**Problem of the Week**

__2/27/17__

**House Prices**

The
average house price in Boomtown rose 30 percent each year for the last five
years. If the average house price is currently $250,000, what was the average
house price five years ago?

__2/6/17__

**Upright Integers**

An
integer is defined as upright if the sum of its first two digits equals its
third digit. For example, 145 is an upright integer since
1 + 4 = 5. How many positive three-digit integers are
upright?

**Solution**

45. From the
definition, the first and second digits of an upright integer automatically
determine the third digit, which is the sum of the first two digits. Consider
those upright integers beginning with 1: 101, 112, 123, 134, 145, 156, 167,
178, and 189; there is a total of 9 such numbers. (Note that the second digit
may not be 9; otherwise, the last “digit” would be
1 + 9 = 10.) Beginning with 2, the upright integers are
202, 213, 224, 235, 246, 257, 268, and 279; there is a total of 8 such numbers.
We may continue this pattern of analysis to show that the numbers of upright
integers beginning with a digit of 3, 4, 5, 6, 7, 8, or 9 are 7, 6, 5, 4, 3, 2,
and 1, respectively. Therefore, there is a total of
9+8+7+6+5+4+3+2+1 = 45 three-digit upright integers.

__1/30/17__

**e-traders**

Three
students open e-trade accounts and become day traders. Although they all work
hard, they achieve the following steady rates of losing money: The first
student loses $1000 in one hour, the second student loses $1000 in two hours,
and the third student loses $1000 in three hours. Find the number of minutes it
takes for the three students together to lose a total of $2000.

**Solution**

About 65
minutes (to the nearest minute). The losing rates are $1000/hr., $500/hr., and
$333.33/hr. The combined losing rate is $1833.33/hr. Thus,

1833/1 = 2000/x so x = 1 and 1/11 hours or 65 minutes

__1/23/17__

**Investments**

A man
has $10,000 to invest. He invests $4,000 at 5 percent and $3,500 at 4 percent.
To have a yearly income of $500 from the investment, at what rate must he
invest the remainder of the money?

**Solution**

At
6.4 percent or higher.

The
man wants to earn $500 in interest each year. Investing $4000 at 5 percent
yields yearly interest of 0.05 x $4000, or $200; whereas $3500 invested at 4
percent yields 0.04 x $3500, or $140; so the remaining $2500 needs to be
invested at a rate that yields interest of $160 per year (to have a total
interest of $500 each year). If *r* is
the annual percentage rate for the $2500, then

so the man
needs to invest the remaining $2500 at 6.4 percent or higher to guarantee an
interest income of at least $500 a year.

__1/16/17__

**What Is the Price?**

Matthew
and Matilda want to buy a set of DVDs. Matthew has $47 less than the purchase
price, and Matilda has $2 less. If they pool their money, they still do not
have enough to buy the DVDs. If the set costs a whole number of dollars, what
is its price?

**Solution**

$48.

If *P* is the price of the DVDs, Matthew has *P* - 47 dollars and Matilda has *P* - 2 dollars.

We
know that (*P* - 47) + (*P* - 2) < *P*, since the amount of their combined savings is still less than
the price of the DVDs.

Solving for *P*, we have *P* < 49; but *P* > 47,
since Matthew had some money. Therefore, *P*
is a whole number such that 47 < *P*
< 49, so the price of the DVDs is $48.

__1/9/17__

**Tom and Bill**

Tom
is standing in a hole that is 4 feet deep. Bill asks him how much deeper he is
going to dig the hole. Tom replies that he will dig 4 feet 2 inches deeper and
that the top of his head will then be the same distance below ground level that
it is now above ground level. How tall is Tom?

**Solution**

6 feet 1
inch. The top of his head will go down 4 feet 2 inches with the additional
digging. Half that distance was above the hole before the additional digging,
so he is 4 feet + 2 feet 1 inch, or 6 feet 1 inch, tall.

__1/2/17__

**Wire Length**

In the diagram, the rectangular wire grid contains
15 identical squares. The length of the grid is 10. What is the length of wire
needed to construct the grid?

**Solution**

76. Since the length of the rectangular grid is 10, the side of each square in the grid is 10 ÷ 5 = 2. The height of the grid is therefore 6 (3 squares). There are four horizontal wires, each of length 10, and six vertical wires, each of length 6, for a total length of wire of 4 • 10 + 6 • 6 = 40 + 36 = 76.

__12/19/16__

**How Tall Was the Tree?**

Lightning hit a tree
one-fourth of the distance up the trunk from the ground and broke the tree so
that its top landed 60 feet from its base, thus creating a triangle. How tall
was the tree before it broke?

**Solution**

Approximately 84.85 feet tall.

Let *h* be the height, in feet, of the tree
before it fell. Then (*h*/4)^{2} + 60^{2} = (3*h*/4)^{2},
so *h*^{2}/16 + 3600 = 9*h*^{2}/16.

Solving for *h* reveals that h^{2}/2 =
3600, so *h* = , and the original height of the
tree equals , or approximately 84.85 feet.

__12/12/16__

**Facetious**

In how many possible
arrangements of the letters in the word FACETIOUS are the vowels in
alphabetical order?

**Solution**

3024 ways.

A total of 9! (9x8x7x6x5x4x3x2x1) ways exist to
arrange the letters in the word FACETIOUS (nine ways of picking the first
letter exist, eight ways to pick the second exist, and so on).

For each arrangement of the consonants in the word,
the vowels can be arranged in 5! (5x4x3x2x1) ways. In only one of them are the
vowels in alphabetical order.

Hence, 1/5! of the arrangements have the vowels in
alphabetical order, for a total of (9!/5!), or 3024.

__12/5/16__

**Townhouses**

A contractor is asked to build a new set of
townhouses in attached clusters of different sizes. He created plans for one-,
two-, and three-house clusters, as shown in the diagram below. The builder used
computer software to draw the line segments used to represent the houses. How
many line segments are needed to draw 4 houses? 10 houses? 47 houses? *n*
houses? Explain your strategy.

**Solution**

21; 51; 236; 5*n* + 1. One house takes 6
segments. Each additional house only requires 5 additional segments. Therefore,
the sequence for the number of segments is
6, 11, 16, 21, 26, 31, 36, . . . .
If n is the number of houses, then the *n*th term for this sequence is 5*n* + 1,
or 5 times the number of houses plus 1. The 5th house would require
5(5) + 1 = 26 segments; the 10th house would require
5(10) + 1 = 51 segments; and the 47th house would require
5(47) + 1 = 236 segments. Another method is to begin with 6
segments for the first house and notice that each additional house adds 5
segments to the total. Therefore, the total number of segments is
6 + 5(*n* – 1) = 5*n* + 1.

__11/28/16__

**Classmates' Birthdays**

Your
mathematics teacher displays a monthly calendar, which highlights all the
birthdays of your classmates. Your mathematics class has 25 students. What is
the probability that 3 or more students were born in the same month?

**Solution**

1, or 100
percent. In a class of 24 students, it is possible for exactly 2 students to
have been born in each of the 12 months. For 25 or more students, it is certain
that there would be at least 1 month containing 3 or more birthdays. Therefore,
the probability is 1.

__11/14/16__

**Cards in a Hat**

Cards
numbered 1 to 50 are put in a hat. What is the probability that the first two
cards chosen at random have prime numbers on them?

**Solution**

3/35, or
about .086. There are 15 prime numbers in the range of 1 to 50: 2, 3, 5, 7, 11,
13, 17, 19, 23, 29, 31, 37, 41, 43, and 47. Remember that 1 is not a prime
number. The probability of selecting a prime on the first pick is 15 out of 50;
for the second pick, the probability is 14/49. The probabilities are combined
by multiplying them, so 15/50 × 14/49 = 210/2450, or 3/35, or about .086.

__10/31/16__

**How Many?**

How
many two-digit numbers exist such that when the product of its digits is added
to the sum of its digits, the result is equal to the original two-digit number?

**Solution**

Nine
two-digit numbers: 19, 29, 39, 49, 59, 69, 79, 89, and 99. Any two-digit number
can be represented by 10*a + b*. The
product of the digits is *ab*, and the
sum is *a + b*. When does *ab* + (*a + b*) = 10*a + b*?
Solving, when *ab* = 9*a*, or when *b* = 9.

__10/24/16__

**Old Books**

A
mathematics teacher collects old books. One day, a student asked him how many
old math books he has. He replied, “If I divide the books into two unequal
whole numbers, then 64 times the difference between the two numbers equals the
difference between the squares of the two numbers.” How many old math books
does the teacher have?

**Solution**

64. Let *a* and *b* represent the two unequal whole numbers.
Then 64(*a* – *b*) = *a*^{2} – *b*^{2}.
Factoring the right side yields 64(*a* – *b*) = (*a* – *b*)(*a* + *b*).
Since *a* and *b* are not equal, (*a* – *b*) does not
equal zero.
Therefore, you can divide both sides by (*a* – *b*),
which
yields 64 = *a* + *b*.
To solve the problem,
you do not
have to know the values of *a* and *b*,
just the sum of *a* and *b*,
which is 64,
the total number of books.

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