[November Calendar]Author(s): Jean McGivney-Burelle and Janet A. WhiteSource: The Mathematics Teacher, Vol. 100, No. 4, Making Mathematics Meaningful(NOVEMBER 2006), pp. 264-269Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27972217 .

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lOVEMBER

NCTM

NATIONAL COUNCIL OF

TEACHERS OF MATHEMATICS

A woman has $2.15 in change in her purse, made up entirely of dimes and quar

ters. Given that there are more quarters

than dimes in her purse, what is the total

number of dimes?

A digit is placed in each empty square in the grid so that each row and each column contains the

digits 1, 2, 3, 4, 5 . What

digit is placed in the square

at the bottom

right corner?

5

What is the last digit d of the 9-digit number 19700019d, given that the

number is prime?

How many 5-digit numbers with all digits nonzero and no digit repeated are divisible

by 25?

8

Suppose that/(x) is a function such that

3/t? + 2/(1 - ) = 2x + 9 for every real number x.

What is the value of/(2)?

A right tri?ngle with integer side lengths a, b, and c satisfies a<b<c and a + c - 49.

What is the area of the right triangle?

10

There are positive integers a, b, and c that

satisfy this system of two equations:

fc2-a2-i2=101

What is

the value of a + b + c?

11

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In the figure, OAB and ODC are sectors of two concentric circles centered at O. The

length of OD is one-third the length of OA. What is the ratio of the area of the region ABCD to the area of the sector ODC?

16

A group of boys and girls took a test Ex actly 2/3 of the boys and exactly 3/4 of the. girls passed the test. If an equal number of boys and girls passed the test, what frac

tion of the entire group passed?

17

How many different real-valued pairs (x,y) satisfy this system of two equations? X

What is the value

of the following product? tan5? ? tanl5? ? tan25? ? tan35? ? tan45?

tan55? ? tan65? ? tan75? ? tan85p

24

25

What is the value of

1 ^3?1<%45?1 ^7? 1 ^3.1 ^5?1 ^7'

18

How many positive real numbers are

solutions to yfx = I jc4 - li ?

19

Each car of a five-car train must be painted a solid color. The only color choices are red, blue, and yellow. If each of these col ors must be used for at least one car, in

how many ways can this train be painted?

26

What is the number of distinct real numbers that have the property that the median of

the five numbers x, 6,4,1,9 is equal to their

mean?

27

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1. 4 dimes. There are at most 8 quar ters. Since there are more quarters than

dimes, there are at most 7 dimes. Thus, the dimes contribute at most 70 cents to the total; so the quarters contribute at

least 215 - 70 = 145 cents to the total. As 215 ends with the digit 5, we also see that there are an odd number of

quarters. It follows that there must be 7

quarters and, hence, 4 dimes.

2. 3. In the left-most column, the digit 5 must be placed in the bottom square. Then there is only one possibility for where the digit 3 can be placed in the bottom row, namely, in the bottom right corner.

3.1. Let N= 19700019d. We know that d cannot be even or 5. Also, if d is equal to either 3 or 9, then will be divisible

by 3, which can be checked by summing its digits. As 197 is an apparent divisor of Nif d = 7, we are left with only the

possibility that d = 1.

4.12. The altitudes drawn from A in AABD and AACD are the same. The

product of the areas is maximized when the product of the lengths of the bases, that is, BD ? CD, is maximized. Letting = BD, we have 4 - = CD. Thus, we

want to know where x(4 -

x) = 4 -

(x -

2)2 is maximized. As 4 - (x -

2)2 cannot be greater than 4 and the value of 4 is attained at = 2, we deduce that we want = 2. One checks that S1 = S2 =

2>/3andS1-52 = 12.

(1/2)(4)(2V3) = S, + S2. Thus the prod uct Sj

? S2 expressed as a function of S^.

p(^)=^(W5-sJ This quadratic will attain its maximum value midway between its x-intercepts. Thus, Sl = S2 = 2V3 and S^S2 = 12.

5. + 3. If r(x) is the remainder, then there is a quotient q(x) such that x2006 - x2005 + ( + l)2 = (x2

- l)q{x) +

r(x), where r(x) is at most first degree. Substitution of = ?1 leads to r(l) =

4 and r(-l) = 2. This leads to a linear

equation and a remainder expressed as

r(x)=x + 3.

Alt?rnate Solution.****-x^ + fr+l)^ x*??s(x

- 1) +

2 + 2x + 1. (Synthetic) division by (jc + 1) yields

The remainder when dividing the quo tient by the remaining divisor of (x

- 1),

using the synthetic division method, will be 1. This is easily shown realizing that there are an odd number of coefficients

equal to 0 between x2005 and x\ Thus, the rational expressi?ns representing the two remainders are

x + 1 and

x'-?

which leads to r(x) - + 3.

6. 36ff. Let r denote the radius of the smaller circle and R the radius of the,

larger circle, as shown. The area of the

nonshaded region can be expressed as

kR2 -Ttr2- n(R2 -

r2). By the Pythago rean theorem, R2 - r2 =

36. Hence, the area

of the shaded region ie*(?2-/)='36*

4 - D B

Alternate solution. The area of AABC -

266 MATHEMATICS TEACHER | Vol. 100, No. 4 ? November 2006

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7. 38. Observe that 86 = 3b2 + 2b+l, so

3b2 + 2b - 85 = 0. Hence, b = 5, and the

required number can be represented as

b2 + 2b + 3 = 25 + 10 + 3 = 38. Alterna

tively, the sum of the base b numbers 321 and 123 is (3b2 + 2b + 1) + (b2 + 22? + 3) =

4(2?2 + 2> + 1). Note that b2 + b = b(b+ 1) is the product of two consecutive integers and, hence, even. Thus, b2 + b + 1 is odd. In base ten, this means that the sum of 86 and the number we want must be divis ible by 4 and not 8. The only choice that satisfies this is 38.

8.420. The 5-digit number must end with 25 or 75. The other digits can be arbitrary but must be distinct from each other, the two right-most digits, and zero. This leaves 7 ? 6 ? 5 = 210 choices for the remaining digits. The answer is 2 ? 210 = 420.

9. 5. Setting = 2 and then = -1 gives 3/(2) + 2/t-l) = 13 and 3/(-l) + 2/(2) = 7. Multiplying through in the first equa tion by 3 and in the second equation by 2 and then subtracting gives 5/(2)

= 25.

Hence,/(2) = 5.

10. 210. By the given information and the Pythagorean theorem, a2 + b2 = c2 =

(49 -

a)2 = 492 - 98a + a2, so that b2 =

49(49 -

2a). We deduce that 49 -2a is a

square. As it is also odd, we must have

so that

49-2a e {l, 9y 25}

a e {24, 20, 12}.

If a = 24, a < b < c, and a + c = 49, which

implies that b is nonintegral, a contra diction. If a = 20, then b = 21. If a = 12, then b = 35. In either case, the area of the triangle is (ab)/2 = 210.

11. 43. As a, b, and c are positive, the number c + a + b is larger than the abso lute value of c - a - b. From

[c + a + b){c - a - b) = c2 - (a + b)2

= c2-a2-b2- 2ab = 101-2-72 = -43

we deduce that c + a + b = 43 and c - a - b = -1.

12. .26. If one box has exactly a white marbles and the other box has exactly b white marbles, then we deduce that

1^20

so that ab = .21 ? 400 = 84. As a and b are integers that are at most 20, we de duce that the set {a, b} is either {6,14} or {7,12}. Since the total number of black marbles is different from the total number of white marbles, we know a +

b 20. Thus, the probability that both marbles drawn are black is

(20 - ?)(20

- b) (20

- 7)(20 -12)

400 ~

400

_(13)(8)_ 400

= .26.

13. 16. Since the remainder is 6, we have > 6. If q is the quotient, we have 2006 =

nq + 6, so that nq = 2000. Thus, is a divisor of 2000 that is greater than 6. It is not hard to see that this is both necessary and sufficient for to satisfy the condi tions in the problem. As 2000 = 24 ? 53, it has (4 + 1)(3 + 1) = 20 positive integral divisors. This includes the divisors 1, 2, 4, and 5, which are all less than or equal to 6. Hence, the answer is 20 - 4 = 16.

14. 6. In each of the nine codes, one of the digits a, b, c, d appears in a correct

position. As there are exactly four pos sibilities for a correct digit in a correct

position, by the Pigeonhole Principle, there must be at least three codes that contain one of a, b, c, and d in a correct

position. As 2 is the only digit that oc curs three times in the same position and this happens in the third position, we deduce that c = 2. Of the six codes that do not have 2 in the third position, there are no more than two occurrences

of the same digit in the same position. It follows that each of a, b, and d must oc cur in its correct position exactly twice

among these six codes. In particular, this gives d = 6.

15. 9. Since the two triangles are similar, we get

DC AC _ AD

DA ~

AB ~

DB

Since AC = 15 and AB = 20, we get

DC = -DA and DB = -DA. 4 3

The last two equations imply

DB-DC = ?DA. 12

Now, DB - DC =CB = 7, so DA = 12 and DC =9.

16. 8. The ratio of the area of the sector O AB to the area of the sector ODC is the

square of the ratio of the length of OA to the length of OD. Hence, the area of sec tor O AB is 9 times as big as the area of sector ODC. Thus, the area of the region

ABCD is 8 times as big as the area of sector ODC.

Alternative Solution. Let OD = and OB = 3x. The area of sector AB can be then expressed as

2 ] 2 and the area of sector ODC can be ex

pressed as 2

2 ^ -\ '

Thus, the area of the region ABCD is the difference of these regions,

80X2

which is 8 times as big as the area of sec tor ODC.

17. 12/17. Let b be the number of boys who took the test and g the number of

girls. We are given that

2b__3g_ 3

~ 4

'

so

b = 9g

Vol. 100, No. 4 ? November 2006 | MATHEMATICS TEACHER 267

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So the desired ratio is

2fc + M 2. M 3 4 J 4

9# 8

3' 21 12

17'

18. 3. Let log denote the logarithm of to any desired fixed base. Then using

the change of base formula and simplify ing, we obtain

log23*log45-log67

log43.1og65.1og87'

log3 t log5 ? log7 log 2 log 4 log 6

'log3 log5 log7 -?-?

^log4 log 6 log 8

log8 = 31og2=3 log2 log2

19. 2. We want to know the number of

positive for which either Vx = 4 - 1

or -Vx = 4 - 1. The graph of y =

4 - 1

intersects the y-axis at (0, -1), and the value of y increases as increases. As

4 grows much quicker than Vx as in

creases, it is not difficult to see that the

graph of y = 4 - 1 intersects the graph

of y = -Vx exactly once. At a larger value of , it intersects the graph of y = Vx ex

actly once. Hence, there are exactly two

solutions to the given equation.

20. 35. As shown, the coordinates of C can be obtained by adding the x-coordi nates of and D and adding the y-coor dinates of and D. Thus, C is the point (30, y + 10). One can compute the area

of the parallelogram as the area of rect

angle XAYC minus the sum of the areas

of the triangles AAYB, AYCB, ACXD, and AXAD. We deduce that the area of the parallelogram is

3 (# + 1 ) \( 30.10 + 10(^ + 10)

2^ +30.10 + 10(?/ + 10)

= 20^-100.

(30,#+l?) C

21. 35/16. Since 2 + a + b = (

- sinl5?)

?

( -

cosl5?), we deduce that a = - sinl50 -

cosl5? and b = sinl5? ? cosl5?. Observe that

a2 = sin215? + 2 sinl5? ? cosl5? + cos215? = l + sin30?=3/2

and

2?=(l/2).2sinl5?cosl5c = (1/2)

? sin30? = 1/4.

We can then deduce that

b2= -

22. 1. Let/(x) = 5 - 6x4 + Ax3 + Bx2 +

Cx + D. The sum of the roots (counted to their multiplicity) of a polynomial of

degree is the opposite of the coefficient of xn~1 divided by the coefficient of xn.

Hence, the sum of the roots of/(x) is 6. Note that/(x) has exactly five com

plex roots (counting again a root to

its multiplicity). The coefficients are

integers and, hence, real; the imaginary roots come in conjugate pairs. Thus, 1 +

i, 1 -

/, 1 + 2i, and 1 - 2i are all roots of

f(x). The sum of the roots of/(x) being 6 implies that the fifth root of/(x) is 2.

As/(x) has a leading coefficient of 1, we

obtain

f{x) = (

_ 2)

? (x - 1 - /)

- ( - 1 + i)

(x-l-2?)-(x-l + 2?).

Thus,

-5+A + ? + C + D: :/(!) : -(-0 - i - (-2?) .

:-4.

(20

Alternate Solution. Given that/( ) has real coefficients, the complex solutions will occur in conjugate pairs. Thus,

fix) = (x4

- 4x3 + llx2 - 14x + 10)(x -

k).

The fourth degree term of/(x) = -6x4

will be formed by the sum of the prod ucts of (-4x3)(x) and (x4)(-fe). Thus, k = 2, leading toA + ? + C + D=l.

23. 12. We have DA + CD = 14; and from the Pythagorean theorem, CD2+BD? =

169 and DA2+BD2 = 225. Subtracting the two equations leads to DA2 - CD2 = 56. Since DA2 - CD2 = (DA + CD){DA-CD),

by substitution, DA2 - CD2 = (14) (DA -

CD) = 56. So, DA-CD = 4, DA = 9, and CD = 5. Therefore, BD = 12.

Alternate solution. AABC is a Hero nian triangle, a triangle having rational side lengths and rational area. Hero's

formula,

Since the area is 600, we obtain y = 35. Therefore, A+? + C + D = 5- 4=l.

Area = yjs(s -a)(s- b)(s

- c),

where 5 = ? (a + b + c), 2

can therefore be used to obtain the area

of AABC and thus the altitude of BD.

24. 4. From the second equation we get that either = 0 or y = ?1. Substituting = 0 in the first equation, we get the

solutions (0, 1) and (0, -1). Next, when

y = 1 we get the solutions (0,1) and

(2,1). Finally, when y = -1 we get the

solutions (0, -1), (-2, -1). So the system has four distinct solutions.

25. 1. Since sin(90? -

x) = cosx and

cos(90?- x) = sinx, we get tan(90?- x) ?

tanx = 1. Applying the above formula for = 5?, 15?, 25?, and 35?, and taking

into account tan 45? = 1, the value of the product is 1.

26. 150. One can paint 5 cars using 3

colors in 35 = 243 ways. Also, one can

paint 5 cars using only 2 out of 3 colors

in 3 ? 25 = 96 ways (there are 3 ways to

pick 2 out of 3 colors). Finally, one can

paint 5 cars using only 1 out of 3 colors

in 3 ? l5 = 3 ways. By the inclusion

exclusion principle, the answer is 243 -

96 + 3 = 150.

268 MATHEMATICS TEACHER | Vol. 100, No. 4 ? November 2006

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27. 3. The median cannot be 1 (there are

at least three numbers in the set larger than 1). For similar reasons, the median is not 9. The median could be 6. Then

4 = 20 + x

6 = x+6+4+1+9

so = 10. Since 6 is the median of the set

{1,4, 6, 9,10} and the mean of the set

is also 6, the number 6 has the described

property. Further, if the median is 4, then

4 = x+6+4+1+9

so = 0. Again, one checks that = 0 has the desired property. Finally, if the median is x, then

x+6+4+1+9 -'

so = 5, and one checks that = 5 has the desired property, too. Thus, there are 3 distinct values for that work.

Alternate Solution. Order the values that are known: {1, 4, 6, 9}. Consider the separate cases, where < 4, > 6, and 4 < < 6. For < 4,

leads to = 0. For > 6,

n 20 + x 6 =

5

leads to = 10. For 4 < < 6,

20 + x x =

5

leads to = 5. Thus, there are 3 distinct values for that work.

28. Nine. Since * 0 and y 0. Solving for y, we get

4x A 16 y

=-= 4 +-. x-4 x-4

So, (x -

4) divides 16, and since 0 and y * 0, (

- 4) must be in the set

{-16, -8, -2, -1, 1, 2, 4, 8, 16}. So, there are nine integer solutions: (-12, 3), (-4, 2), (2, -4), (3, -12), (5, 20), (6, 12), (8, 8), (12, 6), and (20, 5).

29.260. There are two cases to consider: when the north and south plots have differ ent kinds of flowers and when they have

the same kind of flower. There are (5) (4) =

20 ways to choose distinct kinds of flowers for the north and south plots. Also, if dis tinct kinds of flowers are planted on plots north and south, there are 3 kinds of flow ers left to plant in plots west and east. So, in the case when distinct kinds of flowers are planted on plots north and south, there are a total of (20) (3) (3) = 180 ways to plant flowers in the plots, so that flowers in plots sharing a common edge are different. Next, there are 5 ways to select the same kind of flower for the north and south plots. In this case, there are 4 kinds of flowers left to

plant in plots west and east. So, in the case

when the same kind of flowers are planted on plots north and south, there are a total of (5) (4) (4) = 80 ways to plant flowers in the plots, so that flowers in plots which share a common edge are different. So, the total number of choices is 260.

30. 396. 605 has prime factorization 2103555. There are 11 powers of 2

(2? through 210) that are among the factors of 605, 6 powers of 3 and 6 powers of 5.

Thus, the total number of factors is repre sented by the product 11x6x6 = 396. ?>

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