Download Workout 1

Workout 1
Answers
1.
Danny
2. 24
(E, G, M, T)
5. 2008
(C, F, T)
3. 17.30
(C, T)
4. 2100
(C)
(C, F, G, P, T)
6. 1*
(C, F, T)
7. 145
(C, F, P)
8. 4.3
9.
(C, F, T)
(C, E, M, P, T)
10. 9256
(C, F)
Solution - Problem #2
We can let the figure to the right illustrate the
scenario in the problem. Notice that Joe’s points and
Frank’s points make up + = of the total points in the
game. Therefore, the remaining four points scored by Ken
and Mike are one-sixth of the game’s total points. There
must have been 6´4 = 24 points scored in the game.
Solution/Multiple Representations - Problem #6
One way to approach this problem is to see it as a “Rate ´ Time = Distance” problem. We know
= of an hour. (Remember that if
the rate is 15 mph, and he is biking for four minutes, which is our rate is in miles per hour, then our time must be in hours, and our distance will be in miles.) Now we
can calculate that 15 mph ´ hr = 1 mile.
We also could view this as a proportional reasoning problem. We know that he travels
15 miles in 60 minutes, and we want to know the number of miles he travels in four minutes. This leads
= [ . Setting the cross products equal, we have 60x = 4(15). Dividing both sides
to the proportion by 60 then gets us to [ = = = mile.
Connection to... Geometric Sequences (Problem #7)
This problem refers to an arithmetic sequence though you could probably solve the problem
without knowing what an arithmetic sequence is. An arithmetic sequence is a sequence of terms in
which the difference of the n th and n +1st term remains constant. (Simply put, the difference of
consecutive terms remains the same throughout the sequence.) A geometric sequence is similar, but
rather than every pair of consecutive terms having the same difference, each pair has the same
quotient. An example of a geometric sequence is 2, 6, 18, 54, ... . Can you see how the quotient of
consecutive terms is no matter which two consecutive terms we use? (You also could describe this
as a geometric sequence in which each term is multiplied by 3 to get the next term.) It’s important to
remember that arithmetic sequences may have terms that increase in value (11, 13, 15, 17, ...) or
decrease in value (11, 9, 7, 5, ...) or stay the same in value (11, 11, 11, 11, ...). The same is true for
geometric sequences. Using 8 as the first term, can you show the next three terms of a geometric
sequence with terms that decrease in value?
* The plural form of the units will always be provided in the answer blank, even if the
answer appears to require the singular form of the units.
26
MATHCOUNTS 2004-2005
Workout 2
Answers
1.
19
(P, T)
5. 215
2. 2830
(C)
6. 4
3. 86.4
(C)
7. 840
4. 32
(C)
8. 350
(C)
(C, E, G, T)
9. 49
(C)
(C, F)
10. 15
(C, G, M, P, T)
(C, P, T)
Solution - Problem #4
We know that we are numbering houses from 4 through 120, but using only the multiples of
four. We could list these out, but let’s see if we can count them quickly. Since 4 = 4 ´1 and
120 = 4´30, we are working with the first 30 multiples of 4. Unfortunately, some are one-digit
integers, others are two-digit integers and still others have three digits. Let’s group them accordingly.
(It helps to know that 100 is the first three-digit multiple of 4.)
2 mult.
3-digit
4´25 = 100
•••
2-digit
4´3 = 12
•••
1-digit
4´1 = 4
4´2 = 8
4´24 = 96
4´30 = 120
24 - 2 = 22 mult.
30 - 24 = 6 mult.
The two one-digit multiples account for a total of two digits; there are 22 two-digit multiples,
resulting in 22´2 = 44 digits; and there are six three-digit multiples yielding 6´3 = 18 digits. This is a
total of 2 + 44 + 18 = 64 digits, which will cost 64 ´ $0.50 = $32.
Solution/Multiple Representations - Problem #10
The figure included with this problem is extremely important because we see that the product
of the number of rows and the number of columns (or band members in each row) is the total number
of band members, or 105. In Formation A with m rows and n columns, we know m ´n = 105.
Formation B has m + 6 rows and n - 2 columns (or band members per row), and we know
(m + 6)(n - 2) = 105.
From this point we see that we have two equations with two unknowns. The first equation can
be rewritten as m = Q . Substituting this into the second equation, we get ( Q + 6)(n - 2) = 105.
Multiplying both sides of the equation by n gets us to (105 + 6n )(n - 2) = 105n ;
6n 2 + 105n - 12n - 210 = 105n ; 6n 2 - 12n - 210 = 0; n 2 - 2n - 35 = 0; (n - 7)(n + 5) = 0; and
finally n = 7 or -5. Since n is the number of band members per row, the value of n must be positive,
and therefore n = 7. If there are seven members per row, there are 105 ¸ 7 = 15 rows in Formation A.
Let’s go back to the original two equations m ´n = 105 and (m + 6 )(n - 2) = 105. Rather than
jumping into the algebraic solution above, let’s examine the factor pairs for 105: 1 & 105, 3 & 35,
5 & 21 and 7 & 15. By inspecting these four factor pairs, we can see that if we let (m , n ) be the
factor pair (15, 7), then (m + 6, n - 2) matches with the factor pair (21, 5). Therefore, m = 15, and
there are 15 rows in Formation A.
32
MATHCOUNTS 2004-2005
Workout 3
Answers
1.
17
(C, P, S, T)
2. 40
(C)
3. 3
(C, E, G, M, T)
4. 2,970,000
5. 312
(C, F, M)
8. 24
6. 26
(C, F, S)
9. 12.7
(C)
10. 7.2
(F)
7. 46.01
(C)
(C, F, G, M, P)
(C)
Solution - Problem #6
We could calculate the volumes of both Mars and Pluto since we are given each planet’s radius,
and then divide the values of the two volumes. However, relying on the property that if a :b is the
ratio of the radii of similar figures, then a 3 :b 3 is the ratio of their volumes, we can skip right to our
answer by simplifying 21063 :7073, which is approximately 26, to the nearest whole number.
Solution/Multiple Representations - Problem #8
y
First, we can add some variables to our picture. We don’t know that
y
the fencing between the two gardens splits the entire area in half, but we
x
know xy = 336 and 3y + 2x = 100. We now need to solve for y. We have
y
two equations and two unknowns, so it’s possible to solve at this point.
However, even after getting these merged into one equation with one
variable, it is going to be challenging to solve. Starting with xy = 336, we can divide both sides by y
2
2
to see that x = \ . Now we have 3y + 2( \ ) = 100; 3y + 2(336) = 100y ; 3y - 100y + 672 = 0;
(3y - 28)(y - 24) = 0. From here, we see that y = or 24. Since we were told the dimensions were
integers, we have y = 24.
Alternatively, we could use our calculator with some Guess, Check & Revise and perhaps get to
the answer quicker. Since 3y + 2x = 100, we know that both x and y must be less than 50. (Since
they’re both positive integers, 50 is a good estimate cap since 3(0) + 2(50) = 100.) We also know that
xy = 336. We don’t need to check factor pairs of 336 with extremely large factors like 336 & 1 or
168 & 2 because of the cap we’ve already established. Our factor pair is going to have to consist of
numbers that are both closer to the square root of 336 (which is approximately 18.3) than the extreme
values. The prime factorization of 336 is 24 ´ 3 ´ 7. We can’t form the factors 18, 19 or 20 from
these prime factors, but let’s look at a pair of factors involving 21. The factor pair is 21 & 16. This
pair must satisfy 3(21) + 2(16) = 100 or 3(16) + 2(21) = 100. It fails, but both are close. We can’t
form 22 or 23 from the factorization of 336, but let’s examine 24. The factor pair would be 24 & 14.
Upon testing 3(24) + 2(14), we see the result is 100, and so y = 24.
We also could use the TABLE feature of a graphing calculator to solve this problem. We will
use the relationships y = and 3y + 2x = 100. First, we will enter the following equations in the
[
“Y=” field: Y1 = [ and Y2 = 3( [ ) + 2x. When we go to the TABLE feature, the first column will
show a value of x , the Y1 column will show the corresponding value of y (notice the product of every
pair of values in these first two columns is 336), and the value in the Y 2 column is the corresponding
value of 3y + 2x. If we scroll down until we have 100 in the Y2 column, we can see that this happens
only when x = 14 and y = 24.
38
MATHCOUNTS 2004-2005
Workout 4
Answers
1.
1880
2. 3
(C)
(C, E, G, P, T)
3. 27
(C, F)
4. 8145.06
(C, F)
5. 0
6. 40
7. 22,832
(C, F)
8. 114
(C, F, P)
(C, F, M)
9. 400
(C, F, M)
(C, F)
10. 98
(C, F)
Solution/Multiple Representations - Problem #5
Perhaps your first thought is to plug in the value of for each x. In doing this, we have
− + − . Simplifying this isn’t too difficult if we reduce fractions along the way:
− + − = ××× − ××× + × − . Notice the difference of the first two
49 49 49
4 94 94 9 4 94 9 4 9
terms is 0, as is the difference of the last two terms, so the total value of the expression is 0, and we
didn’t have to work with any large numbers or complicated fractions in the process!
Since this is a Workout problem, we could have used our calculator to do all of the work for
us. If we enter the expression 8x 3 – 6x 2 + 12x – 9 into the “Y=” field of a graphing calculator, we
can then find the value of this expression for any value of x. Since we’re particularly interested in
the value of the expression when x = , go to TBLSET and set TblStart equal to . If we now go to
TABLE, we will see that when x = 0.75 (or ), the y-value (or value of the expression we entered into
“Y=”) is 0.
Though probably not the most obvious or efficient solution, we also can take the original
expression and use factoring. By changing the order of the terms, we have the expression
8x 3 + 12x – 6x 2 – 9, which is equivalent to 4x (2x 2 + 3) – 3(2x 2 + 3) or (4x – 3)(2x 2 + 3). By plugging
in for x, we see that the value of the first binomial factor is 0, and therefore the product of the
two binomial factors (and the original expression) is 0.
Solution/Multiple Representations - Problem #9
We don’t know much about George’s first trip other than it was 300 miles. If we relate this
problem to the common formula Rate ´ Time = Distance, we have r1 ´ t1 = 300 miles. For his second
trip, we know that the rate and time of the first trip changed. Since his time is doubled, we’ll multiply
the left side of our equation by two, and therefore, we’ll multiply the right side of the equation by
two, as well. Now we have r1 ´ (2 ´ t1 ) = 2 ´ 300. Similarly, the rate (or speed) for the second trip
was only two-thirds that of the first trip. This means we should multiply both sides of the equation by
two-thirds, which results in ( ´ r1 ) ´ (2 ´ t1 ) = ´ 2 ´ 300. By simplifying the right side of the
equation, we see that the product of the new rate and the new time yield a distance of ´ 2 ´ 300 =
400 miles.
Since we weren’t given too many constraints for the first trip, notice that we could solve this
problem by just choosing values for the rate and time of the first trip that give us a distance of
300 miles. We could choose 30 mph & 10 hrs, 50 mph & 6 hrs, 60 mph & 5 hrs, etc. Doubling any of
these times for the second trip will not be difficult, but since we have to take two-thirds of the rate,
choosing a scenario with a rate that is divisible by three will make this easier. Let’s go with the first
trip being 60 mph and 5 hrs. We now know that the second trip would be 40 mph and 10 hrs, which
results in a 40 ´ 10 = 400-mile trip.
44
MATHCOUNTS 2004-2005
Workout 5
Answers
1.
5
2. 0.0235
3. 3.18 ´ 10
-3
4. 30
(C, G, T)
5. 1140
(C, F, T)
6. 29
(C)
(C, F, M)
8. 20
(C, F, P, S, T)
(C, F, G, S, T)
9. 9.1
(C, F, M)
10. 63
(C, F, G, T)
7. 494,550
(C, F, P, T)
(C, F, M, P, T)
Solution - Problem #4
One efficient solution to this problem uses the formula for determining the number of ways to
choose m items from a group of n items, assuming that the order in which the m items are chosen does
Q
not matter. We call this “n choose m ” and the expression for determining this number is P Q − P .
For this problem, we first need to figure out how many different pairs of girls can be chosen for the
=
= 3 pairs. (Since
team. There are three girls to choose from, so we’ll use 3 choose 2 =
−
there were only three girls to start with, you may have been able to see that each one of the girls
could have been chosen to be left off the team, which also results in three different selections for
the female team members.) Now we also need to determine the number of different pairs of boys that
can be chosen. There are five boys to choose from, so we’ll use 5 choose 2 =
= 10 pairs. Each of
the three possible pairs of girls could be matched with each of the 10 possible pairs of boys, which
results in 3 ´ 10 = 30 possible four-member teams consisting of two girls and two boys.
Solution/Multiple Representations - Problem #10
We see that x can be any value from 5 to 8, but once we choose the value, we must use it in
both expressions. Since we are trying to find the greatest possible difference, it would make sense to
test the extremes of our options for x, mainly 5 and 8. Using x = 5, we get 3(5) - 4 = 11 and
5 - 6(5) = -25, for a positive difference of 11 - (-25) = 36. Using x = 8, we get 3(8) - 4 = 20 and
5 - 6(8) = -43, for a positive difference of 20 - (-43) = 63. The difference of 63 is the greatest of
the two. We can see that as x increases, the value of the first expression increases while the value of
the second expression decreases. This creates a greater difference between the two expressions, so
it is correct to choose the greatest possible value of x.
Consider the compound inequality 5 £ x £ 8 and
5£ x £8
5£ x £8
alter it to match each of the two original expressions (as
15 £ 3x £ 24
-30 ³ -6x ³ -48
shown here). We again see that the values of 11 and -25
11 £ 3x - 4 £ 20
-25 ³ -6x + 5 ³ -43
are attained from x = 5, and the values of 20 and -43
-25 ³ 5 - 6x ³ -43
are attained from x = 8. The positive differences for
the two expressions then range from 36 to 63, with the
greatest being 63. Since the expressions are each linear (x to the first power), we also can assume
that choosing the middle value of 6.5 for x will result in a positive difference of 49.5 (which is the
average of the differences found at x = 5 and x = 8).
Using the two expressions, we could graph the equations y = 3x - 4 and y = 5 - 6x. Examining
the graph in the interval 5 £ x £ 8, we see that the lines are furthest apart at x = 8. This is the point
of the greatest difference. The difference of the two y -values at x = 8 is again 63.
Additionally, using the “Y=” field of a graphing calculator, we can enter
“Y=” abs((3x - 4) - (5 - 6x )) and view the TABLE starting at x = 5. We can quickly see in the table
that this value (the positive difference of the two expressions) increases as x increases, and the
difference of 63 at x = 8 is the greatest for our interval.
50
MATHCOUNTS 2004-2005
Workout 6
Answers
1.
3
(E, G)
2. 3214
(G, P, T)
3. 56.5
(C, F)
4. 8
(C, E, G, S, T)
5. 11,926
6.
7. 6
(E, G, M, P)
8. 3300
(C, F, M)
(C, F, M, S)
9. 2232
10.
(E, G, T)
(C, F, G, T)
(C, M, T)
Solution - Problem #5
First, we must understand that we are trying to fill the
squares so that the dotted line shown in Fig. 1 is a line of symmetry.
The squares “occupied” by the dotted line can contain any number
since they are on the line of symmetry, but the numbers in the other
Fig. 1
squares must have the symmetry demonstrated here. The shapes/
shading in Fig. 1 indicate corresponding squares.
The key to solving this problem is to consider the shapes
and sizes of the seven tiles. Cutting out the pieces and moving them
around the grid is extremely helpful! The sizes of the tiles
eliminate many possibilities. The tiles to pay particular attention to
are the two 3x1 tiles, the 1x3 tile and the two 1x2 tiles. The two
Fig. 3
3x1 tiles must be positioned so that there is room for the 1x3 tile
and two 1x2 tiles. Many times after positioning the two 3x1 tiles,
there is only one option for the placement of the 1x3 tile. After
that is placed, there may not be room for both of the 1x2 tiles, as
shown in Fig. 2, where the 1x3 tile can fit across the top, but only
one of the 1x2 tiles will then fit (bottom row).
Fig. 5
If we place the two 3x1 tiles as we did in Fig. 3, the
1x3 tile can go in the top row or bottom row. Only placing it in the
top row will preserve the symmetry (Fig. 4). We now need a 2 in the
bottom left square, which we can get only from one of the 1x2 tiles. This
placement still preserves the symmetry even though we have added a 4, too
(Fig. 5). Needing a 5 to finish out the bottom row, we can place the 2x1 tile
(Fig. 6). The remaining two tiles fit nicely (Fig. 7). The sum of the
four-digit numbers in the top two rows is 7452 + 4474 = 11,926.
Fig. 2
Fig. 4
Fig. 6
Fig. 7
The 2004-2005 MATHCOUNTS poster originated from this problem.
On the poster, green = 7, blue = 5, red = 4 and yellow = 2. The bottom row of caps is
yellow, red, blue and yellow.
Solution/Multiple Representations - Problem #4
Every seat of every bus is filled. Since 500 is not divisible by 35 or 45, we
will need a combination of the two bus sizes. In this table, we see the possible
numbers of students for each type of bus, and the pairings that give us a total of
500 students are identified. The combination of four 35-passenger and eight 45passenger buses is the one that uses the fewest number of buses.
We also could use our graphing calculator. We are trying to find solutions to
the equation 35x + 45y = 500, which is really 7x + 9y = 100. Solving for y, we have
y = (100 - 7x ) ¸ 9. If we enter this expression into the “Y=” field of a graphing
calculator, fix TBLSET to start at 1 and increase by 1, we can then select TABLE and
see our (x, y ) pairs that consist of two positive integers. Scrolling down, we see that
(4, 8) and (13, 1) are the only possibilities, with (4, 8) using only 12 buses, eight of
which are 45-passenger buses.
56
35
35
70
105
140
175
210
245
280
315
350
385
420
455
490
45
45
90
135
180
225
270
315
360
405
450
495
MATHCOUNTS 2004-2005
Workout 7
Answers
1.
27.6
2. 5
(C, F, M, T)
5. 13.5
(E, G, M, T)
6. 1332
(C, G, T)
9. -3
7. 14.6
(C, F, M)
10. 144
3. 480
(C, F, M)
4. 7.20
(C)
(C, F, M, S)
8. 0.254
(C, F, M, T)
(C, F, G, M, S)
(C, M, P, T)
Solution - Problem #8
Since there are nine flips and each flip has two possible outcomes, there are a total of
2´2´2´2´2´2´2´2´2 = 29 = 512 possible outcomes. (One such outcome is THTTHTTHH.) The
probability that or more of the nine flips result in Heads is equal to the sum of the probabilities of
flipping nine Heads, eight Heads, seven Heads and six Heads. Obviously, there is only one outcome
with nine Heads (HHHHHHHHH). There are nine outcomes with eight Heads, since we can choose any
of the nine flips to be the single Tails (which explains why 9 choose 8 = 9 choose 1 = × = 9.) How
many outcomes have seven Heads, or in other words, how many ways are there to pick seven of the
nine flips to result in Heads? We get the answer by performing 9 choose 7 = × = 36 ways.
Finally, there are 9 choose 6 = 84 outcomes containing exactly six Heads. This is a total of
1 + 9 + 36 + 84 = 130 outcomes out of 512 that result in at least of the nine flips being Heads. As a
decimal to the nearest thousandth, this is 130 ¸ 512 = 0.254.
Solution/Multiple Representations - Problem #1
If we graph this information, we can see the location of
points (1970, 23.2) and (1998, 26.7) on the graph. We are told
that the linear trend continues, and we are looking for the
y -coordinate of the point (2005, y ). From the previous two
− points we know the slope is − = = = and then
26.7 = ( )(1998) + b ; 26.7 = 249.75 + b ; and finally
b = -223.05. Now using the equation y = x - 223.05 with
(2005, y ), we have y = (2005) - 223.05; and y = 27.575 or
27.6 to the nearest tenth.
Because we are told that the trend continues in a linear fashion, we also can use proportional
reasoning for this problem. The change in the marrying age over the first 28 years was 3.5. We want
= [
to know the change in the marrying age over the next seven years. Therefore, the proportion represents our question. Using the cross-products, we have 28x = 3.5(7); 28x = 24.5; and finally
x = 0.875. Since the previous age was 26.7, if we increase it by our 0.875 for the next seven years,
we get the median marrying age of 26.7 + 0.875 = 27.575 or 27.6 to the nearest tenth, in 2005.
We also can use our graphing calculator to determine the line of best fit, given our initial data.
Under STAT, we choose option 1 (Edit) and enter 1970 and 1998 into L1, and enter 23.2 and 26.7 into
L2. (Be sure that any other data in List 1 or List 2 is cleared out.) Again going to STAT, then scrolling
over to CALC, and choosing option 4 (LinReg(ax + b)) and pressing Enter, we see that the calculator
calculates for us a = .125, which is the slope that we found earlier, and b = -223.05, which is the
y -intercept we found earlier. Now going to the “Y=” screen (be sure any previous equations are
cleared out), and then selecting VARS, and then selecting option 5 (Statistics), and scrolling over to
EQ, we select option 1 (RegEQ). Now the equation of our line is appearing in the “Y=” screen. Going
to TBLSET and entering 2005 for TblStart, we can then go to TABLE and see that the y -value for
x = 2005 on this line is 27.575, or 27.6 to the nearest tenth.
62
MATHCOUNTS 2004-2005
Workout 8
Answers
1.
50
2. August
3. 4836
4. 11
(C, F, G, M, T)
(T, C)
(C, F, S, T)
5. 761
(C, F, P, S, T)
6. 7
(C, E, F, G, T)
7. 486
(C, G, T)
(C, F, T)
8. 10.06
9. 340
10.
(C, F, M)
(C, G, M, S, T)
(C, S, T)
Solution - Problem #1
Assuming our garden is x by y meters, as shown here, we are trying to solve
for integers x and y such that xy = 4y + 2x and xy is as great as possible. Luckily
this is a Workout, and we can use calculators! Notice that xy = 4y + 2x ;
2x = xy - 4y ; 2x = y (x - 4); and \ = [−[ . From the equation, we can see that it is
undefined at x = 4, and we know that y = 0 at x = 0. We’re only concerned with
positive integers x and y, so after inputting \ = [−[ into the “Y=” field of a graphing
calculator and setting TblStart to 1 in the TBLSET field, we can go to TABLE and
scroll down until we see both x and y as positive integers. This occurs at (5, 10),
(6, 6), (8, 4) and (12, 3), and then there are a lot of entries before we get to y = 2 or y = 1.
verify that these first four ordered pairs are solutions.
5(10) = 4(10) + 2(5)
50 = 40 + 10
6(6) = 4(6) + 2(6)
36 = 24 + 12
8(4) = 4(4) + 2(8)
32 = 16 + 16
y
y
x
y
y
We can
12(3) = 4(3) + 2(12)
36 = 12 + 24
At y = 2, we have 2x = 8 + 2x, which leads to 0 = 8, which is not possible. At y = 1, we have
x = 4 + 2x, which leads to -x = 4 and x = -4, which is not a solution for this situation. Of the four
possible (x, y ) dimensions, (5, 10) has the greatest area of 50 square meters.
Solution/Multiple Representations - Problem #5
The first three formations of cubes are shown on the previous
page, and the information is given in this table. Is there a pattern
forming? By drawing Formation 4, we know the fourth row of values in
the table. Notice that the numbers of squares being added each time
are consecutive multiples of four. The number of cubes in Formation
20 is then 1 + 4 + 2(4) + 3(4) + ... + 19(4) = 1 + 4(1 + 2 + 3 + ... + 19) =
1 + 4(190) = 1 + 760 = 761 cubes.
Formation
# of Cubes
1
1
2
5
3
13
4
25
+4
+8
+12
Let’s approach this differently. Assume each formation started as a square formation of
cubes with an odd number of rows and columns. If we had numbered the cubes in the original square
formations, we would see that the even-numbered cubes have been removed, and only the oddnumbered cubes remain in the final formations for this problem. Notice that the final Formation 2
came from three rows and three columns of cubes, but now cubes 2, 4, 6 and 8 are missing. This is
shown in the figure below. Since the initial square regions have odd dimensions, they will have an area
with an odd number of square units. For Formation 2, the original square formation fit nine cubes
when they were all there. Since only the odd-numbered cubes are there now, there are = 4.5 or
five cubes in the formation (and four cubes missing). When we divide the area of one of our square
formations by two, we will round up for the number of cubes that actually remain in the formation and
round down for the number of cubes “missing” from the
square formation. In Formation 3 that measures 5 by 5, there
are = 12.5 or 13 cubes in the formation (and 12 cubes
missing). In Formation 20, the dimensions of the square
formation will be 39 by 39 (since 39 is the 20th odd number),
and there will be × = 760.5 or 761 cubes in the final formation.
68
MATHCOUNTS 2004-2005
Workout 9
Answers
1.
(C, F, M)
2. 272
(C, E, F, M, T)
3. 801
4.
(E, G, P)
5. 384
(E, G, P)
6. 377
(C, P, S, T)
7. 3
(C, E, G, P, T)
8. 157,000
(C, F, M)
9. 30.6
10.
(C, F, M)
(C, F, T)
(C, M, T)
Solution - Problem #6
We’ll define a “good bit string” as a 12-digit bit string with no consecutive 1s. Each good bit
string must have fewer than seven 1s or it will be impossible for no two 1s to be consecutive.
Therefore, we are counting the number of good bit strings with no 1s, one 1, two 1s, three 1s, four 1s,
five 1s and six 1s. Counting the number of good bit strings with no 1s is easy. There is only one
(000000000000). Counting the number of good bit strings with a single 1 is also not too bad. The 1
can be in any of the 12 spots, so there are 12 possible good bit strings. Notice this is 12 choose 1 = 12.
Let’s now determine the number of good bit strings with two 1s. We can’t simply perform
12 choose 2 = 66 because these 66 bit strings include bit strings with consecutive 1s such as
001100000000. There must be ten 0s in a good bit string with two 1s. So a good bit string is
_0_0_0_0_0_0_0_0_0_0_ with a 1 inserted into two of these spaces. Notice there are 11 places to
put the two 1s. The two 1s can’t be placed on the same end or between the same consecutive 0s. This
means we have to pick two of our 11 spaces for our 1s, and there are 11 choose 2 = 55 pairs of spaces
to choose. Each pair of spaces we choose creates a unique good bit string. When we go to the
scenario with three 1s, there are then nine 0s and 10 spots to insert the three 1s. This leads to
10 choose 3 = 120 good bit strings. Similarly, there are 9 choose 4 = 126 good bit strings with four 1s,
8 choose 5 = 56 good bit strings with five 1s, and 7 choose 6 = 7 good bit strings with six 1s. This is a
total of 1 + 12 + 55 + 120 + 126 + 56 + 7 = 377 good bit strings.
Solution/Multiple Representations - Problem #9
The shaded regions of the original problem are congruent
isosceles trapezoids. Only one of these trapezoids is shown here.
x
Each interior angle of the regular octagon is 135°, so the trapezoid
shown has two angles measuring 135° and two angles measuring 45°.
If an altitude is dropped, creating a 45-45-90 triangle (shaded in the
figure to the right) with a hypotenuse of 2 feet, we know that the height x of the trapezoid is
feet and the length of the bottom base of the trapezoid is 4 + 2 feet. The area of the
trapezoid is thus (4 + (4 + 2 ))( ) = (8 + 2 )( ) = (4 + )( ) = 4 + 2 square
feet. There are four of these trapezoids in the original figure, so the total area of the four shaded
regions is 4(4 + 2) » 30.6 square feet, to the nearest tenth.
The total area of the shaded regions in Problem #9 is equal to the
area of the octagon minus the area of the unshaded square. The area of a
regular polygon is ap, where p is the perimeter and a is the distance
a
from the center of the polygon to the midpoint of any of its sides. For
this octagon, p = 8(4) = 32 feet. A central angle is = 45°, 34 bisects
this angle, PQ = a and 34 is the perpendicular bisector of the side it
intersects. Using our trigonometry ratios, we know (tan 22.5) = D or a = WDQ . Therefore,
a » 4.8284 feet. The area of the octagon is (4.8284)(32) square feet. The area of a rhombus or
square is d1d2, and for the unshaded square in this problem, d1 = d2 = 2a . The area of this unshaded
square is thus [2(4.8284) ´ 2(4.8284)] square feet. The difference of these two areas is
(4.8284)(32) - [2(4.8284) ´ 2(4.8284)] » 30.6 square feet, to the nearest tenth.
74
MATHCOUNTS 2004-2005