**dan's math@home - problem of the week - archives****Problem Archives**page 10**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 index91- A Sinking Storm 92 - Tiling Me Softly 93- One for the Ages 94 - Hole In The Ball 95 - Above Avg. Art 96 - Missing Dollar ? 97-Three Funny Nos. 98 - The Dinner Date 99- Birds on the Run 100- Parents To Bee? - Problem #91 - Posted Tuesday, August 22, 2000
- That Sinking Storm ! (back to top)
- Fishing boat captains, when big storms form, behave as follows: 50% of the time they head for home,
- 30% of the time they choose to ride it out, and 20% of the time they head away from the storm.
- In a severe storm, the probability of being sunk is 97% if they head for home, 44% if they ride it
- out, and 17% if they head away. (i) What is the probability of being sunk? (ii) What is the probability
- that a boat was headed for home, given that it was sunk?

**Solution:**This is a "conditional probability" type question. Let's look more closely:**i)**A boat can be sunk in**three ways**: heading home (H), riding out the storm (R), or going away from it (A).- In each situation the prob of sinking
is given; we
**multiply**by the given cond. probs to get the**partial**probs, - then
**add 'em up**! . . . **Prob(sinking)**= P(SH) + P(SR) + P(SA) = (97% of 50%) + (44% of 30%) + (17% of 20%)- = (.97)(.50) + (.44)(.30) + (.17)(.20)
= 0.485 + 0.132 + 0.034 = 0.651 =
**65.1%**prob. of being sunk. **ii)**The condition is that the boat sunk, so**Prob(home if sunk)**= P(home and sunk) / P(sunk)- = 0.485 / 0.651 = 0.74500768049 =
**74.5%**prob. a sunk boat was headed home.

**WINNERS - Problem 91 .**(back to top)**.**leader board . . . Welcome new DVC contestants!**Pat Gremban**.**Lisa Schechner**. . . . 8 pts - Extra point; you only missed first place by 11 minutes.**Arthur Morris**.**Kia Davidson**(new) . .5 pts - Nice way of describing problem; "I put 100 boats in..."**Patrik Petersson**. . . 4 pts - Thanks for proper terminology: tree diagram, Baye's Law.**Phil Sayre**.**Irina Kolesnik**(new) . . 3 pts - Welcome to the contest; good answer!**Filipe Gonçalves**. . . 2 pts - Lots of people (including me originally) think 50/97.**Tom Esser**(new) . . . . . 2 pts - Hi Tom! First part ok, second part needs division.**Megan Spillane**(new) . 2 pts - Good try; half perfect; thanks for the website support!**Miho Koto**(new) . . . . . 2 pts - Welcome! First % good, second part: divide by first.**Sue B**. .**Tu Truong**(new). . . . . . 2 pts - Thanks for entering; the 65.1% was right!- Problem #92 - Posted Friday, September 1, 2000
- Last of the 1999/2000 season! NEW CONTEST starts with Problem 93!
- Tiling Me Softly (back to top)
- Roberta and Lauryn want to tile their 20-ft by 21-ft floor, and they insist on using square tiles. The tiles are uncuttable, non-overlapping, and with integer sides. What is the smallest (total) number of tiles they can use, and what's the exact arrangement?*

- *You can attach a simple GIF file, or simply list the size of each square and the coords of its lower-left corner, starting with (0,0). More than one tile of the same size may be used, but tiles cannot hang over the edge.

**Solution:**There are a couple of ways to do it with**10**tiles: squares of sizes:- {7, 7, 7, 13, 8, 5, 3, 2, 1, 1} or {15, 5, 5, 5, 6, 6, 6, 2, 2, 2}.
- I thought the best tiling was
**9**squares, {16, 5, 5, 5, 5, 4, 4, 4, 4} as shown, but I - (and at least one of you) have done
better:
**8 squares, {14, 7, 7, 6, 6, 6, 3, 3}.**Arthur sent in: - 14X14 tile at (0,0) ; 7X7 tiles at (14,0), (14,7) ; 6X6 tiles at (0,14), (6,14),(12,14) ; 3X3 tiles at (18,14), (18,17)
- Click here to see Sue B's picture of the winning solution.

**WINNERS - Problem 92 .**(back to top)**.**leader board . . . LAST PROBLEM OF 1999/2000 CONTEST!**Phil Sayre**.**(8 sqs)**Good job; you can steal my GIF if you want.**Arthur Morris**. .**(8 sqs)**You were first, but didn't count the squares for me.**Sue B**. .**(8 sqs)**That's the right answer - another visual learner!**Irina Kolesnik**. . . . . .5 pts -**(8 sqs)**Yes, nice clear simple picture, once I opened the .rtf**Tom Esser**. . . . . . . . . 3 pts -**(9 sqs)**I'm glad this problem bugged you; there's the ans!**Miho Koto**. . . . . . . . . 2 pts -**(18 sqs)**Good try; coulda done 12 if you combined some sqs**Filipe Gonçalves**. . . 2 pts -**(20 sqs)**You could have had the 8 if you combined smaller sqs.**Montta**. (new) . . . . . . . . 1 pt -**(420 sqs)**The number of squares goes down with bigger squares**Adam Levy**. . . . . . . . 1 pt - Nice way of looking at it numerically: 420 = a^2 + b^2 + c^2 + . . .- Problem #93 - Posted Sunday, September 10, 2000
- One For The Ages (back to top) (First problem of the 2000/01 year - our 4th season!)
- Macy and Tracy are friends whose combined ages are forty-four years. Macy is twice as old as Tracy
- was when Macy was half as old as Tracy will be when Tracy is three times as old as Macy was when
- Macy was three times as old as Tracy. How old is Macy? Be sure to include all your steps and reasoning!

**Solution:**I stole this from my college Math Lab tutor desk; it's a classic "Mary and Ann" problem- by the famous Sam Loyd. Here's
the solution table and steps our DVC Math Lab coordinator
**Jean** **Phillips**found: Follow along number by steps from (1) to (7), then see the equations below.**was**(b)**was**(a)**now****will be****Macy**(6) . . **2 y / 3**(4) . . **y**(1) . . **x**- - - **Tracy**(7) . . **2 y / 9**(3) . . **x / 2**(2) . . **44 - x**(5) . . **2 y**- (1)
**Let Macy's age now = x**. (2) Then Tracy's age now = 44 - x. (3) Tracy's age was (a) : x / 2 . - (4) Macy's age was (a) : y . . . (5) Tracy's age will be : 2y .
- (6) Macy's age was (b) : 2y / 3 . . . (7) Tracy's age was (b) : 2 y / 9 (= 1/3 of #6)
- Age difference = constant = x - (44 - x) = 2x - 44. This stays the same at every point in time.
- At point (a) , age diff = y - (x/2) = 2x - 44 ==>
**y = (5/2)x - 44.** - At point (b) , diff = (2y/3) - (2y/9) = 2x - 44 ==>
**y = (9/2)x - 99.** - Equate these and get (5/2)x - 44 = (9/2)x - 99 ; - 2x = - 55 ; x = 27.5
**So Macy is 27 1/2 years old**(and Tracy is 16 1/2). Thanks Jean!

**WINNERS - Problem 93 .**(back to top)**.**leader board . . . FIRST PROBLEM OF 2000-01 CONTEST!**Irina Kolesnik**. . . . . . . 10 pts - Good system of equations; well-explained!**Ludwig Deruyck**(new) . 7 pts - Welcome to the contest; nice 'story' answer!**Al B**. . . . . . . . . . . . . . . . . 5 pts - Fun problem, good answer, glad you're back!**Phil Sayre**.**Allen Druze**. . . . . . . . . 3 pts - Good idea, working backwards!**Sue B**. .**Sarah Mahdavi**. . . . . . 1 pt - m = 29.4 years is close; try checking it with t = 14.6.- Problem #94 - Posted Tuesday, September 19, 2000
- Hole in the Ball (back to top)
**Another classic problem from my puzzling childhood!**- Did you see that wooden ball? It had a hole drilled clean through the center!
- The hole was five inches long; I dunno how big the sphere was, but I can cal-
- culate how much wood was left. What volume was left, and how'd I figure it?
- Be sure to include all your steps and reasoning for maximum ranking!

**Solution:**Arthur said it well: "Since I trust you, Dan, I assumed that since you didn't tell us the diameter- of the hole, it mustn't matter." (Ok, Arthur, I paraphrased. Besides, this is a classic problem I wish I'd invented.)
- That's correct, so if the hole is infinitesimally thin, the ball is a nearly-solid sphere of wood with diameter 5 inches.
- So we can just figure the volume of wood to be V = (4/3) pi * r^3 , where the radius = r = 5/2 here.
- The amount of wood is
**V**= (4/3) pi (5/2)^3**= 125 pi / 6**, which is roughly**65.449847 cubic inches.** - Using Calculus (as Phil and Irina did), you can prove this is the correct volume for a sphere of radius R and any hole radius r, showing we have r^2 + (5/2)^2 = R^2 , looking at a cross section, then doing an integral to calculate the volume of the appropriate solid of revolution, getting an exact answer of 125 pi / 6 cubic inches! . . .
'Give 'em dx, dx, dx.' - UC Berk math cheer ;-}

**WINNERS - Problem 94 .**(back to top)**.**leader board**Arthur Morris**. .**Irina Kolesnik**. . . . . . . 7 pts - Impressive use of calculus! Wait until you study surfaces!**Al B**. . . . . . . . . . . . . . . . 5 pts - Glad you got to discover the hole diam. is irrelevant!**Lisa Schechner**. . . . . . 4 pts - Good answer although the sphere diam can't be less than 5.**Phil Sayre**.**Sue B**. .**George Perry**(new). . . . 3 pts - Quite a poetic answer, and with a fixed-up formula!**Ludwig Deruyck**. . . . . 2 pts - The volume is how much wood is left, but good try!**Sarah Mahdavi**. . . . . . 2 pts - That's true about using half of the wood left and integrating**Tu Truong**. . . . . . . . . . . 2 pts - Thanks for entering; it was hole length not sphere diam.**Caroline Hung**(new). . . 2 pts - It was marked #95 so I saw it late; hole was 5" not sphere.**Jenee Donner**(new). . . . 2 pts - Good way to say diam of hole doesn't matter; formula a bit off- Problem #95 - Posted Friday, September 29, 2000
- Above Average Art! (back to top)
- The graphics department at
dansmath.com has one artist and
**A**production assistants. - Each assistant can make
**B**illustrations per day, but the artist can make**C**more - per day than the daily average of the whole department (including the artist).
**1.**Find A, B, and C , given these clues:- . .
**A + B + C + 3 = (A!) / (5!) = (A^2 + AB) / 3 = 2C.** **2.**What's the total production of dansmath.com, in illustrations per day?- . . A, B, C, and your answer should all be whole numbers.

**Solution: 1.**Let's figure out A, B, and C: Sue sez: A! / 5! = 2 C ; tried A = 6, then 7 ;**A = 7**gave**C = 21**.- . . Phil: A + B + C + 3 =
2 C gives C = A + B + 3. Arthur:
A = 7 means B = 5 + 6 n ; trial
gives
**B = 11**. **2.**Now solve the problem:**1**artist,**7**assts. Each asst makes**11**ill per day, let's define**M**= average (in- . . ill
per day) of whole dept; M + C = M + 21 = artist's ill per day;
total ill
**T**can be written in two ways: - . . T = 7 * 11 + (M + 21) = 8 M ==> 98 + M = 8 M ==> 7 M = 98 ==> M = 14 ==> now get T:
- . . T
= 8 M = 112 ; My graphics dept. produces
**112 illustrations per day.**

**WINNERS - Problem 95 .**(back to top)**.**leader board**Arthur Morris**. .**Phil Sayre**.**Sue B**.**Ludwig Deruyck**. . . . 4 pts - You can arrange the equations in any old order. Good ans.**Al B**. . . . . . . . . . . . . . . . 3 pts - Right that A = 6 yields an impossibly small C.**Irina Kolesnik**. . . . . . 3 pts - Fun equation: A! = 720A/(A - 6); T = 112 is right.**Patrik Petersson**. . . . 3 pts - Nice to hear from you again! A = 7 but show all your attempts!- Here are some points for good attempts but not quite perfect:
**Filipe Gonçalves**.**Montta**. . . . . . . . . . . . . 2 pts - Your numbers were correct but I need to see reasons!**Sarah Mahdavi**. . . . . . 1 pt - Those seem like good A, B, and C but try checking em.- Problem #96 - Posted Sunday, October 8, 2000
- The Missing Dollar (back to top)
- Three guys walk into a hotel. The clerk says, "A room is thirty dollars." Each guy gives the clerk ten dollars. The men go to their room; the clerk remembers the room is only twenty-five dollars. She gives the bellboy five singles to give back to the men. The bellboy can't decide how to split five among three men, so he gives each man one dollar and keeps the other two for himself. Now: The men have payed nine dollars each for the room. Three times nine is only twenty- seven, and with the two the bellboy has, that makes twenty-nine. Explain clearly: Where did the other dollar go?

**Solution:**Many of you got this classic problem:**there is no missing dollar.**The secret is that you don't do 27 + 2 = 29 and compare with 30, the hotel has collected $25 (= 30 - 5), the men have $3 , and the bellboy has 2. That adds up to 30. It's not like anybody's holding $27! This problem can be found in many sources!

**Winners**(back to top)**.**leader board**Irina Kolesnik**. . . . . . . . . . . .10 pts- Right; I created the paradox by 'adding the unaddable.'**Sue B**.**Lisa Schechner**. . . . . . . . . . . . 5 pts - You didn't fall for my mystery addition trap.**Ludwig Deruyck**. . . . . . . . . . 4 pts - That's it; nothing's wrong except my logic.**Arthur Morris**. .**Amy Maris**(new). . . . . . . . 3 pts**Kristen Hehr**(new). . . . . . . . . 3 pts**Al B**. . . . . . . . . . . . . . . . . . 3 pts**Chris North**(new). . . . . . . . . . 3 pts**Cari Richardson**new. . . . . 2 pts**Phil Sayre**.**Exotic Nut**(new). . . 2 pts**Branden Bedwell**. 2 pts**Tom Esser**.**Miho Koto**. .**Nilesh Patel**(new). . 2 pts**Megan Spillane**.**Kevin Ho**.**Quoctan Nguyen**. . 2 pts**Chen Sin Tak**. . . 2 pts**Tim Nelson**. . .**Jesse Goldlink**(new) 2 pts**Yvette Bitar**(new) 2 pts**Paige Campbell**new 2 pts**Montta**. . . . . . . . . . . 2 pts**Filipe Gonçalves**.**Brita Groves**(new) . 1 pt**Slanky**(new) . . . . . . . 1 pt**John DeLaurentis**new 1 pt**Terry Wood**(new). . 1 pt- Problem #97 - Posted Tuesday, October 17, 2000
- Three Funny Numbers (back to top)
- The first funny number: If a 1 is placed after this five-digit number, the result is triple the number you'd get by putting the 1 in front. The second funny number: It's the second-smallest odd number greater than 1 that's a perfect cube and also a perfect square. The third funny number is the smallest with the property that if any prime between 10 and 20 is divided into it, the remainder is 1. What's the exact product of the three funny numbers? Tell the 'whole' story: give the entire answer and show how you got each step.

**Solution: ***Let**n**= the first f.n. Then 10**n**+ 1 = the number with a 1 after it, and it's 3 times as big as 100,000 +**n**= the number with a 1 put in front of it. So we solve 10**n**+ 1 = 3 (100000 +**n**) ; 10**n**+ 1 = 300000 + 3**n**; 7**n**= 299999 ;**n = 42857.*******Now let**m**= the second f.n. We know**m**=**a**^3 =**b**^2 ; any prime**p**in the prime factorization (p.f.) of**m**is also in the p.f. of**a**and in the p.f. of**b**. But the power of**p**in**m**has to be a multiple of 3 because of**a**^3 and a multiple of 2 bec. of the**b**^2. The simplest case is**m**=**p**^6; the smallest odd cases are 3^6 = 729 and**5^6 =****15625.*******The third f.n. let's call**w**. Well,**w**is 1 more than a multiple of 11, of 13, of 17, and of 19. By the famous Chinese Remainder Theorem, we must have**w**= 1 + k (11 * 13 * 17 * 19). With k = 1 we get**w**= 1 + 46189 =**46190.**(You might want to allow k = 0, but that gives**w**= 1 , which isn't very funny, and primes don't really go 'into' it! Some of you thought "divided by" not "divided into.")*****The**product**of these is then 42857 * 15625 * 46190 =**30,930,700,468,750**; over thirty trillion! (I used Mathematica for this; others did it by hand or said that Excel does the trick. It overflowed my regular calculator and also my 'Dan's Talking Calculator'.)

**WINNERS - Problem 97 .**(back to top)**.**leader board**Arthur Morris**. .**Ludwig Deruyck**. . . . 7 pts - Nice explanations to go with the correct answers!**Slanky**. . . . . . . . . . . . . 5 pts - That was my original method, one digit at a time. >whew!<**Irina Kolesnik**. . . . . . 4 pts - Your second way on (3) was right. I didn't dock you a pt; I moved others up.**Lisa Schechner**. . . . . . 4 pts - Thanks for the 'whole story'; you got the disputed bonus pt!**Sue B**.**Tim Nelson**.**Phil Sayre**.**Al B**. . . . . . . . . . . . . . . . 3 pts - Take the square root of cibes; get a whole (odd) number.- Here are points for good but not perfect attempts:
**Filipe Gonçalves**.**Chris North**. .**Chen Sin Tak**.**Jaewon Yoon**(new) . . . 2 pts - Same thing; do 5^6 not 3^6 and divide the primes into n.**Marowen Ng**(new) . . . .2 pts - Good try, but 62485506 was using 42857 and 729 and 2.**Nilesh Patel**.**Megan Spillane**. . . .**Montta**. . . . . . . . . . . . . 1 pt - Good expl on the last two numbers; keep entering!- Problem #98 - Posted Wednesday, October 25, 2000
- The Dinner Date (back to top)
- That 'Math Council Dinner' is in December, but I forgot which night, so I asked around. Alan said that the date was an odd number; Brenda claimed it was greater than 13. Carly declared it was not a perfect square, while Dara swore it was a perfect cube. Finally, Edward told me the date was less than one-fourth his age, which I know to be 68. Yesterday I learned that only one of them had told the truth! What is the date of the dinner?

**Solution:**One way to do it is to make a chart of numbers 1 - 31 and make check marks for each person- (or letters A,B,C,D,E like Slanky did). Here's Montta's solution:
- . .Person . . . . Statement . . . .
. . . Opposite

Alan . . . . . .Odd number . <-> Even number

Brenda . . . .X > 13 . . . . . <-> X <= 13

Carly . . . . .not X^2 . . . . .<-> X^2 . . . . . (meaning X not a square ; X is a square)

Dara . . . . . X^3 . . . . . . . .<-> not X^3 . . (cube, not a cube)

Edward X < 17 . . . . <-> X >= 17 - 1) If Alan's claim is true, the number is odd, X<13, X^2, not X^3 and X>17. But there is no available date.
- 2) If Brenda claim is true and the others' claims are false, the number is even, X.13, X^2, not X^3 and X>17.
- . . . but there is no available date.
- 3) If Carly's claim is true, the number have to be even, X<13, not X^s, not X^3 and X>17. But there is no available date.
- 4) If Dare's claim is true, the number have to be even, X<13, X^2, X^3 and X>17. But there is no available date.
- 5) If Edward's claim is true, the number have to be even,
X<13, X^2, not X^3 and X<17. Date 4 is suitable for all
conditions.

Therefore,**December 4 is the dinner date.**

**WINNERS - Problem 98 .**(back to top)**.**leader board**Irina Kolesnik**. . . . . 10 pts - Nice method; consider 'each truth' one at a time.**Arthur Morris**.**Kia Davidson**.**Sue B**.**Ludwig Deruyck**. . . . 3 pts - I didn't hold your place in line but you got the same score.**Lisa Schechner**. . . . . . 3 pts - Two kinds of people: paranoids and liars... no that's another joke.**Tim Nelson**.**Slanky**. . . . . . . . . . . . . 3 pts - "Ok, Dan, I'm hooked" is like music to my ears! o-}**Al B**. . . . . . . . . . . . . . . . 3 pts - Constant whittling down of the list 1, 2, ..., 31. Good!- The rest were good answers but came in after the first bunch.
**Marowen Ng**. . . . 2 pts - Not 27th**Filipe Gonçalves**. . 2 pts - Hi Filipe!**Nilesh Patel**.**Monsan**(new) . . . . . . 2 pts - Welcome!**Phil Sayre**.**Jesse Goldlink**. . . . 2 pts - Yep, A=sq.**Default**(new) . . . . . .2 pts - 'Given' name?**Montta**. . . . . . . . . . . 2 pts - You're on!**Yvette Bitar**.**Nick Perri**. (new) . . . 2 pts - P ranger.**Chris North**.**John Trott**(new) . . . 2 pts - Words work!

**Katie Nechayev**(new) 2 pts - Hi Katie.**Anthony Leung**(new) 1 pt - Give reasons!- Problem #99 - Posted Friday, November 3, 2000
- Birds on the Run? (back to top)
- Someone opened up the cages at the pet shop, and over 100 birds got away! There were exactly 300 birds to begin with. The local paper (the Daily Poop) reported: "Of the birds that remained, a third were finches, a quarter were budgies, a fifth were canaries, a seventh were mynah birds, and a ninth were parrots. The original number of canaries was three times the number of parrots that stayed." The reporter got just one of the fractions wrong. How many canaries escaped?

**Solution**: This was a fun problem; I wish I'd thought it up but it's stolen! Here's a nice answer by**Monsan**of Japan**:****"**Because one of the fractions is wrong, the number of the birds that escaped is a common multiple of the right 4 numbers.

The common multiple of 3,4,5 and 7 is 420 ; 3,4,5 and 9 is**180**; 3,4,7 and 9 is 252 ; 3,5,7 and 9 is 315 ; 4,5,7 and 9 is 1260.

There were exactly 300 birds, and over 100 birds escaped. So the number of the birds that remained is less than 200.

Only 180 is less than 200. So the number of parrots that stayed is 20, and the original number of canaries is 60.

The number of canaries that stayed is 36. So the number of canaries that escaped is 60 - 36 =**24 canaries escaped."**

**WINNERS - Problem 99 .**(back to top)**.**leader board**Slanky**. . . . . . . . . . . . 10 pts - Right, and 19 mynahs left; the reporter does need math help!**Lisa Schechner**. . . . . . 7 pts - Yes, the main LCM is 1260, and 1260/7 = 180 < 200.**Ludwig Deruyck**. . . . 5 pts - Nice; solve 1/a + 1/b + 1/c + 1/d + x/e = 1 , a,b,c,d in {3,4,5,7,9}.**Monsan**.**Irina Kolesnik**. . . . . . 3 pts - You knew the mynahs for above are 31, 19, 41, 67, 373 too!**Al B**. . . . . . . . . . . . . . . . 3 pts - Right you are; there are 60 finches and 45 budgies left too.**Sue B**.- These were nearly perfect answers or came in later than the others:
**Jesse Goldlink**. . . . . . 2 pts - Fun to watch you discover the answer during your letter!**Phil Sayre**.**Tim Nelson**.**Sudipta Das**(new) . . . . 2 pts - Thanks for entering and explaining yourself so well!- Problem #100 - Posted Saturday, November 11, 2000
- Parents To Bee? (back to top)
- Every male bee has just a female parent, while every female bee has both a male and a
- female parent. How many ancestors does a male bee have, counting ten generations back?
- Twenty generations? Don't count this male as a generation or include him in the 'final tally.'

**Solution**: Arthur Morris started us out on the right bee-line of reasoning:- "The number of female bees in
the preceding generation is equal to number is equal to the total
number of bees in this generation.

The number of male bees in the preceeding generation is equal to the number of female bees in this generation. The number of female and male bees in a series of generations are therefore both**Fibonacci****series**where each number is the sum of the preceding two numbers ( 1 1 2 3 5 8 13 ...)." Good explanation! - Now, going back
**10 generations**, the number of ancestors is the**sum**of a Fibonacci series of length 10 : 1 + 2 + 3 + 5 + . . . + 55 + 89 =**231**. Similarly, going back**20****generations**, the number of ancestors is 1 + 2 + 3 + 5 + . . . + 10946 =**28656**.

**WINNERS - Problem 100 .**(back to top)**.**leader board . . . > whew! there 'bee' lotsa contestants! <**Ludwig Deruyck**. . . 10 pts - Right, it's recursive; 1/2/3/5/8/13/21/34/55/89 adds to 233 - 2.**Tim Nelson**.**Lisa Schechner**. . . . . . 5 pts - Total of Fib. nos. up to F(n) is F(n+2) - 2 ; F(22) - 2 = 28655.**Al B**. . . . . . . . . . . . . . . . 4 pts - Count all the branches on the 'family tree'; 231 + 28424 = 28655.**Jesse Goldlink**. . . . . . 3 pts - Nice use of tables and displaying values, as in 'reform calculus'.**Sudipta Das**. . . . . . . . 3 pts - Ok, the next f is this m + f, and the next m is this f. I like it!**Chris North**.**Montta**. .**Robert Hussey**(new). . 3 pts - Welcome! 28424 was very close, bonus pt for early entry!- These were largely correct answers or came in later than the others:
**Irina Kolesnik**. . . . . . 2 pts - You could include one more generation and get it just right.**Arthur Morris**. . . . . . 2 pts - 89+144-2=231; the 28656 was almost right without the rest.**Slanky**. . . . . . . . . . . . . 2 pts - Recursive sequences can lead to interesting general formulas!**Adam Levy**. . . . . . . . . 2 pts - The 233 is definitely on the right track, fibanachi man!**Dan Lowenthal**(new). . 2 pts - Welcome to the contest! (Add em up for total noi of ancestors)**Nilesh Patel**.**Amy Maris**.**John De Laurentis**. . 2 pts - Good work but your 8, 13, 22, threw the rest off a bit.**Filipe Gonçalves**.**Marowen Ng**. . . . . . . 1 pt - Not sure if the factorials work out, should add not multiply.**Phil Sayre**.**;-}****THANKS to all of you who have entered, or even just clicked and looked.****My site is now in its fifth season - OVER 25,000 HITS so far!**(Not factorial.)**Help it grow by telling your friends, teachers, and family about it.**YOU CAN ALWAYS FIND ME AT **- Dan the Man Bach**- 3*23*29 A.D.

**Problem Archives Index****Probs**&**answers . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90**Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 **Probs**&**answers . . . 91-100 . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151+**Problems only . . . . . 91-100 . 101-110 . 111-120 . 121-