4 ≈ 0.518 and the probability of getting at least one double 6 in 24
rolls of a pair of dice is 1 − (35/36)24 ≈ 0.491.
Part II
16. Given that −1 < q < 1, express the following infinite product
(1 + q)(1 + q 2 )(1 + q 4 )(1 + q 8 )(1 + q 16 ) · · ·
in the simplest algebraic form.
1
1−q
Let P = (1 + q)(1 + q 2 )(1 + q 4 )(1 + q 8 )(1 + q 16 ) · · · denote the infinite product. Then,
(1 − q)P
1
Thus, P =
.
1−q
=
=
=
=
..
.
=
(1 − q)(1 + q)(1 + q 2 )(1 + q 4 )(1 + q 8 )(1 + q 16 ) · · ·
(1 − q 2 )(1 + q 2 )(1 + q 4 )(1 + q 8 )(1 + q 16 ) · · ·
(1 − q 4 )(1 + q 4 )(1 + q 8 )(1 + q 16 ) · · ·
(1 − q 8 )(1 + q 8 )(1 + q 16 ) · · ·
1.
17. Given that p(x) is a polynomial of degree six such that p(n) = 1/n for n = 1, 2, 3, . . . , 7, find p(8).
1
4
Consider the polynomial f (x) = xp(x) − 1, which is of degree seven. Then, for n = 1, 2, . . . , 7, we have
f (n) = np(n) − 1 = n · (1/n) − 1 = 0.
Thus, we know that
f (x) = C(x − 1)(x − 2)(x − 3)(x − 4)(x − 5)(x − 6)(x − 7),
for some constant C.
Now, on one hand, f (0) = 0 · p(0) − 1 = −1. On the other hand,
f (0) = C(−1)(−2)(−3)(−4)(−5)(−6)(−7) = −C · 7!
So, C =
1
and
7!
f (8) =
1
(8 − 1)(8 − 2)(8 − 3)(8 − 4)(8 − 5)(8 − 6)(8 − 7) = 1.
7!
Then,
p(8) =
f (8) + 1
1+1
1
=
= .
8
8
4
18. Find two different numbers, each of which is the square of the other.
√
1
3
i
− ±
2
2
Let x and y be two such numbers. Then, x and y satisfy the equations
x2
y2
= y
= x
and
(1)
(2)
Substituting (2) into (1) yields
(y 2 )2 = y =⇒ y 4 − y = 0 =⇒ y(y − 1)(y 2 + y + 1) = 0.
If y = 0, then x = 0, which is not allowed. Similarly, if y = 1, then x = 1; again, not allowed. Thus,
to find y, we solve
√
−1 ± 3i
2
.
y + y + 1 = 0 =⇒ y =
2
√
−1 + 3i
, we find that
Choosing y =
2
√ !2
√
−1
+
3i
−1 − 3i
2
x=y =
=
.
2
2
Note: choosing y =
√
√
−1 − 3i
−1 + 3i
yields x = y 2 =
, as expected.
2
2
√
1+ 5
19. Given that cos(36 ) =
, show (without using a calculator) that tan2 (54◦ ) tan2 (18◦ ) is a rational
4
number.
◦
tan2 (54◦ ) tan2 (18◦ ) =
1
5
First, note that
◦
◦
2
◦
cos(72 ) = cos(2 · 36 ) = 2 cos (36 ) − 1 = 2
√ !2
√
1+ 5
−1 + 5
−1=
.
4
4
Next using the facts
cos(a − b) − cos(a + b) = 2 sin a sin b
and
cos(a − b) + cos(a + b) = 2 cos a cos b,
we find that
tan2 (54◦ ) tan2 (18◦ )
=
=
2
sin(54◦ ) sin(18◦ )
cos(54◦ ) cos(18◦ )
2
(cos(36◦ ) − cos(72◦ ))/2
(cos(36◦ ) + cos(72◦ ))/2
√
√ !2 2
2
1 + 5 − (−1 + 5)
1
√
√
√
=
= .
5
1 + 5 + (−1 + 5)
2 5
=
20. Let P be a point inside a square such that the distances from P to the four vertices (in order) are 6,
30, 42, and x. What is the value of x?
30
Consider the figure above in which we have dropped the perpendiculars from P to the sides of the
square. Then, by repeatedly using the Pythagorean Theorem, we have
a2 + b2
b2 + c2
c2 + d2
a2 + d2
= 302
= 62
= x2
= 422
Then, (1)-(4) yields b2 − d2 = 302 − 422 and (2)-(3) yields b2 − d2 = 62 − x2 . Hence,
302 − 422 = 62 − x2 =⇒ x2 = 422 − 302 + 62 = 900 =⇒ x = 30.
(3)
(4)
(5)
(6)