Download Ithaca College Math Day Competition March 31, 2006 Solutions Part I

Ithaca College Math Day Competition
March 31, 2006
Solutions
Part I
1. In a cross-country race, Jane finished exactly in the middle of all participants. Dan placed lower, in
tenth place, and Jim finished sixteenth. How many runners took part in the race?
17
First, since Jane finished exactly in the middle of all runners, there must be an odd number of runners.
Since Dan finished in 10th place, Jane placed better than 10th. Since Jim placed 16th, there are at
least 17 runners and Jane must then have placed 9th. If there were more than 17 runners, Jane would
have placed 10th or lower, which is impossible.
2. The average of 40 numbers is 50 and the average of 60 other numbers is 30. What is the average of all
100 numbers?
38
The average of all 100 numbers is
40 · 50 + 60 · 30
= 38.
100
3. A positive number n has the property that n% of n is 16. What is n?
40
We have
n
· n = 16. So, n2 = 1600, and hence, n = 40.
100
4. For how many positive integers m is m2 − 5m + 6 a prime number?
2
Notice that m2 − 5m + 6 = (m − 2)(m − 3). So, if m ≥ 5, then m2 − 5m + 6 = (m − 2)(m − 3) factors
as a product of two integers greater than 1, and hence, is not prime. Check m = 1, 2, 3, and 4, we find
that m = 1 and m = 4 both yield a prime value.
5. What is the average of all 4 digit numbers that can be formed by using the digits 1, 3, 5, and 7 exactly
once?
4444
Each digit appears the same number of times in each place value of the 4 digit number. So, the average
1
of the digits in each place is (1+3+5+7) = 4. Hence, the average of all the numbers is 4·1111 = 4444.
4
6. Consider all nine-digit zipcodes abcde − f ghi where a through i are distinct digits (0 through 9) and
in increasing order. How many different zipcodes of this form are there?
10
There are 9 places to fill in in the zipcode and there are 10 digits to do so with. For each digit we do
NOT use, there is exactly one way to order the remaining 9 in increasing order. Hence, the number of
zipcodes of this form is 10.
7. For how many positive integers n less than or equal to 20 is n! evenly divisible by 1 + 2 + · · · + n?
(Note: n! = n · (n − 1) · (n − 2) · · · 2 · 1)
13
First note that 1 + 2 + · · · + n =
n(n + 1)
. Then, n! is divisible by 1 + 2 + · · · + n means that
2
n!
n(n+1)
2
is an integer.
But,
n!
n(n+1)
2
=
2(n − 1)!
n+1
2(n − 1)!
is an integer unless n + 1 is an odd prime. There are 7 odd primes less than 21, and so,
n+1
there are 20 − 7 = 13 positive integers less than 20 satisfying the condition.
and
8. For real numbers x and y, define the operation x ? y =
p
x2 + y 2 . What is the value of
(4 ? 3) ? ((−3) ? (−4))?
√
5 2
p
p
p
√
(4 ? 3) ? ((−3) ? (−4)) = ( 42 + 32 ) ? ( (−3)2 + (−4)2 ) = 5 ? 5 = 52 + 52 = 5 2.
9. Find a string of 100 consecutive composite natural numbers. You should only write down the first
number in the string. (Note: a composite number is a number that has divisors other than 1 and
itself.)
101! + 2
The string 101! + 2, 101! + 3, 101! + 4, . . . , 101! + 101 is a string of consecutive composite natural
numbers since for each 2 ≤ k ≤ 101,
101! + k
which is composite.
= 101 · 100 · 99 · · · (k + 1)(k)(k − 1) · · · 3 · 2 · 1) + k
= [101 · 100 · 99 · · · (k + 1)(k − 1) · · · 3 · 2 · 1 + 1] · k,
10. Ten poker chips are numbered as follows and placed in a bag. One is numbered “1”, two are numbered
“2”, three are numbered “3”, and four are numbered “4.” Three chips are drawn at random from the
bag, without replacement. What is the probability that the sum of numbers showing on the drawn
chips is a multiple of 5?
5
24
The possible draws of three coins yielding a sum that is divisible by 5 are such that the coins are
numbered:
1, 2, 2 in any order or 2, 4, 4 in any order or 3, 3, 4 in any order.
There are three possible orderings of each of the possible draws, and so, the probability is is given by
1 2 1
2 4 3
3 2 4
5
3·
· · +
· · +
· ·
=
.
10 9 8 10 9 8 10 9 8
24
11. The sum of a man’s age and his monkey’s age is 100 years. The man is three times as old as the
monkey was when the man was ten years older than the monkey is now. How old is the monkey?
37
Let x be the man’s age, let y be the monkey’s age, and let z be the number of years in the past as stated
in the second sentence in the problem. The statement of the problem then gives us three equations
relating these quantities:
x + y = 100
x = 3(y − z)
x − z = 10 + y
Solving this system of equations, we find that x = 63, y = 37, and z = 16. Thus, the monkey is 37
years old.
12. A rectangle with a diagonal of length ` is twice as long as it is wide. What is the area of the rectangle?
2 2
`
5
If we let w be the width of the rectangle, then its length is 2w. By the Pythagorean Theorem,
1
4w2 + w2 = `2 , or 5w2 = `2 . So, w2 = `2 . Thus, the area of the rectangle is w · 2w = 2w2 =
5
we have
2 2
` .
5
13. The product of three consecutive positive integers is 16 times their sum. What is the sum of the
squares of these integers?
149
Let n − 1, n, and n + 1 be the three consecutive positive integers. Their sum, S, and product, P , are
given by
S = (n − 1) + n + (n + 1) = 3n and P = (n − 1)n(n + 1) = n3 − n.
We are given P = 16S, which we solve for n:
P = 16S =⇒ n3 − n = 48n =⇒ n3 − 49n = 0 =⇒ n(n2 − 49) = 0 =⇒ n = 7.
Hence, the three consecutive positive integers are 6, 7, and 8. Finally, 62 + 72 + 82 = 149.
14. Suppose every couple continues having children until they have a boy and then they stop. What
percentage of male births will result? (Assume that the probability that a couple has a girl is 1/2.)
50%
15. How many numbers between 1 and 2006 are integer multiples of 3 or 5 but not of 15?
803
There are 668 multiples of 3 between 1 and 2006, 401 multiples of 5, and 133 multiples of 15. Thus,
the number of numbers between 1 and 2006 that are integer multiples of 3 or 5 but not of 15 is given
by
(668 − 133) + (401 − 133) = 803.
Part II
In Part II, write your final answer in the box provided. Also, include your work below your answer.
16. Let a and b be two digit integers such that b is obtained by reversing the digits of a. The integers a
and b satisfy a2 − b2 = n2 for some positive integer n. What is a + b + n?
154
First note that a2 − b2 = n2 implies that a > b. If we set a = 10x + y, then b = 10y + x, with
9 ≥ x > y ≥ 1. Now,
n2 = (10x + y)2 − (10y + x)2 = 99x2 − 99y 2 = 99(x2 − y 2 ).
This means that 99(x2 − y 2 ) is a perfect square. Since 99 = 32 · 11, we must have x2 − y 2 = 11k 2 for
some integer k. Then,
(x + y)(x − y) = x2 − y 2 = 11k 2 .
But, x − y ≤ 9 − 1 = 8 and x + y ≤ 9 + 8 = 17. This implies that x + y = 11 and x − y = k 2 . This
forces the value of k to be either 1 or 2. If k = 2, then x = 15/2 and y = 7/2. This is impossible since
since x and y must be integers. Hence, k = 1, giving us x = 6 and y = 5. Thus, a = 65, b = 56, and
n = 33. So, a + b + n = 154.
17. Let f (x) = x2 + 10x + 7 and let R be the set of points (x, y) in the plane satisfying
f (x) + f (y) ≤ 0
and
f (x) − f (y) ≤ 0.
Find the area of R.
18π
First,
f (x) + f (y) ≤ 0 ⇐⇒
⇐⇒
⇐⇒
⇐⇒
x2 + 10x + 7 + y 2 + 10y + 7 ≤ 0
(x2 + 10x + 25) + (y 2 + 10y + 25) − 36 ≤ 0
(x + 5)2 + (y + 5)2 − 36 ≤ 0
(x + 5)2 + (y + 5)2 ≤ 36.
So, the set of points satisfying f (x) + f (y) ≤ 0 is the set of points (x, y) on and inside thie circle
centered at (−5, −5) with radius 6.
Next,
f (x) − f (y) ≤ 0 ⇐⇒
⇐⇒
⇐⇒
⇐⇒
x2 + 10x − y 2 − 10y ≤ 0
x2 − y 2 − 10(x − y) ≤ 0
(x + y)(x − y) − 10(x − y) ≤ 0
(x − y)(x + y + 10) ≤ 0.
This last inequality holds if
(x − y ≤ 0 and x + y + 10 ≥ 0) or (x − y ≥ 0 and x + y + 10 ≤ 0).
In the first case, we get the set of points satisfying y ≥ x and y ≥ −x − 10. That is, these are the
points on and above the lines y = x and y = −x − 10, which intersect at (−5, 5). In the second case,
we get the set of points satisfying y ≤ x and y ≤ −x − 10. That is, these are the points on and below
the lines y = x and y = −x − 10, which intersect at (−5, 5).
Thus, the set of points satisfying the conditions of the problem are those points on and inside the
circle centered at (−5, −5) with radius 6, which also satisfy the second set of conditions found the last
paragraph. This defines a region R which is half of the circle and its interior. This is the shaded area
in the figure below. Thus, the area of R is (1/2)π(62 ) = 18π.
18. Let 4ABC be a right triangle with m∠ABC = 90◦ . Let M be the midpoint of side BA and let N be
the midpoint of side BC. If AN = 19 and CM = 22, then find AC.
26
If BM = a and BN = b, then BA = 2a and BC = 2b. Applying the Pythagorean Theorem to 4ABN,
we have (2a)2 + b2 = 192 . Applying the Pythagorean Theorem to 4M BC, we have a2 + (2b)2 = 222 .
Adding these two equations, we find that
5a2 + 5b2 = 845 =⇒ a2 + b2 = 169 = 132 .
Then,
AC =
19. Consider f (x) =
p
p
√
(2a)2 + (2b)2 = 4a2 + 4b2 = 2 169 = 26.
1
. Let f1 (x) = f (x) and for k = 2, 3, . . ., let fk (x) = f (fk−1 (x)). Find the value
1−x
of f2006 (2006).
2005
2006
First, we find that
f1 (x)
=
1
1−x
f2 (x)
= f (f1 (x)) = f
f3 (x)
= f (f2 (x)) = f
1
1−x
1
x
1−
=
1
=
1
1 − 1−x
=
1
1
1 = 1 =x
1 − (1 − x )
x
1
1−x−1
1−x
=
1
1−x
=1−
−x
x
So, for every j ≥ 1, we have
f3j (x) = x,
f3j+1 (x) = f1 (x),
and f3j+2 (x) = f2 (x).
Note that 2006 = 3 · 668 + 2, and so,
f2006 (x) = f3·668+2 (x) = f2 (x) = 1 −
Hence,
f2006 (2006) = 1 −
1
2005
=
.
2006
2006
1
.
x
20. A point P is randomly chosen from the rectangular region with vertices (0, 0), (4, 0), (4, 2), and (0, 2).
What is the probability that P is closer to the point (6, 2) than it is to the origin?
1
4
Let R be the region within the rectangle defined by the set of points that are closer to (6, 2) than to
the origin. Then, the probability that a point in the rectangle is closer to the point (6, 2) than it is
to the origin is the ratio of the area of R to the area of the entire rectangle. Since the area of the
rectangle is 8, the probability is (1/8)(area of R).
The set of points in the plane equidistant from the origin and (6, 2) is the perpendicular bisector
of the line segment connecting the origin to (6, 2). Note that this bisector passes through the point
(3, 1). Then, R is the set of points in the rectangle that lies to the right of this bisector. This
defines a trapezoidal region and the area of this trapezoid is half the area of the square with vertices
(2, 0), (4, 0), (4, 2), and (2, 2). Thus, the area of R is (1/2)4 = 2. Thus the probability is (1/8)(2) = 1/4.
R is the shaded region in the figure below.