Men and/or Women?

In a certain algebra lecture class, Chris and Pat count the students and compare notes.

"Hmm, 12/17 of my classmates in here are women," notes Pat.
"Funny," recounts Chris, "5/7 of my classmates are women."

They were both right. How many students, men and women, were in the class, and what were the genders of Chris and Pat?

(men and women are mutually exclusive sets for the purposes of this problem. explain your reasoning.)

Solution:

Let n be the total number of students in the class, w = the number of women.

Now 12 / 17 < 5 / 7 , so Pat has to be a woman, as Chris sees more women classmates from his perspective. Then Pat sees that (w-1) / (n-1) = 12 / 17 , and Chris sees w / (n-1) = 5 / 7.

The denominator (n-1) can be the product 7*17 = 119, because 12/17 = 84/119 and 5/7 = 85/119 , so
since 84 and 85 are 1 apart, we're there. Pat has 84 women classmates, and Chris has 85.
Don't forget the 119 doesn't count the observer, so there are 120 students in the class.

How Much Space?

A wire belt is wound tightly around the equator of the earth (circumference 25,000 miles), then 25 feet of wire is added and the belt is propped up at an equal height all the way around the planet.
How much space will there be under the wire?   (Please explain your reasoning.)

a)  Not enough for an ant to crawl under            b)  Enough room for an ant, but not a mouse,

c)  A Siamese cat can just squeeze under it       d)  Dan the math teacher could limbo under it!

Solution:

This problem is the same whether the original wire is wound around the earth, the moon, or an orange!

Let d = the space under the wire, and r = the radius, in feet, of the earth (or an orange, for that matter).
The circumference of the earth is 2 Pi r feet, so the circumference of the wire is 2 Pi r + 25 feet.

The radius of the loop of wire is (2 π r + 25) / 2π = r + (25 / 2π), so the extra radius (the space under the wire) is 25 / 2π feet, which works out to about 25 / 6.2831853 ; approx. 3.9789 feet.

(even Dan the Math teacher should be able to get under that... wanna see?)

Fermat's First Theorem?

You've heard of Fermat's Last Theorem? As stated in 1637, and proved (by Andrew Wiles) in 1995:
a^n + b^n = c^n has no integer solutions for n > 2.
But 6^3 + 8^3 is very close to 9^3 . . . 216 + 512 = 728, not 729.)

What is the smallest number that equals the sum of two (positive) perfect cubes in two different ways?  (For example, 65 is the sum of two squares in two different ways: 65 = 8^2 + 1^2 = 7^2 + 4^2.)

Solution: This is the famous "Ramanujan taxicab number."

Prof. J.E. Littlewood, on a visit to Ramanujan at Trinity College in England, remarked that his taxicab number, 1729, was a rather dull number. "Oh, no," replied Ramanujan, "it is a very interesting number. It is the smallest number that is the sum of two cubes in two different ways!"

(dan's question: Does this contradict Fermat's Last Theorem?)

That's Sum Product!

Pat and Chris are gambling in Las Vegas. I asked them to 'put their winnings together.'

"Nine factorial," said Pat product-ively."

How much money did each person win?

(give exact reasons and explain as well as you can.)

Solution:  (partly by Andy Murdock)

This can be done by looking for two numbers m and n , whose PRODUCT (as Pat says) is
9! = 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362880 , and whose SUM (Chris adds) is 38^2 = 1444. So:

m n = 362880 and m + n = 1444, or n = 1444 - m. Substituting, we get
m(1444 - m) = 362880 ; m^2 - 1444 m + 362880 = 0.

Factoring this is the same as looking for two numbers whose sum is 1444 and product is 362880.
The Mathematica command (copy this and paste into Mathematica) is of course
Factor[m^2 - 1444 m + 362880]

Complete the square, says Andy Murdock, and get
m^2 - 1444 m + 722^2 - 722^2 + 362880 = 0
(m - 722)^2 - 521284 + 362880 = (m - 722)^2 - 158404 = 0 ;
m - 722 = + or - Sqrt[158404] = + - 398 ; Pat and Chris had winnings of
m = 722 + 398 = \$1120 and n = 722 - 398 = \$324 dollars, irrespectively.

And if you want to do it without algebra, I'd suggest making a table of pairs of numbers that add up
to 1444, with their products:
1 * 1443 = 443 ≠ 362880
2 * 1442 = 2884 ≠ 362880
3 * 1441 = 4323 ≠ 362880
. . .
323 * 1121 = 362083 ≠ 362880
324 * 1120 = 362880 . . . YES!

To Seven-Eleven?

Solve this system of equations for x and y :

x + √y = 7
y + √x = 11

That is, find a pair of real numbers x and y making both equations true.

(Here √n means square root of n)

Solution: (partly by Trevor Bird)

Well, this is a case where using algebra on the problem makes it worse.
Just figure that if x and y are whole numbers, they have to be perfect squares and
trying small ones (1, 4, 9) yields x = 4 , y = 9 .
(Check: 4 + √9 = 4 + 3 = 7 and 9 + √4 = 9 + 2 = 11.)

Speaking of algebra, if you substitute x = (7 - √y)  you get . . . y + √(7 - √y) = 11 ,
√(7 - √y) = 11 - y ; (7 - √y) = (11 - y)^2 = 121 - 22y + y^2 ;
- √y = 114 - 22y + y^2 ; y = (114 - 22y + y^2 )^2 = y^4 - 44 y^3 + 712 y^2 - 5016 y + 12996 ;
y^4 - 44 y^3 + 712 y^2 - 5017 y + 12996 = 0 .
To look for roots other than y = 9, you could divide the left by (y - 9) . . .

Another way (thanx Trevor) is to plot the two curves x + √y = 7 and y + √x = 11 and see where they intersect.

The Two Elephants: Tons of Tens, or Tens of Tons?

Tenny and Tonny are two elephants. Every winter, Tenny gains ten percent of his body weight, and Tonny loses ten percent. Every summer, the opposite happens: Tenny loses ten percent and Tonny gains ten percent. This has gone on for ten years, and now they each weigh ten tons.

How many tons did Tenny and Tonny weigh ten years ago?

(explain steps, round to nearest pound.)

Solution:

Along my favorite theme: "Things are often not what they seem," the answer is NOT TEN TONS.

See, 90% of 110% of any number n is (.90)(1.10)(n) = 0.99 n. Same goes for 110% of 90%.

So each year the elephants lose 1%; we multiply by 0.99 ten times, and we get (if n = orig. wt):
[(0.99)^10] * n = 10 tons ; n = 10 / (0.99)^10 = 10 / 0.904382075 = 11.05727355 tons.

Assuming regular 2000-lb tons, that's 22,114.54711, or roughly 22,115 lbs each.

How Many Tiles?

A rectangular floor is composed of whole square tiles. A diagonal line is drawn and ruins some of the tiles. (See that on a 2 x 5 floor, 6 tiles are ruined, on a 2 x 4, only 4 are ruined.)

a)  How many tiles are ruined on a 4 by 6 floor?

b)  How about a 63 by 81 floor?

c)  Generalize to an m by n floor.

Solution:  (by Trevor Bird "as told to" Dan Bach)

Let's define a "tile function" T(m,n) = number of ruined tiles in an m by n rectangle.
First let's look at the case where m < n have no common factors; every time the diagonal
line crosses a vertical or horizontal line, a new tile is ruined, and there are m-1 vertical
and n-1 horizontal lines; add the initial tile and get T(m,n) = m+n-1.

If m and n have a common factor g then there will be a pattern that repeats g times.
The size of the repeating pattern will be m/g by n/g . If g is the greatest common factor
(the GCD of m and n) then we have a total of T(m,n) = g(m/g + n/g - 1) ruined tiles.
Another way to write this is T(m,n) = m + n - g .

Apply this formula to the 4 x 6 case: g = 2 , so T(4, 6) = 4 + 6 - 2 = 8 ,
. . . . . . . . and now the 63 x 81 case: g = 9 , so T(63, 81) = 63 + 81 - 9 = 135.

(Andy M. also observed that if m goes into n evenly then there are n ruined tiles.)

Where On Earth?

I start running at 12:00, go 2 miles south by 12:15, 2 miles west by 12:30, and 2 miles north by 12:45. After 45 minutes I'm right back where I started!

Where on Earth am I?  (there's more than one possibility)

Solution: (partly from new contestant Beth Wilson.)

The North Pole is the "top" answer, but there are an infinite number of circles near the South Pole where I could have started. One example is two miles north of any point at which the circumference of the circle that parallels the Equator is 2 miles (the radius would be 1/π) so that after running two miles you will be back where you started.

The other points follow this same pattern, except that the circumference of the circle is 2/2, 2/3, 2/4, 2/5, 2/6 ..., and so on!

Fridays the Thirteenth

What is the maximum number of Friday-the-13ths that there can be in a single year?

What is the minimum number?

Solution:

A regular year can start with Jan 1 on any of the 7 days of the week, and so can a leap year.
So there are 14 possible "year-patterns"; these are the days for the 13th of each month:
Notice the 13th is delayed the next month by the number of days above 28; 3 for Jan (31) etc.
Fr Mo Mo Th Sa Tu Th Su We Fr Mo We. (reg) , and Fr Mo Tu Fr Su We Fr Mo Th Sa Tu Th (leap).

As Beth Wilson notes, the max number of times a day appears is 3, and the min is 1.
A thorough check yields the fact that indeed in either a normal year or a leap year there are
a max of three Friday the 13ths and a min of one Friday the 13th.

(dan's incorrect note: i thought i remembered a year when there were no friday the 13ths!)

Turk-o-nacci Sequence!

Like the Fibonacci sequence  1, 1, 2, 3, 5, 8, . . .  a certain turkey flock has as many turkeys on a given day as the sum of the number of turkeys on the previous two days.

If there were 79 turkeys on November 7-th, and 542 turkeys on November 11-th, how many turkeys were there on November 18th?

Solution:

Try to work with two unknowns, say the A = 5th number and B = 6th . Then A + B = 79.
B + 79 = B + A + B = A + 2B = 8th, 2A +3B = 9th, 3A + 5B = 10th, and 5A + 8B = 11th = 542.

So solving the system {A + B = 79 , 5 A + 8 B = 542} gives 5 A + 5 B = 395 , 3 B = 147, B = 49 ,
then A = 30 ; we get 30, 49, 79, 128, 207, 335, 542, 877, 1419, 2296, 3715, 6011, 9726, 15737.

There were 15,737 turkeys on the 18th.

(dan's note: could they have had a "million-turkey-march" by thanksgiving 1998 ?)