Download 2012 Contest with solutions

SU Math/Computer Day: Math Contest Solutions
1. How many consecutive 0’s are at the end of (525)2 (28)7 ?
Solution: e. By factoring the number into primes, we find that 104 is a factor and that 105 is
not.
2. A list of integers has mode 32 and mean 22. The smallest number in the list is 10. The median
m of the list is a member of the list. If the list member m were replaced by m + 10, the mean
and median of the new list would be 24 and m + 10, respectively. If m were instead replaced
by m 8, the median of the new list would be m 4. What is m?
Solution: e. First consider the following facts: the mean of the list of integers is 22, the
median is a member of the list, and the mean is 24. From these we can determine that there
are 5 integers in the list. So far we know that 10 and m are in the list. Next, since m 4 is the
median if m is replaced by m 8 and the mode is 32, we know that the list is 10, m 4, m,
32, and 32. Knowing that the mean of these number is 22, we conclude that m = 20.
3. For any positive integer n let
f (n) =
What is
2012
X
(
if log8 n is rational ,
otherwise.
log8 n
0
f (n)?
n=1
Solution: c. Since log8 n = 13 (log2 n), it follows that log8 n is rational if and only if log2 n is
rational. The nonzero numbers in the sum will therefore be all numbers of the form log8 n,
where n is an integral power of 2. The highest power of 2 that does not exceed 2012 is 210 ,
so the sum is:
log8 1 + log8 2 + log8 22 + log8 23 + · · · + log8 210 =
1 2 3
1
55
0 + + + + ··· + = .
3 3 3
3
3
4. If a, b > 0 and the triangle in the first quadrant bounded by the coordinate axes and the
graph of ax + by = 6 has area 6, then ab =
Solution: a. The x-intercept and y-intercept of ax + by = 6 are
we have
✓ ◆✓ ◆✓ ◆
1
6
6
= 6,
2
a
b
and so ab = 3.
6
a
and
6
b
respectively. Thus
5. A function f from the integers to the integers is defined as follows:
f (n) =
(
n+3
n
2
if n is odd,
if n is even.
Suppose k is odd and f (f (f (k))) = 27. What is the sum of the digits of k?
k+3
Solution: b. Since k is odd, f (k) = k + 3. This is even, so f (f (k)) = k+3
2 . If 2 is odd, then
k+3
27 = 2 + 3. So k = 45. However, f (f (f (45))) = 12 and we conclude that k+3
2 cannot be
k+3
k+3
odd. If 2 is even, then 27 = 4 and k = 105. Furthermore, f (f (f (105))) = 27. Thus, the
sum of the digits is 1 + 0 + 5 = 6.
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6. Let f (x) be a function with the two properties:
(i) for any two real numbers x and y, f (x + y) = x + f (y), and
(ii) f (0) = 2.
Solution: e. f (2012) = f (2012 + 0) = 2012 + f (0) = 2012 + 2 = 2014.
7. The graph below shows a portion of the curve defined by the quartic polynomial
P (x) = x4 + ax3 + bx2 + cx + d. Which of the following is the smallest?
Solution: c. First note that the quartic polynomial cannot have more real zeros than the
two shown. We have that P ( 1) is greater
than 4. The product of the zeros of P is the
constant term d, which is the y-intercept and
is greater than 5. The sum of the coefficients
of P is P (1), which is greater than 3. The
sum of the real zeros of P is greater than 4.5.
Since the product of the zeros of P is less
than 6 and the product of the real zeros of
P is greater than 4.5, the product of the nonreal zeros of P is less than 2, making this the
smallest value.
8. Points M and N are the midpoints of the sides P A and P B of P AB. As P moves horizontally along a line that is parallel to the side AB, how many of the four quantities listed below
change?
• the length of the segment M N
• the perimeter of
• the area of
P AB
P AB
• the area of the trapezoid ABM N
Solution: b. By the Triangle Midsegment Theorem, M N = AB/2. Since the base AB and
the altitude to AB of 4ABP do not change, the area does not change. The altitude of the
trapezoid is half that of the trangle, and the bses do not change as P changes, so tha area
of the trapezoid does not change. Only the perimeter changes (reaching a minimum when
4ABP is isoceles.
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9. Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last
initials) will be in alphabetical order with no letters repeated. How many such monograms
are possible?
Solution: b. Since Z is the last initial, there are 25 letters remaining. There are 25 · 24 = 600
pairs of distinct letters. For each pair, there is also the reverse of the pair. One will be in
alphabetical order and the other will not. Thus there are 6000/2 = 300 such monograms.
10. Which of the following is largest?
(a)
(b)
(c)
(d)
(e)
qp
3
5·6
q p
3
6 5
q p
3
5 6
q p
3
5 6
q p
3
6 5
Solution: b. By considering the square of the (a), (b), and (c), we can see that (b) is the larger
of the three. By cubing (d) and (e), we can see that (e) is the larger of the two. Comparing (b)
and (e), we have 61/2 · 51/6 and 61/3 · 51/6 respectively. We see that (b) is the larger of the two.
11. Which of the following is equal to e
Solution: d. We have e
3⇡i
2
=
3⇡i
2
cos( 3⇡
2 )
?
+ i sin( 3⇡
2 )=
i.
12. Which of the following triangles cannot exist?
(a) an acute isosceles triangle
(b) an isosceles right triangle
(c) an obtuse right triangle
(d) a scalene right triangle
(e) a scalene obtuse triangle
Solution: c. An obtuse right triangle would have one angle greater than 90 and one equal
to 90 . Thus, the sum of the angles would be greater than 180 .
13. Al and Barb start their new jobs on the same day. Al’s schedule is 3 work-days followed by
1 rest-day. Barb’s schedule is 7 work-days followed by 3 rest-days. On how many of their
first 1000 days do both have rest-days on the same day?
Solution: e. Al is on a 4 day cycle, and Barb is on a 10 day cycle. Since the least common
multiple of 4 and 10 is 20, they will have the same number of rest-days together every 20
days. In the first 20 days, they share 2 rest-days: the 8th day and the 20th day. Thus, in the
first 1000 days, they share 2 · 50 = 100 rest-days.
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14. How many solutions in the real numbers does the following equation have?
x
= sin x
100
x
Solution: c. We have 100
= sin x if and only if 100x = sin( x); thus, the given equation has
an equal number of positive and negative solutions. Also x = 0 is a solution. Furthermore,
all positive solutions are less than or equal to 100 (why?). There is one solution to given
equation between 0 and ⇡, and two solutions in each of the intervals from (2k 1)⇡ to
(2k + 1)⇡, 1  k  15. The total number of solutions is, therefore, 1 + 2(1 + 2 · 15) = 63.
15. How many positive integers between 1 and 100 are one more than the sum of the squares of
their base ten digits? For example, 35 = 1 + 32 + 52 .
Solution: a. First 100 6= 1 + 12 + 02 + 02 . Now we have
10x + y = 1 + x2 + y 2 ,
where x and y are integers between 0 and 9. By resolving the equation, we obtain
x(10
x) = y 2
y + 1.
By trial and error, we have that 35 and 75 are the only two numbers having the stated property.
16. The figure shown is the union of a circle and two semicircles of diameters a and b, all of
whose centers are collinear. The ratio of the area of the shaded region to that of the unshaded
region is
Solution: b. By drawing in the diameter of
the circle, it can be seen that the area of the
shaded region is
✓
◆
✓ ◆
✓ ◆ !
✓
◆
✓ ◆
✓ ◆2 !
⇡
a+b 2
a 2
b 2
+
2
2
2
2
✓
◆
⇡a+b a+b a b
=
+
2 2
2
2
⇡(a + b)a
=
4
and the area of the unshaded region is
⇡
a+b 2
a 2
b
+
2
2
2
2
✓
◆
⇡a+b a+b b a
=
+
2 2
2
2
⇡(a + b)b
=
4
Their ratio is a/b.
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17. Suppose you repeatedly toss a fair coin until you get two heads in a row. What is the probability that you stop on the 8th toss?
Solution: a. There are 28 = 256 possible outcomes when one flips a coin 8 times. Let Ti be
the result of the ith flip. We want to compute the number of possible outcomes which have
T7 = T8 = H and no adjacent H’s in T1 , T2 , · · · , T7 . Clearly, we must have T6 = T . We now
wish to know how many sequences of length 5 have no adjacent H’s. These sequences can be
explicitly listed, or we can find a recursive relationship: Let Sn be the number of sequences of
length n without 2 consecutive occurrrenes of H. EAch such sequence must either start with
HT followed by a sequence of length n 2 with no consecutive H 0 s or start with T followed
by a sequence of length n 1 wit no consecutieve H’s. Therefore Sn = Sn 1 + Sn 2 , and also
S1 = 2, S2 = 3. Thus we have S5 = 13, the 7th Fibonacci number. The probability that such
an outcome occurs is then 13/256.
18. A woman canoes upstream, against the current, between two markers in 10 minutes. She
turns around and canoes downstream, with the current, and makes it back to the starting
point in just 4 minutes. If the woman had instead decided to simply float downstream
instead, how long would it have taken her to reach the first marker?
Solution: d. Let d be the distance between the markers, r her rowing speed, and c the
current speed. Then d = (r c)10 = (r + c)4. From this we conclude that r = (7/3)c, and
d
that d = (40/3)c. The time it takes to float downstream is t = d/c = (3/40)d
= 40/3 minutes.
19. A very long hallway has 1000 doors numbered 1 to 1000; all the doors are initially closed.
One by one, 1000 people go down the hall: the first person opens each door, the second
person closes all doors with even numbers, the third person closes door 3, opens door 6,
closes door 9, opens door 12, etc. That is, the nth person changes all doors whose numbers
are divisible by n. After all 1000 people have gone done the hall, which collection of doors
consists of the ones that are open?
Solution: c. Door D is changed by the nth person if and only if n divides D. Doors that have
an even number of divisors are closed, those with an odd number of divisors are open. If n
divides D, so does D/n, thus D has an even number of divisors unless D is a perfect square.
Thus, doors with a number that is a perfect square will be open.
20. In the graph below, how many paths that never go up connect node A to node B? (Paths
must follow edges indicated.)
Solution: d. Every path from A to B goes along 9 edges, 5 slanting right, 4 slanting left and
any such path joins A and B. Thus there are 95 = 126 paths.
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