**Problem of the Week**

__5/22/2017__

**Fair Die**

A fair die is tossed four
times. What is the probability that it lands with either 5 or 6 on top at least
once?

**Solution**

65/81 or approximately .80247.

The number of possible ways to roll four dice is 6^{4},
or 1296: six choices for each of the four rolls.

There are four ways to roll a single die once and
not get a 5 or 6 (that is, to get a 1, 2, 3, or 4), so the number of ways to
roll a die four times and not get a 5 or a 6 is 4^{4}, or 256.

__5/15/2017__

*AD*** and ***BD*

What
is the sum of the distances *AD* and *BD* in the figure shown?

**Solution**

27. By the Pythagorean theorem, *BD* = 10. Let *E* be a point
on *AB* with ∠*DEA* = 90°. Then *DE* = 8, and *AE* = 21 – 6 =
15. By the Pythagorean theorem, *AD* =
17. Hence, 10 + 17 = 27.

__5/8/2017__

**Dog Show**

Of
the animals entered in a dog show, the number of poodles is at least one-fifth
of the number of beagles and at most one-sixth the number of collies. The
number of dogs that are poodles or beagles is at least 23. Find the minimum
number of collies entered in the show.

**Solution**

*c* ≥ 24. Let *p*, *b*, and *c* represent poodles, beagles, and
collies, respectively. Then

(1/6)*c* ≥ *p* ≥ (1/5)*b*

and

*p + b*
≥ 23.

Since *p* ≥
(1/5)*b*, 5*p* ≥ *b*, and 6*p* ≥ *b
+ p* ≥ 23,

6*p* ≥ 23 and *p* ≥ 23/6.

But *p* must be
a natural number, so *p* ≥ 4. Since

(1/6)*c* ≥ *p* ≥ 4

we have
(1/6)*c* ≥ 4 and *c* ≥ 24.

__5/1/2017__

**Less than 2005°**

The
sum of the interior angles of a polygon is less than 2005°. What is the largest
possible number of sides of the polygon?

**Solution**

13. The sum
of the interior angles of a polygon = 180°(*n*
– 2) < 2005° → 180°*n* < 2365° → *n* < 13.139. Since *n* must be an integer, the largest value
for *n* is 13.

__4/24/2017__

**One or Ten?**

Given
that they are made of the same material, which is heavier: a ball with a radius
of 10 inches or 10 balls each with a radius of 1 inch?

**Solution**

The ball with a radius of 10 inches is heavier. Since
the volume of a sphere is (4/3)π*r*^{3}, one ball with a radius of 10 inches
would have a volume of (4/3)π⋅10^{3} = (4000/3)π in^{3}. Thus 10 balls of radius 1 inch
would have an accumulated volume of only

10⋅[(4/3)π⋅1^{3}]
= (40/3)π in^{3}.

__4/3/2017__

**Tamika > Carlos?**

Tamika
selects two different numbers at random from the set {8, 9, 10} and adds them.
Carlos takes two different numbers at random from the set {3, 5, 6} and
multiplies them. What is the probability that Tamika's result is greater than
Carlos's result?

**Solution**

4/9. Tamika
can get the numbers 8 + 9 = 17, 8 + 10 = 18, or 9 + 10 = 19. Carlos can get 3 ×
5 = 15, 3 × 6 = 18, or 5 × 6 = 30. The possible ways to pair these numbers are
(17, 15), (17, 18), (17, 30), (18, 15), (18, 18), (18, 30), (19, 15), (19, 18),
and (19, 30). Four of these nine pairs show Tamika with a higher result, so the
probability is 4/9.

__3/27/2017__

**A New Mathematical Operation**

If *a***b*
= *a*^{b}*
– b*, find (2*3)*4

**Solution**

621

(2*3)*4
= (2^{3} – 3)*4 = (8 – 3)*4 = 5*4 = 5^{4} – 4 = 625 – 4 = 621.

__3/20/2017__

**A Quarter Pounder**

A
quarter-pound hamburger contains approximately 80 calories per ounce of meat,
an average french fry contains about 14 calories, a cola contains about 10
calories per ounce, and a bun contains 200 calories. Suppose you have a
quarter-pound hamburger with a bun and six ounces of cola. How many french
fries can you eat and still keep your meal below 800 calories?

**Solution**

15. We have
80*h* + 14*f* + 10*c* + 200 < 800.
Using the information in the problem we can write 80(4) + 14*f* + 10(6) + 200 < 800. This implies
that 14*f* < 220 and *f* < 16. Therefore, you can eat 15
french fries.

__3/13/17__

**Perfect Squares and Cubes...NOT**

How
many numbers from 1 to 1 million, inclusive, are not perfect squares or perfect
cubes?

**Solution**

998,910.
There are 1000 perfect squares between 1 and 1 million; these are the squares
of the first 1000 integers. Similarly, there are 100 perfect cubes—the cubes of
the numbers from 1 to 100. Subtract the squares and the cubes from 1 million to
get 998,900. However, every number that is a perfect sixth power has been
subtracted twice (the largest of these is 10^{6} = 1,000,000). Adding these back in gives 998,910.

__3/6/17__

**999 Coins**

Starting
with a single pile of 999 coins, a person does the following in a series of
steps: In step one, he splits the pile into two nonempty piles. Thereafter, at
each step, he chooses a pile with 3 or more coins and splits this pile into two
piles. What is the largest number of steps that is possible?

**Solution**

997. The
number of steps is one less than the number of piles, and 998 is the largest
number of piles, 997 with 1 coin and 1 with two coins.

__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?

**Solution**

Approximately
$67,332. If P was the average house price five years ago, then the current
average price is 1.3^{5P }or
3.71293P. Thus, 250,000 = 3.71293P, so P=250,000/3.71293, which equals
$67,332.27

__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.

##