Children Sharing Money

Al, Bee, Cecil, and Di have \$16,\$24,\$32, and \$48, respectively. Their father proposed that Al and Bee share their wealth equally, and then Bee and Cecil do likewise, and then Cecil and Di. Their mother’s plan is the same except that Di and Cecil begin by sharing equally, then Cecil and Bee, and then Bee and Al. Determine the number of children who end up with more money under their father’s plan than under their mother’s plan.
Source: NCTM Mathematics Teacher, August 2006

Solution
Father’s plan
(\mathrm{Al}+\mathrm{Bee})/2=(16+24)/2=20
(\mathrm{Bee}+\mathrm{Cecil})/2=(20+32)/2=26
(\mathrm{Cecil}+\mathrm{Di})/2=(26+48)/2=37
Al, Bee, and Cecil end up with more money.
Mother’s plan
(\mathrm{Di}+\mathrm{Cecil})/2=(48+32)/2=40
(\mathrm{Cecil}+\mathrm{Bee})/2=(40+24)/2=32
(\mathrm{Bee}+\mathrm{Al})/2=(32+16)/2=24
Only Al ends up with more money.
Two children end up with more money under father’s plan than under mother’s plan.

Answer: 2

Advertisements
Posted in Problem solving | Tagged , | Leave a comment

Cottages on a Straight Road

There are four cottages on a straight road. The distance between Ted’s and Alice’s cottages is 3 km. Both Bob’s and Carol’s cottages are twice as far from Alice’s as they are from Ted’s. Find the distance between Bob’s and Carol’s cottages in kilometers.
Source: NCTM Mathematics Teacher, August 2006

Solution
image
Suppose the four cottages of Bob, Ted, Carol, and Alice are located on a straight line with respective coordinates  b,t,c, and a such that 0<b<t<c<a.
Recall that |x-y|= distance between real numbers x and y on the number line.
image
|t-a|= distance between Ted’s and Alice’s cottages
|b-a|= distance between Bob’s and Alice’
s cottages
|c-a|= distance between Carol’s and Alice’s cottages
|c-t|= distance between Carol’s and Ted’s cottages
Given that |t-a|=3,|b-a|=2|b-t|, and |c-a|=2|c-t|, we want to find |b-c| the distance between Bob’s and Carol’s cottages.
|t-a|=3 implies a-t=3\qquad\qquad\:\: (1)
Calculate |b-t| the distance between Bob’s and Ted’s cottages
|b-a|=2|b-t| implies a-b=2(t-b)
Add (b-t) to both sides of the equation
a-b+(b-t)=2t-2b+(b-t)
a-t=t-b
Substitute the value of (a-t) from Eq. (1)
3=t-b which implies |b-t|=3\qquad (2)
Calculate |t-c| the distance between Ted’s and Carol’s cottages
|t-c|+|c-a|=3
|t-c|+2|c-t|=3
3|t-c|=3
|t-c|=1\qquad\qquad (3)
Calculate |b-c| the distance between Bob’s and Carol’s cottages. From Eqs. (2) and (3)
|b-c|=|b-t|+|t-c|=3+1=4 km

Answer: 4 km

Posted in Problem solving | Tagged , , | Leave a comment

Partners in Integers

Two integers are said to be partners if both are divisible by the same set of prime numbers. Find the number of positive integer less than 25 that have no partners less than 25.
Source: NCTM Mathematics Teacher, August 2006

Solution
24=2^3\!\cdot\! 3
18=2\!\cdot\! 3^2
12=2^2\!\cdot\! 3
6=2\!\cdot\! 3
24,18,12,6 are partners; all are divisible by \{2,3\}
20=2^2\!\cdot\! 5
10=2\!\cdot\! 5
20,10 are partners; all are divisible by \{2,5\}
16=2^4
8=2^3
4=2^2
2=2
16,8,4,2 are partners; all are divisible by \{2\}
9=3^2
3=3
9,3 are partners; all are divisible by \{3\}

12 integers have no partners: 1,5,7,11,13,14,15,17,19,21,22,23

Answer: 12

Posted in Problem solving | Tagged , , , , | Leave a comment

Water in Glass Box

A glass box 7\times 12\times 18 cm, closed on all six sides, is partially filled with colored water. When the box is placed on one of its 7\times 12 sides, the water level is 15 cm above the table. If the box is placed on one of its 7\times 18 sides, what will be the water level above the table, in centimeters?
Source: NCTM Mathematics Teacher, August 2006

Solution
When the box is placed on one of on its 7\times 12 sides, the volume of water = 7\times 12\times 15. If x represents the level of water above the table when the box is placed on one of its 7\times 18 sides, the same volume 7\times 12\times 15=x\times 7\times 18. Hence the water level x=10 cm.

Answer: 10 cm

Alternative solution
If we filled the box completely full with water, the box will look full no matter how we set it on the table. The height of the water level equals the height of the box 7,12, or 18 cm depending on which side it is placed on. Similarly, if we filled the box half full, the box will look half full no matter how we set it on the table. The height of the water level equals half the height of the box 7/2,12/2, or 18/2 depending on which side it is placed on. Since the water level equals 15 when the box height equals 18, the ratio of water level to box height is 15/18=5/6. So when the box is placed on one of its 7\times 18 sides, the water level equals (5/6)12=10 cm.

Posted in Problem solving | Tagged , , , | Leave a comment

Making Triangles

You have six sticks of lengths 10,20,30,40,50, and 60 cm. Find the number of noncongruent triangles that can be formed using three of these sticks as sides.
Source: NCTM Mathematics Teacher, August 2006

Solution
Three sticks make a triangle. There are \dbinom{6}{3}=20 ways to choose three sticks out of six. Of these 20 ways only 7 satisfy the Triangle Inequality theorem to make 7 triangles, namely
20,30,40
20,40,50
20,50,60
30,40,50
30,40,60
30,50,60
40,50,60
and none of them are congruent.

Answer: 7

Posted in Problem solving | Tagged , , , , , , , | Leave a comment

U.S. Senate Committee

There are 100 members in the U.S. Senate (2 from each state). In how many ways can a committee of 5 senators be formed if no state may be represented more than once?
Source: NCTM Mathematics Teacher, August 2006

Solution
We want to organize the 100 senators into 5-member committees with each state sending only one senator to the committees.
Ways to choose 5 states from 50 states = \dbinom{50}{5}
Ways to send one of two senators from each of the five states = 2^5
Ways to form committees = \dbinom{50}{5}\times 2^5=67,\!800,\!320

Answer: 67,\!800,\!320

Alternative solution
We select 5 senators from a group of 100 to form 5-member committees. First, we pick one senator from 100; second, we pick one from the remaining 98 (not 99 because we cannot have 2 senators from the same state); third, we pick one from 96, etc. Hence the number of ways to pick 5 senators = 100\times 98\times 96\times 94\times 92. Each way is an ordering (permutation) of the senators and the number of orderings of the objects in a set of 5 objects equals 5!. Since the order of selection is not important, we divide the number of ways by 5! to get \dfrac{100\times 98\times 96\times 94\times 92}{5!}=67,\!800,\!320

Posted in Problem solving | Tagged , , , , , , , | Leave a comment

Interest Earned

At the end of every month, Elle deposits \$500 into a savings account with an annual interest rate of 6 percent, compounded monthly. How much interest will be earned at the end of four years?
Source: NCTM Mathematics Teacher, August 2006

Solution
To compute the total interest earned at the end of four years we compute the interest earned by each individual deposit during that time. Since some of the later deposits are invested in less than a year and the interest is compounded monthly, we calculate 
the interest earned in units of months. The interest rate per month is .06/12=.005. The first \$500 invested for 47 months grows to \$500(1.005)^{47}, the second \$500 invested for 46 months grows to \$500(1.005)^{46}, etc. The interests earned by each individual deposit are listed below
Month 1:500(1.005)^{47}-500
Month 2:500(1.005)^{46}-500
Month 3:500(1.005)^{45}-500
\cdots
Month 46:500(1.005)^2-500
Month 47:500(1.005)^1-500
Month 48:500(1.005)^0-500
Total interest earned
500(1.005)^{47}-500+500(1.005)^{46}-500+\cdots+500(1.005)^0-500
=500(1.005^0+1.005^1+\cdots+1.005^{47}-48)
1.005^0,1.005^1,\cdots,1.005^{47} is a geometric sequence with first term 1.005^0=1, common ratio 1.005, and number of terms 48.
=500\left (1\dfrac{1-1.005^{48}}{1-1.005}-48\right )=3048.92

Answer: \$3048.92

Posted in Problem solving | Tagged , , , , | Leave a comment

Sum of Three Integers

For certain positive integers a,b, and c, a^5+b^2+c^2=2010. If b and c are prime, find b+c.
Source: SCVMA Math Olympiad 2010

Solution
a^5+b^2+c^2=2010
Since b and c are odd, b^2 and c^2 are odd which implies that b^2+c^2 is even. Hence a^5 is even because the sum a^5+b^2+c^2 is even.
a^5=2010-b^2-c^2 implies that a^5 is an even perfect fifth power less that 2010.
2^5=32<2010
4^5=1024<2010
6^5=7776>2010
Case 1: a^5=2^5=32
32+b^2+c^2=2010
Trial and error method: looking for prime integers b and c such that 32+b^2+c^2=2010
32+3^2+7^2=90 too low
32+3^2+17^2=330 still too low
32+13^2++37^2=1570 better
32+13^2+43^2=2050 too high
Maybe we should try b and c closer to each other like 23 and 29
32+23^2+29^2=1402
32+29^2+31^2=1834
32+31^2+31^2=1954
32+31^2+37^2=2362
Case 2: Time to try a^5=4^5=1024
1024+3^2+23^2=1562
1024+3^2+29^2=1874
1024+3^2+31^2=1994
1024+3^2+37^2=2402
Try b^2=5^2
1024+5^2+29^2=1890
1024+5^2+31^2=2010 Bingo! b=5 and c=31

Answer: 5+31=36

Posted in Problem solving | Tagged , , , , , , , , | Leave a comment

Cot(C) and Sine(2C)

If \mathrm{cot\,C}=k (a positive integer), express \mathrm{sin\,2C} as a fraction involving k.
Source: SCVMA Math Olympiad 2010

Solution
image
\mathrm{cot\,C}=a/b=k thus a=kb
\mathrm{sin\,2C}=2\,\mathrm{sin\,C\,cos\,C}
=2\times\dfrac{b}{c}\times \dfrac{a}{c}=2\times\dfrac{b}{c}\times \dfrac{kb}{c}=2\times\dfrac{kb^2}{c^2}
=\dfrac{2kb^2}{a^2+b^2}=\dfrac{2kb^2}{k^2b^2+b^2}=\dfrac{2k}{k^2+1}

Answer: \dfrac{2k}{k^2+1}

Posted in Problem solving | Tagged , , , | Leave a comment

Sum of Infinite Geometric Series

The sum of the infinite geometric series a+ar+ar^2+ar^3+\cdots is 4, and the sum of the series whose terms are the squares of the terms of this series is 6.  Find the sum of the infinite geometric series a-ar+ar^2-ar^3+ar^4-ar^5+\cdots
Source: SCVMA Math Olympiad 2010

Solution
a+ar+ar^2+ar^3+\cdots=4
a(1+r+r^2+r^3+\cdots)=4
1+r+r^2+r^3+\cdots is the sum of an infinite geometric series with first term = 1, common ratio = |r|\!<1, and sum = 1/(1-r).
a\left (\dfrac{1}{1-r}\right )=4
a=4(1-r)\qquad\qquad (1)

a^2+(ar)^2+(ar^2)^2+(ar^3)^2+\cdots=6
a^2+a^2r^2+a^2r^4+a^2r^6+\cdots=6
a^2(1+r^2+r^4+r^6+\cdots)=6
1+r^2+r^4+r^6+\cdots is the sum of an infinite geometric series with first term = 1, common ratio = |r^2|\!<1, and sum = 1/(1-r^2).
a^2\left (\dfrac{1}{1-r^2}\right )=6
a^2=6(1-r^2)\qquad\quad (2)

Substitute the value of a from Eq. (1) into Eq. (2)
16(1-r)^2=6(1-r^2)
8(1-r)^2=3(1-r^2)
8(1-2r+r^2)=3-3r^2
8-16r+8r^2=3-3r^2
11r^2-16r+5=0
Solving for r using the quadratic formula yields r=1 (not possible) or r=5/11.
Substitute the value of r into Eq. (1)
a=4(1-5/11)=4(6/11)=24/11

Let S represent the sum of the infinite geometric series
S=a-ar+ar^2-ar^3+ar^4-ar^5+\cdots
=a(1-r+r^2-r^3+r^4-r^5+\cdots)
Collecting the positive and negative terms
S=a[(1+r^2+r^4+\cdots)-(r+r^3+r^5+\cdots)]
1+r^2+r^4+\cdots is the sum of an infinite geometric series with first term = 1, common ratio = |r^2|\!<1, and sum = 1/(1-r^2).
r+r^3+r^5+\cdots is the sum of an infinite geometric series with first term= r, common ratio = |r^2|\!<1, and sum = r/(1-r^2).
S=a\left (\dfrac{1}{1-r^2}-\dfrac{r}{1-r^2}\right )
=a\left (\dfrac{1-r}{1-r^2}\right )
=\dfrac{a}{1+r}
Substitute the values of a=24/11 and r=5/11
S=\dfrac{\dfrac{24}{11}}{1+\dfrac{5}{11}}=\dfrac{\dfrac{24}{11}}{\dfrac{16}{11}}=\dfrac{24}{16}=\dfrac{3}{2}

Answer: 3/2

Posted in Problem solving | Tagged , , | Leave a comment