**dan's math@home - problem of the week - archives****Problem Archives**page 3**with Answers and Winners.**For Problems Only, click here.**1-10****. 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100****101-110****. 111-120 . 121-130 . 131-140 . 141-150 . 151+ .**prob index21-Men and/or Women 22 - How Much Space? 23-Fermat's First Thm 24:That's Sum Product 25 - To Seven-Eleven? 26-The Two Elephants 27 - How Many Tiles? 28 - Where On Earth? 29 - Fridays the 13 th 30 -Turk-o-nacci Nos. - Problem #21 - Posted Wednesday, August 5, 1998
- 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 denom (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.

**WINNERS - Problem 21 . . .**(back to top)**. . .**leader board**Trevor Bird**. . . .- 8 pts - Correct answer (but I'd like more explanatory steps shown, please)**Andy Murdock**- 3 pts - You forgot to make Pat & Chris count themselves- Problem #22 - Posted Thursday, August 20, 1998
- 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 circumf of the wire is 2 Pi r + 25 feet.
- The radius of the loop of wire is (2 Pi r + 25) / (2 Pi) = r + (25 / (2 Pi)), so the extra radius (the space under the wire)
- is 25 / (2 Pi) 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?)

**WINNERS - Problem 22 . . .**(back to top)**. . .**leader board**Andy Murdock**. . . .- 10 pts - 3.978859124 ft is approximately right! 25 / (2 Pi) is exact.**Maurice Turrieta**. . .- 6 pts - Rounded off severely (slight penalty imposed by the "point guard.")**Joao Sousa**. . . . . . . .- 5 pts - Cool, you used calculus to do it! d(area) = (circumf) * d(radius)**Cliff Young**. . . . . . .- 4 pts - I'm glad the answer surprised you!**Greg Hess**. . . . . . . . - 3 pts - Correct answer (but a few people beat you to it.)**Arthur Thung**. . . . . .- 2 pts - You too!**Candace Brant**. . . . - 2 pts - It's natural to think that there's hardly any room at all!- All of you used the given circumference 25000 miles in your calculations. You didn't need to!
- Problem #23 - Posted Tuesday, Sept 1, 1998
- 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?)

**WINNERS - Problem 23 . . .**(back to top)**. . .**leader board**Trevor Bird**. . . . . . .- 10 pts - And you even knew it was Ramanujan !**Greg Hess**. . . . . . . . - 7 pts - You got it. Go for the bonus point!***Joao Sousa**. . . . . . . - 5 pts - That's right. What, no calculus this time?- (Dan's note: 1729 = 12^3 + 1^3 = 10^3 + 9^3.
- What's the next such number after 1729? e-mail me first for a bonus point!)
- * Response by Greg Hess, 9/19/98:
- "Dan - The next #......87,539,319 = 167^3 + 436^3 = 228^3 + 423^3 = 255^3 + 414^3."
- "I admit I cheated . . . I did a search of "Ramanujan" and came up with a list of 'taxi cab' #'s."
- (Dan's note: I didn't know this one! Everyone still try to find the second number (after 1729)
- that's the sum of two cubes in two different ways. It's a lot smaller than this. +1 pt anyway, Greg.)
- ** A nother esponse from Greg, some days later, using a calculator, 2^3 + 16^3 = 9^3 + 15^3
- (since 8 + 4096 = 729 + 3375 = 4104) and as a bonus, 11^3 + 14^3 is 4105. +1 nother pt . &-}
- Problem #24 - Posted Saturday, September 12, 1998
- 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."Thirty-eight squared," added Chris.
- How much money did each person win? (Give exact reasons and explain as well as you can.)
**Solution**: 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 . . . 1443 =/= 362880
- 2 . . . . 1442 . . . 2884 =/= 362880
- 3 . . . . 1441 . . . 4323 =/= 362880
- . . . . . . . . . . . . . . . . . . . . . . . . . .
- 323 . . 1121 . . 362083 =/= 362880
- 324 . . 1120 . . 362880 . . YES!
**WINNERS - Problem 24 . . .**(back to top)**. . .**leader board**Andy Murdock**. . . 10 pts , he said completely squarely. (see above) ; - }**Joao Sousa**. . . . . . . . 7 pts - did you use Mathematica?**Trevor Bird**. . . . . . . 4 pts - You're welcome!**Greg Hess**. . . . . . . . 2 pts - See the above solution**Cathy Bird**. . . . . . . . 1 pt - Thanks for taking part (and point)!- Problem #25 - Posted Thursday, September 24, 1998
- 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**: Well, this is a case where using algebra on the problem makes it worse.- Just figure that if x and y are whole munbers, 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.
**WINNERS - Problem 25 . . . all were correct!**(back to top)**. . .**leader board**Greg Hess**. . . . . . . 10 pts - Good initial search conditions**Andy Murdock**. . . . 7 pts - Correct to eschew algebra (and spit it out) on this one. ; - }**Trevor Bird**. . . . . . . 5 pts - Nice table approach , and thanks for the idea of graphing!- Problem #26 - Posted Friday, October 9, 1998
- 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, show steps, find each weight to the nearest pound.)
**Solution**: Along my favorite theme: "Things are often not what they seem," the answer is NOT TEN TONS.- 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.** **WINNERS - Problem 26 . . .**(back to top)**. . .**leader board**Andy Murdock**. . . 10 pts - Nice job; don't be stingy with decimals!**Trevor Bird**. . . . . . . 7 pts - Good order of ops : x[(1.1)(0.9)]^10 = 10 tons...**Greg Hess**. . . . . . . . 2 pts - Correct answer to slightly different question- Problem #27 - Posted Wednesday, October 21, 1998
- How Many Tiles?
- A rectangular floor is composed of whole square tiles.
- A diagonal line is drawn and ruins some of the tiles.
- (on a 2 x5 floor, 6 tiles are ruined, on a 2 x4, 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**horiz 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. **WINNERS - Problem 27 . . .**(back to top)**. . .**leader board**Trevor Bird**. . . . . 10 pts - Very nice explanation, I'll e-mail you more graph paper!**Andy Murdock**. . . 2 pts - Rounding off is dangerous, in gymnastics and in integer arithmetic.- Problem #28 - Posted Monday, November 2, 1998
- 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/Pi) 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!

**WINNERS - Problem 28 . . .**(back to top)**. . .**leader board**Trevor Bird**. . . . . 10 pts - Actually an infinite number of points, for this problem!**Beth Wilson**. . . . . 5 pts - Very close, but you left out the circ of 1 mile.**Andy Murdock**. . . 2 pts - For not exploding near the South Pole**Greg Hess**. . . . . . . 2 pts - 2-mile spherical triangle is too small to have 360 degrees.- Problem #29 - Posted Friday, November 13, 1998
- 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!)

**WINNERS - Problem 29 . . .**(back to top)**. . .**leader board**Beth Wilson**. . . . . 9 pts - Right you are, and first by 90 minutes and 2 time zones!**Andy Murdock**. . . 6 pts - Your paradigm's worth more than 20 sense!- Problem #30 - Posted Tuesday, November 24, 1998
- 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 7th, and 542 turkeys on November 11th,
- 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 15737 turkeys on the 18**th**.**- (Dan's note: +1 pt for 1st
correct: Could they have had a "million-turkey-march"
by Thanksgiving 1998 ?)

**WINNERS - Problem 30 . . .**(back to top)**. . .**leader board**Beth Wilson**. . . . . 9 pts - Good use of variables; I received your answer first**Andy Murdock**. . . 9 pts - You sent yours first but I got it much later. (DVC server down)**Heather Jameson**. 5 pts - Welcome to the Contest! (DVC server not always reliable)

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