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Math Contest Round Two Spring 2019 Solutions

Prepared by

Marc Frodyma (Stepping Stones Learning Center, San Marino CA) (Formerly Physics and Math, San Jose City College)

Email: marc.frodyma@gmail.com

Chungwu Ho (Math, Evergreen Valley College)

Gregory Melblom (Math, Evergreen Valley College, Retired)

Sithparran Vanniasegaram (Math, Evergreen Valley College) Problem 1: Solution by MF. This is a problem on conversion of units: (16800 mm) x (1 cm / 10 mm) x ( 1 in. / 2.54 cm) x (1 ft. / 12 in.) = 55.1 ft. Notice that each factor, such as (1 ft. / 12 in.) is being treated as unity and the units are treated as algebraic quantities that can be cancelled. The idea is to cancel every unit except feet. Now since we need slightly more than 55 feet, then the smallest whole number of feet we need is 56, and the answer is (C). Problem 2: Solution by MF. If a set has n elements, then there is a theorem that says the number of subsets of the set is 2n. This theorem can be proven by using mathematical induction. See for example, Discrete Mathematics With Applications, S. Epp, 4th Ed., (Brooks/Cole, 2011) p369. However, one subset of a given set is the set itself, which is not a proper subset, so we have to subtract that one. So the number of proper subsets of the set {1, 2, 3, 4} is 24 – 1 = 15, and we have M = 15. The number of distinct differences is by inspection: ±1,±2,±3, which gives six distinct differences obtained by subtracting distinct elements of the given set. So we have N = 6, M + N = 15 + 6 = 21, and the answer is (C). Problem 3: First solution by MF. See alternate solution by CH below. Let us start by counting the number of ways to break the $50 bill using no tens or twenties, only ones and fives: In the following notation with ordered pairs, let the first element be the number of ones and the second element the number of fives. Then to make up $50 with ones and fives, we have: (50, 0) (45, 1), (40, 2), …, (0, 10). There are 11 ways here. (0 – 10 inclusive).

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Now let us count the number of ways using no twenties and one ten. We use ordered pairs defined as above: (40, 0), (35, 1), (30, 2), …, (0, 8), which is 9 ways. Similarly, with no twenties, and two, three, four and five tens respectively, we have: 7 ways, 5 ways, 3 ways, and 1 way with five tens. So the total number of ways with no twenty is: 11 + 9 + 7 + 5 + 3 + 1 = 36 ways without using a twenty. Now let us use one twenty, and start with no tens: (30, 0), (25, 1), (20, 2), …, (0, 6), and we have 7 ways. The pattern is the same with one, two, and three tens: 5 ways, 3ways, and 1 way with one, two, and three tens respectively. So we have 7 + 5 + 3 + 1 = 16 ways to break the $50 bill using one twenty. Add this to the previous 36 ways to get 36 + 16 = 52 ways. Now we use two twenties, and with no tens we have (10, 0), (5, 1) and (0, 2). Three ways. Finally, with two twenties and one ten, there is only one way. Adding up all the ways, we get 52 + 3 + 1 = 56, and the answer is (D). Problem 3: Solution by CwH. How many ways to make a sum of $50 by using only $20, $10, $5 and $1’s. We will consider different cases. Note that in making up $50, we need only specify how many $20, $10, $5 are used, with the remaining sum made up by $1’s. Suppose two twenties are used. We need to see how many ways to make the remaining $10 by using only $10, $5, and $1’s. There are 4 different ways to do this: 1) a single $10 is used, 2). No $10 is used (there are two ways in this case: one or two $5’s, and 3); no $10 nor $5 ae used (there is only one way in this case: all are $1’s. This shows that a total of 4 different ways in this case. Suppose only a single $20 is used, then we need to make the remaining 30 by using only 10, 5, and 1’s. We could use 1) 3($10) (one way); 2) 2($10) (three ways: zero, one of two $5’s), 3) a single $10 (5 ways: 0, 1, 2, 3, or 4 of $5’,); 4) No $10 is used (7 ways -- using from 0 to 6 $5’s). Total in this case: 16 ways. Suppose no $20 is used. We could use 1) 5($10) (one way); 2) 4($10) (3 ways: 0, 1 or 2 $5’s); 3) 3($10) 5 ways: 0 to 4 $5’s); 3) 2($10) (7 ways: 0 to 6 $5’s); 4) One $10 (9 ways: 0 to 8 $5’s), and 5) No $10 (11 ways: 0 to 10 $5’s). Total in this case. 1+3+5+7+9+11 = 36 ways. Total Number of All the Cases = 4+16+36 = 56 ways. The Answer is (D).

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Problem 4. Solution by CwH. Let ABC be the isosceles triangle with the sides AB and AC of 40 and the side BC 48. L1 and L2 be the perpendiculars of the sides AB and AC, respectively. Then L1 and L2 will intersect at the line AD at the center O of the circle (without prejudging the issue, we have drawn the graphs below for both cases that the center of the circle is both inside and outside the triangle). Since the length AC is 40 = 5x8 and the length DC is 24 = 3x8, by the 3-4-5 triangle, the length AD = 4x8 = 32. Now use the similar triangles ADC and AEO, we see that !"

!"= !"

!" . Thus, the radius

𝐴𝑂 = 𝐴𝐸 ∙ !"!"= 20 ∙ !"

!"= 25. The answer is (E).

A

C

D O

L L

E

1 2

B 24

20

20

A

C

DO

L L

E

1 2

24

20

20

24

24

B

Note that since the radius of the circle is shorter than the distance AD, the center of the circle is inside the triangle and the first graph shows what will happen. Problem 5: Solution by CwH. Let M be the number of digits when the largest prime factor of 2019 is written in the binary expansion and N be the number of digits when 2019 is written in the base 16. Find M – N. First, note that 2019 = 3 x 673 and that 673 is a prime. We may see this as follows: 673 is clearly not divisible by 3 and 5. Using a calculator we can easily show that it is not divisible by 7, 11, 13, 17, and 23 either. (Since 673 ≈ 25.94 we need only check primes through 23). Now use the standard method to write 673 in the binary form: 673 = 1,010,100,001 in binary. Thus, M = 10. In base 16, we can write: 2019 = 7 x 162 + 14 x 16 + 3. Since (14) in base 10 is (E) in base 16, we have 2019 = 7E3 in base 16 and N = 3. So M – N = 7, and the answer is (A). Problem 6: Solution by CwH. Triangle ABC has vertices (8,8), (6,4), and (10,7). Find the sum of the lengths of the three altitudes of this triangle, rounded to the nearest tenth. From the coordinates of the points A and B, we can easily find an equation for the line through A and B. This can be done as follows: The slope of this line is 𝑚 = !!!

!!!= 2.

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Then use the point-slope formula to get the equation as 𝑦 − 8 = 2(𝑥 − 8) or 2𝑥 − 𝑦 −8 = 0. We can then use the formula 𝑑 = |!!!!!!!!!|

!!!!! for the distance d from a point (x1,

y1) to a line 𝐴𝑥 + 𝐵𝑦 + 𝐶 = 0 to compute the length of the altitude from the vertex C(10, 7) to the line 2𝑥 − 𝑦 − 8 = 0 through AB. This length is |! !" !!!!|

!!!(!!)!= !

!= 5.

Similarly, the line through AC has slope 𝑚 = !!!

!!!"= −1/2, and the line is 𝑦 − 8 =

− !!(𝑥 − 8) or

𝑥 + 2𝑦 − 24 = 0. The altitude on this line has the length |!!! ! !!"|!!!!!

= !"!= 2 5.

Finally, the line through BC has slope 𝑚 = !!!

!"!!= !

! and the line is 𝑦 − 4 = !

!(𝑥 − 6) or

3𝑥 − 4𝑦 − 2 = 0, and the altitude on this line has the length |! ! !! ! !!|!!!(!!)!

= !"!= 2.

Thus, the sum of the three altitudes is 3 5+ 2 ≈ 8.71. The answer is (A).

x0 2 4 6 8 10

2

4

6

8

y

A

B

C

Problem 7: First solution by MF. Since (x + 1) is a factor of the given polynomial, then x = -1 is a root. When the given polynomial is divided by (x + 1), the quotient will be a quadratic, whose two roots are the desired non-real roots. Recall from the quadratic formula that the sum of the two roots of a quadratic equation is –(b/a), where a is the coefficient of the quadratic term, and b is the coefficient of the linear one. So we start a long division of the given polynomial by (x + 1) in order to find the coefficients a and b:

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The

long division need not be completed, because we see from the figure that a = 2 and b = -1. This gives the sum of roots: –(b/a) = ½, and the answer is (D). Strangely, the condition d + c = 29 seems not to be needed. Problem7:SolutionbyGM.Polynomial: 2x3 + x2 + cx + d. The roots of this polynomial is the same as the roots of x3 + (1/2)x2 + (c/2)x + (d/2) Note that: (x - a)·(x - b)·(x - c) = x3 - (a + b + c)·x2 + (a·b + a·c + b·c)·x - a·b·c where a, b, and c are the three roots of the cubic equation. Since the coefficient of x2 is minus one-half, the sum of all three roots of this polynomial is (-1/2). Since the polynomial is divisible by (x + 1), then -1 is a root of the cubic. Therefore the sum of the remaining two roots is 1/2. Answer is D: 1/2. Problem 8: Solution by CwH. Let N be the smallest integer greater than 2 for which NN-1 is not a square. Trying N = 3, 4, 5, and 6, we see that 6 is the first such integer. So the given polynomial takes the form: 5x4 + bx3 + cx2 + dx + 6. By the rational root theorem, the possible roots are:

±1,±2,±3,±6,±15 ,±

25 ,±

35 ,±

65 (1)

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Since b, c, and d can be any integers, we will show that for each rational number in the list above, there is a choice of b, c, and d such that the number could be a root of the polynomial. Recall by the rational root theorem that each number in the list above has the form p/q, where p is a positive factor of 6 and q = +/- 1 or +/- 5. Substitute x = p/q into the equation 5x4 + bx3 + cx2 + dx + 6 = 0, and the result can be put into the form:

5𝑝! + 𝑏𝑝!𝑞 + 𝑐𝑝!𝑞! + 𝑑𝑞𝑞! = −6𝑞! (2) where all denominators have been cleared by multiplying both sides by q4. Since b, c, and d are arbitrary, we will show that for every rational candidate in (1) above, there is a choice of b, c, and d that allows for it to be a root. Equation (2) can be re-written as:

𝑝! 5𝑝 + 𝑏𝑞 + 𝑐𝑝!𝑞! + 𝑑𝑝𝑞! = −6𝑞! (3) Since q = +/- 1 or +/- 5, we can choose b to make the coefficient of p3, (5p + bq) in equation (3) zero. Now let c = 0. Since p is a factor of 6, we can also choose a value of d such that dpq3 = -6q4. With these choices of b, c, and d, we see that all of the numbers in the list in (1) above are possible rational solutions to equations (2) or (3). Since each number in the list (1) is in +/- pairs, and there is an even number of terms in the list, the product of them all should be positive. So we just compute the product as: (1 x 2 x 3 x 6 x (1/5) x (2/5) x (3/5) x (6/5))2 = (1296 / 625)2, with the approximate value of 4.30, and the answer is (D). Problem 9: Solution by MF. Note that it is not possible for X to be a knight, because X claims to be a spy, which would be a false statement. So answers C and D are immediately eliminated. Suppose X is a spy: Y says “X is telling the truth”, which is a true statement under current assumptions, so Y must be the knight. So Z must be the knave. But Z says “I am not a spy”, which is a true statement, contradicting the result that Z is a knave. So X is not a spy, and answers A and B are eliminated, leaving answer (E). But let us check that answer E does not also lead to a contradiction: Assume X is a knave. X says “I am a spy”, a false statement that is consistent with X being a knave. Y says “X is telling the truth”, which is a false statement, forcing Y to be a spy, since X already has the honor of being the knave. So Z must be the knight. Z says: “I am not a spy”, which is true and consistent with Z being the knight, no contradiction and the answer is (E).

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Note that it could have turned out that all five answers gave contradictions, so that they were all wrong, but then there would have been a mistake in the formulation of the problem. Such mistakes have occurred. Problem 10: Solution by CwH. Let the first transmission, ∙−−∙− , be denoted F, and suppose that transmission B is correct. Then transmissions C and D disagree with B in three places, so they both have three errors. This is not allowed, so B cannot be correct. Similarly, if we assume that transmission C is correct, then both B and F would have three errors, so C cannot be correct. Also transmission D cannot be correct, because then both B and C would have three errors, as we saw above by assuming B was correct. This leaves transmission A. If A is correct, then B has four errors, C has one error, D has three errors, and F has two errors. These are all distinct numbers of errors and as required, no transmission has five errors, so A is correction and the answer is (A). Problem 11: First solution by CH. Let A, B, and C be the three forbidden squares on the 5 x 5 checkerboard shown in the problem. All paths must end at the square in the lower left corner, which is the target square. We are only allowed to move one square at a time either down or left. The checkerboard is shown in the figure below:

First, let us ignore the forbidden squares and allow paths through them. From the beginning at the upper right corner, we need to go down 4 squares and to the left by 4 squares to reach the target. This can be done if we go down (D) and left (L) four times each in any order. For example, the sequence of steps LLDDLDLD would be one way to reach the target, ignoring the forbidden squares, and DDLLDLDL would be another way. The number of ways to reach the target can be computed by choosing four steps out of the eight steps needed to be D in any order. This number is called the combination of eight objects, chosen four at a time in any order, and it has the notation and formula:

𝐶 8,4 =8 ∙ 7 ∙ 6 ∙ 51 ∙ 2 ∙ 3 ∙ 4 = 70

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We need to subtract from this total the routes which go through one or more forbidden squares. Note that there are two ways to go from the starting square to square A in the figure, either LD or DL. Also, the number of paths from square A to the target square can be computed by the same reasoning as above: We need three D steps and three L ones out of six total steps. The number of ways to do this is C(6, 3):

𝐶 6,3 =6 ∙ 5 ∙ 41 ∙ 2 ∙ 3 = 20

But since there are two ways to reach square A, the total number of paths we must subtract from 70 is 2 x 20 = 40. So we have 30 paths that avoid square A. Some of these paths go through one or both of forbidden squares B and C. Now we count the number of paths going through square B only. There are four ways to reach square B from the start: LLLDD, LLDLD, LLDDL, and DDLLL. Also, there are three ways to get from square B to the target at the lower left corner: DDL, LDD, and DLD. So there are 4 x 3 = 12 paths going through square B only. We subtract these from the 30 paths above and get 18 remaining paths. Now we subtract the number of paths that go through point C only. There are only two ways to reach square C from the start at the upper right: LDDD and DDDL. To get from square C to the target square, we need one D step, selected in any order from a group of four steps. This is the combination C(4, 1) = 4. Since there are two ways to reach square C from the start, we have 2 x 4 = 8 paths that go through square C only. Finally, note from inspection that it is not possible, using only D and L steps, to go through both squares B and C. It follows that the total number of paths avoiding the three forbidden squares is: 70 – 40 – 12 – 8 = 10 paths, and the answer is (D). Problem 11: Solution by GM. Consider the figure below: 1 1 1 1 1

3 2 1 x 1

3 x 2 1 1

5 2 2 x 1

10 5 3 1 1 Place a 1 in the upper right corner of the 5x5 square. From there, use this rule. The number in a given square will be the sum of the numbers immediately above and to the right of the given square. If there is no number above or two to the right of the given square, then use what number is available above or two to the right. After the numbers are placed in the 5x5 square, the number in the lower left is

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the answer. This works because the number in a given square is the sum of the number of paths leading to it from above and to the right. Answer is D: 10 Problem 12: Solution by CwH. There is a theorem known as De Moivre’s Theorem, according to which, we have the equation:

cos 𝑡 + 𝑖 sin 𝑡 ! = cos𝑛𝑡 + 𝑖 sin𝑛𝑡 where i is sqrt (-1). If we expand the left side of this equation using the Binomial Theorem, then the cos(nt) will be the real part of the expansion and sin(nt) the imaginary part. So to obtain the coefficient of (cos(t))3 in the expansion of cos(8t), we use n = 8 in the above equation and take the real part of the expansion, which becomes: cos 8𝑡 = cos 𝑡 ! − cos 𝑡 ! sin 𝑡 ! + cos 𝑡 ! sin 𝑡 ! − cos 𝑡 ! sin 𝑡 ! + sin 𝑡 !

Note that all powers in the expansion are even. Since (sin(t))2 = 1 – (cos(t))2, we see that any even power of sin(t) will also yield only even powers of cos(t) when the sines in the expansion of cos(8t) are replaced by cosines. Therefore, we cannot generate any odd powers of cos(t) in the expansion of cos(8t), and the coefficient of (cos(t))3 must be zero. The answer is (C). Problem 13: Solution by CwH. First Note that the line through the point (7,8) and parallel to the vector has slope -4/3. By the point-slope form of a line, we have the equation:

𝑦 − 8 = −43 𝑥 − 7

or

4𝑥 + 3𝑦 − 52 = 0 Now we have to find the minimum distance between this line and a given point. Let us derive a formula for this distance. Consider the figure below, which shows a given line with equation: Ax + By + C = 0, and a given point P not on the line. We wish to find an expression for the distance d, shown as the dashed line in the figure. The point Q is an arbitrary point on the given line, and the vector a points along the line, while vector b points from Q to P.

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We can choose any magnitude we wish for vector a, as long as it points along the line. A convenient choice is: a = . From elementary trigonometry, we see that the desired distance d can be written as:

𝑑 = 𝑄𝑃 sin𝜃 =𝑎 𝑄𝑃 sin𝜃

𝑎

where the vector QP can be written as: . By the definition of the cross product (a x b), and using the Pythagorean Theorem to express the magnitude of vector a, we have:

𝑑 =𝑎×𝑏𝐴! + 𝐵!

Since the given line and point are in the xy-plane, only the z-component of the cross product will be non-zero, by the right hand rule. Also, using the definition of the cross product in terms of the components of the vectors, we have the result:

𝑎×𝑏 = 𝑎!𝑏! − 𝑎!𝑏! = 𝐵 𝑦! − 𝑦! + 𝐴 𝑥! − 𝑥! But recall that point Q is on the given line, so we have: AxQ + ByQ = - C. We make this substitution into the last equation above, with the result:

𝑎×𝑏 = 𝐴𝑥! + 𝐵𝑦! + 𝐶 and then the final result for the distance is:

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𝑑 =𝐴𝑥! + 𝐵𝑦! + 𝐶

𝐴! + 𝐵!

Now back to the problem. Recall that we have the line: 4x + 3y – 52 = 0, and we want to find the shortest distance from this line to the given circle in the problem. From the given equation, the center of the circle is at the point (-3, 2). We use the distance formula above to calculate the distance from the given line to the center of the circle with the result:

𝑑 =4 −3 + 3 2 − 52

4! + 3!=585

Since this distance is greater than the radius (sqrt(3)) of the circle, the perpendicular line from the center of the circle to the given line must cut through the circle at some point which is the point on the circle closest to the given line. To get the distance from the given line to the nearest point on the circle, we subtract the radius of the circle from the distance to the center of the circle. This distance is:

𝑑!"# =585 − 3 ≈ 9.87

and the answer is (B). Problem 14: First solution by MF. Recall from elementary school the concept of division with remainder. For example, if you divide seven by three, we have: 7 = 3 (2) + 1, and we say the quotient is 2, the divisor is 3, and the remainder is 1. Notice that the possible remainders are only 0, 1, or 2 if the divisor is three, because if the remainder was three, we could just increase the quotient by one and the actual remainder is zero, etc. This concept helps us save work on this problem as follows: Using a divisor of three, any number a can be expressed as: a = 3q + r, where q is the quotient and r is the remainder, which, as discussed above, is either zero, one, or two. Now if we square the number, we have: a2 = 9q2 + 6qr + r2. Notice that every term has a factor of three except r2, but since r = 0, 1, or 2, we see that r2 = 0, 1, or 4. But the number 4 can be written as 4 = 3 + 1 and the extra “3” can be incorporated into the quotient. This means that the square of any number can only leave remainder zero or one when divided by three. Now let ra and rb be the remainders when a2 and b2 are divided by three. Also let qa and qb be the corresponding quotients. Then the equation:

𝑎! + 𝑏! = 2019

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can be written as: 3 (qa + qb) + ra + rb = 3 (673) + 0. In order to satisfy this equation, it must be true that: ra + rb = 0 or 3. But as discussed above, the two remainders can each only be zero or one, so we must have: ra = rb = 0. This means that a2 and b2 must be multiples of three. I claim it follows that a and b must then also be multiples of three. Let us show this: Towards contradiction, assume that a and b are not multiples of three, while their squares are such multiples. So we have either a = 3q + 1 or a = 3q + 2. Then we have either a2 = 3 (3q2 + 2q) + 1 or a2 = 3 (3q2 + 4q + 1) + 1. In either case, a2 is not a multiple of three, since it has remainder one, contradicting the assumption that a2 is a multiple of three with remainder zero. It follows that a and b are multiples of three, and we can define m and n such that a = 3m and b = 3n. Then the equation in the problem becomes:

3𝑚 ! + 3𝑛 ! = 3 ∙ 673 ! Now we can divide both sides by 9 with the result:

𝑚! + 𝑛! = 673! and we can look for values of m and n satisfying this equation. Then the desired values for this problem will be a = 3m and b = 3n. We will again use division with remainder with divisor three to try to save some work in the search. Notice that 673 leaves remainder one when divided by three. Since a squared number can only leave remainder zero or one, we must have m leaving remainder one and n leaving remainder zero (or vice versa, since the equation is symmetric with respect to interchange of m and n). Accordingly, let us assume that n leaves remainder zero and write the equation as:

𝑚! = 673! − 𝑛! and we try multiples of three below 673 for n in this equation, beginning with n = 672 and working downwards in multiples of three while looking for a perfect square on the left side. I carried out this procedure and found the following values of m and n that satisfy the last equation above: n = 552, and m = 385. This gives a = 3 (385) = 1155, and b = 3 (552) = 1656, and then a + b = 1155 + 1656 = 2811, and the answer is (E).

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Problem 14: Solution by GM. This one seems to require a brute force technique. Let a and b be positive integers with a2 + b2 = 20192 . Find a + b. When divisor 5 is being used and the terms in the equation are replaced by the remainders they leave when divided by five, we say that the equation is written Modulo 5. So in this case, the equation becomes a2 + b2 = 42 = 1 modulo 5, since 2019 itself leaves remainder four when divided by 5, but 42 = 16 leaves remainder one. This is what is meant by the equation 42 = 1 (It is true modulo 5). Since all squares modulo 5 are 0, 1, and 4, this means the only combination that will work is [a2 = 0 and b2 = 1 modulo 5] or [a2 = 1 and b2 = 0 modulo 5]. Since the problem is symmetric with respect to interchange of a and b, we will assume a2 = 0 and b2 = 1 modulo 5, which means a is a multiple of 5 and b = 1 or 4 modulo 5. So we have established that a must leave remainder zero and b leaves remainder either 1 or 4 when divided by five. Now examine the problem using three as the divisor. Modulo 3, the problem becomes a2 + b2 = 02 = 0 modulo 3. Since all squares modulo 3 are 0 or 1, this means the only combination that will work is [a2 = 0 and b2 = 0 modulo 3], which means a and b are both multiples of 3. But note that we established above that a is also a multiple of five. This means a is a multiple of 15. This can be used to limit the number of values used for a to find the value of b that works. Proceeding in this manner yields the following results for a and b: (a, b) = (1155, 1656) Answer is E: a + b = 2811 Problem 15: Solution by GM. The three given equations are listed below: 2x − 6y − 8z = 15 –8x − 8y + 6z = –65 x − 19y − 17z = 5 Twice the 3rd equation minus five times the 1st equation produces the middle equation. Furthermore the 1st equation and 3rd equation are independent, as seen by comparing ratios of corresponding coefficients. So we have two independent equations for the three variables x, y, and z. Combining these two equations yields an equation of the form: Ax + Bx = C, which is the equation of a line. Therefore the answer is (B), a line.

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Problem 16: Solution by CwH. When the graph of a function is symmetric about the y-axis, it means that f(x) = f(-x), or the function is “even”. It follows that the coefficients of any odd powers of x in the function must vanish. This gives b = 0. Since the graph of a quadratic function is a parabola, we see that the vertex of the function must be at x = 0 on the y-axis. Since we are given that the function has a maximum value, it follows that the parabola must open downwards, and the maximum value occurs at the vertex at x = 0. This yields c = 4. The figure below shows the resulting graph of the function:

Recall that an equilateral triangle has 60 degree angles, so the y-axis divides the triangle into two 30-60-90 ones. From the usual proportions of such a triangle, we can find the value of x, which is one-half the side length of the equilateral triangle, using the Pythagorean Theorem:

𝑥! + 16 = 4𝑥! with the result:

𝑥 =43

So one of the x-intercepts is at this value on the positive x-axis. But with the above results, the function is given by:

𝑓 𝑥 = 𝑎𝑥! + 4 so we have

0 =163 𝑎 + 4

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which gives a = -3/4. It follows that the sum: a + b + c = -3/4 + 0 + 4 = 13/4, and the answer is (D). Problem 17. Solution by CwH. We examine each function in turn. The first one g(x) = ln(ex), is defined for all real numbers since ex cannot be negative or zero. In fact, since the natural log and exponential functions are inverses, g(x) = x, which is manifestly continuous and one-to-one. From the definition of absolute value, we see that h(x) = x2 for x greater than or equal to zero, and h(x) = -x2 for negative x. If we put these two pieces together in a single graph, we have a function that is continuous everywhere and increasing, so it is one-to-one also. We have two functions satisfying the desired conditions so far. The function k(x) = x2 is continuous everywhere but since it is an even function, k(x) = k(-x) and it is not one-to-one. The function m(x) has an infinite discontinuity at x = -1. For the function n(x), note that n(0) = 0 and the n(x) goes positive as x increases. But as x goes to positive infinity, n(x) approaches zero again. This means that for any value of n(x) with x > 0 there must be at least two values of x that satisfy that value of n(x) and n(x) cannot be one-to-one, although it is continuous everywhere. Since the sine function, p(x), oscillates between positive and negative one, it cannot be one-to-one, although it is also continuous everywhere. Forthefunctionq(x),sinceitiswell-knownthatthefunctionf(x)=tanxisacontinuous,increasingfunctionfromtheinterval(-pi/2,pi/2)itsinversefunctionq(x)=arctan(x)isalsoacontinuous,increasingfunctionfromthereallinetotheopeninterval(-pi/2,pi/2).Toseethattheinversefunctionofanincreasingfunctionisincreasing,justobservethatx1<x2ifandonlyify1<y2.Inparticular,q(x)isone-to-one.As for the function r(x) =1/(|x|+1), it is obviously continuous on the entire real line since its denominator is never zero. To see it is one-to-one, first consider the case when x > 0. The function can then be written as r(x) = 1/[ 1 + (1/x)]. As x increases, the denominator decreases, and hence, the value r(x) increases. As for the case x < 0. Since r(- x) = - r(x), the graph of r(x) is symmetric about the origin, an odd function. The graph of r(x) is increasing on the entire real line and thus, it is one-to-one. So we see that four of the eight given functions are continuous everywhere and one-to-one, and the answer is (C).

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Problem 18: Solution by CwH. The arithmetic sequence is constructed by starting with the first term, m and adding the common difference d to each term in succession. This gives: {an}: m, m+d, ..., m+99d; Its sum is then 100m + [99(100)]/2 = 100m + 4950d. The geometric series is constructed, starting with the first term k and then multiplying each term successively by 2, yielding: {gn}: k, 2k, ..., (29)k}; Its sum is k[(210) -1] / (2-1) = 1023k Now these two sums are equal: 100m + 4950d = 1023k, or 100m = 1023k -4950d = 33[31k - 150d]. Notice that the right side of the last equation above is divisible by 33, and so the left side, 100m, must be also. Since 100 is not divisible by 33, then m must be. Thus, m is divisible by 33 and the answer is (E). Problem 19. Solution by CwH. Recall the simple formula: Distance = (Rate) (Time). In this problem we know the total time spent, so it makes sense to break up the hike into the time spent on each type of terrain: Level (L), Uphill (U), and Downhill (D). Since we are given the hiking speed on each type of terrain, we can add up the time spent on each type using: Time = (Distance) / (Speed). We let L, U, and D be the distance of each type of terrain, level, up, and down respectively. So the total time spent on the outward trip is: (L/4 + U/3 + D/6). Now on the return trip, the level terrain remains level, but the uphill terrain becomes downhill and vice versa, but the length of each type of terrain does not change. So for the return trip, we interchange the speeds between the uphill and downhill sections with respect to the outward trip. Then we can write the total round trip time as: (L/4 + U/3 + D/6) + (L/4 + U/6 + D/3) = 8 Clear the denominators by multiplying through by 12, with the result: 3L + 4U + 2D + 3L + 2U + 4D = 96.

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This simplifies to: L + U + D = 16. But since each type of terrain was traveled twice, the total length of the hike was twice sixteen, or 32, and the answer is (B). Problem 20:Solution by SV. There are seven paths to victory for the shooter: (1): 7 or 11 on the first roll, (2): 4 on the first roll and 4 on a later roll, ........., and (7) 10 on the first roll and 10 on a later roll. Path 1: 7 or 11 on the first roll Sum of 7 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} - 6 outcomes Sum of 11 = {(5,6), (6,5)} - 2 outcomes P(win on Path 1) = P(7) + P(11) = 6/36 + 2/36 = 8/36 Path 2: 4 on the first roll and 4 on a later roll Sum of 4 = {(1,3), (2,2), (3,1)} - 3 outcomes P(4 on the first roll) = 3/36. The game continues after the first roll if the shooter doesn't roll a 4 or a 7. P(no 4 or 7) = 1 - (3/36 + 6/36) = 27/36. P(win on Path 2) = 3/36*3/36 + 3/36*27/36*3/36 + 3/36*27/36*27/36*3/36 +..... = 3/36*3/36 *(1 + 27/36 + (27/36)^2 + (27/36)^3 +...........) = 9/1296 *(1/(1 - 27/36)) = 9/1296 / (9/36) = 36/1296 = 1/36 Path 3: 5 on the first roll and 5 on a later roll Sum of 5 = {(1,4), (2,3), (3,2), (4,1)} - 4 outcomes P(5 on the first roll) = 4/36.

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The game continues after the first roll if the shooter doesn't roll a 5 or a 7. P(no 5 or 7) = 1 - (4/36 + 6/36) = 26/36. P(win on Path 3) = 4/36*4/36 + 4/36*26/36*4/36 + 4/36*26/36*26/36*4/36 +..... = 4/36*4/36 *(1 + 26/36 + (26/36)^2 + (26/36)^3 +...........) = (16/1296) *(1/(1 - 26/36)) [sum of infinite geometric series] = (16/1296) / (10/36) = (1/81) / (5/18) = (1/81) *(18/5) = 2 / 45 Path 4: 6 on the first roll and 6 on a later roll Sum of 6 = {(1,5), (2,4), (3,3), (4,2), (5,1)} - 5 outcomes P(6 on the first roll) = 5/36. The game continues after the first roll if the shooter doesn't roll a 6 or a 7. P(no 6 or 7) = 1 - (5/36 + 6/36) = 25/36. The game continues after the first roll if the shooter doesn't roll a 6 or a 7. P(win on Path 4) = 5/36 *5/36 + 5/36*25/36 *5/36 +....... = 5/36*5/36*(1 + 25/36 + (25/36)^2 + (25/36)^3 + .......) = 25/1296*(1/(1-25/36)) = 25/1296*(1/(11/36)) = 25/1296 *(36/11) = 25/396 Path 5: 8 on the first roll and 8 on a later roll P(8) = P(6). So, P(win on Path 5) = P (win on Path 4) = 25/396.

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Path 6: 9 on the first roll and 9 on a later roll P(9) = P(5). So, P(win on Path 6) = P (win on Path 3) = 2 /45. Path 7: 10 on the first roll and 10 on a later roll P(10) = P(4). So, P(win on Path 7) = P (win on Path 2) = 1/36. So, P(shooter wins) = sum of probabilities on the seven paths to victory = 8/36 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36 = 10/36 + 4/45 + 25/198 = 5/18 + 25/198 + 4/45 = 55/198 + 25/198 + 4/45 = 80/198 + 4/45 = 40/99 + 4 / 45 (GCD = 495) = 200/495 + 44/495 = 244/495 So the answer is (C).