dan's math@home - problem of the week - archives
Problem Archives page 10

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 index

91- 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
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.

Pat Gremban . . . . . . 10 pts - Good answer, well explained, just in time!
Lisa Schechner . . . . 8 pts - Extra point; you only missed first place by 11 minutes.
Arthur Morris . . . . . 6 pts - Very clear explanation, good grasp of probability.
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 . . . . . . . . . 3 pts - Multiples of 0.1%, 1000 boats will solve the problem.
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. . . . . . . . . . . . . . 2 pts - Good on first part; the 48.5% was popular for ii).
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.

Phil Sayre . . . . . . . . .10 pts - (8 sqs)Good job; you can steal my GIF if you want.
Arthur Morris. . . . . . 8 pts - (8 sqs)You were first, but didn't count the squares for me.
Sue B. . . . . . . . . . . . . . 7 pts - (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!

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 . . . . . . . . . . . 4 pts - Yes, at a certain point you don't think of half-years!
Allen Druze . . . . . . . . . 3 pts - Good idea, working backwards!
Sue B. . . . . . . . . . . . . . . .2 pts - Some conditions weren't met by all your answers.
Sarah Mahdavi . . . . . . 1 pt - m = 29.4 years is close; try checking it with t = 14.6.

Problem #94 - Posted Tuesday, September 19, 2000
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 ;-}

Arthur Morris. . . . . . 10 pts - Thanks for trusting my problem-stealing ability!
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 . . . . . . . . . . 3 pts - Nice to have you in the spherical cap calculus club!
Sue B. . . . . . . . . . . . . . . 3 pts - Fancy spherical zone formula: V = pi h (3 a^2 + 3 b^2 + h^2)/6
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
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.

Arthur Morris. . . . . . 10 pts - Asks out loud if this is necessarily the only solution. Yes?
Phil Sayre . . . . . . . . . . 7 pts - Yes, and it's true that 2A = 6 - B sqrt[B^2+12B+108].
Sue B . . . . . . . . . . . . . . 5 pts - Yes, A!/5! is whole and even and so A >= 6.
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. . . . 2 pts - First part perfect, second part you left off the excess!
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
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!

Irina Kolesnik . . . . . . . . . . . .10 pts- Right; I created the paradox by 'adding the unaddable.'
Sue B . . . . . . . . . . . . . . . . . . . . 7 pts - Who did it? The bellboy did!
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. . . . . . . . . . . . 3 pts 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 . . . . . . 2 pts Exotic Nut (new). . . 2 pts Branden Bedwell . 2 pts Tom Esser. . . . . . 2 pts Miho Koto. . . . . . . 2 pts Nilesh Patel (new). . 2 pts Megan Spillane. . 2 pts Kevin Ho . . . . . . . . 2 pts Quoctan Nguyen. . 2 pts Chen Sin Tak . . . 2 pts Tim Nelson . . . . . . 2 pts Jesse Goldlink (new) 2 pts Yvette Bitar (new) 2 pts Paige Campbell new 2 pts Montta . . . . . . . . . . . 2 pts
Here are some good attempts but not ideal:
 Filipe Gonçalves. . . . 1 pt 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
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'.)

Arthur Morris. . . . . . 10 pts - Good job; those without 15-bit arithmetic can use pencil & paper!
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 . . . . . . . . . . . . . . 3 pts - Your intuition was right; there's no smaller answer to (3).
Tim Nelson. . . . . . . . . 3 pts - You meant =1 mod p not =0. I rounded your score up.
Phil Sayre . . . . . . . . . . 3 pts - I like the one digit at a time and 25^3 = 125^2 is like #1.
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. . . . 2 pts - The first one was right on; the second two were off. Bonus pt for early answer.
Chris North. . . . . . . . . 2 pts - 729 is the second square cube but the first odd one. Otherwise very good.
Chen Sin Tak . . . . . . . 2 pts - The 729 and divided 'by' 2 traps; good try though.
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 . . . . . . . . 2 pts - The guess and check was easier because 2 is smaller than 46909.
Megan Spillane . . . . . 1 pt - The 42857 and 729 and 2 weren't too far off but what's their product?
Montta . . . . . . . . . . . . . 1 pt - Good expl on the last two numbers; keep entering!

Problem #98 - Posted Wednesday, October 25, 2000
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.

Irina Kolesnik . . . . . 10 pts - Nice method; consider 'each truth' one at a time.
Arthur Morris . . . . . . 7 pts - Thinking 'negative' thoughts definitely paid off here.
Kia Davidson . . . . . . .5 pts - Only one of Edward or Brenda told a lie; good starting point.
Sue B . . . . . . . . . . . . . . 4 pts - People are assumed to be liars unless proved otherwise, eh?
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. . . . . . . . . 3 pts - Glad to supply you with 'thought for food'.
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 . . . . . 2 pts - Good ans. Monsan (new) . . . . . . 2 pts - Welcome! Phil Sayre . . . . . . . 2 pts - No Python? Jesse Goldlink. . . . 2 pts - Yep, A=sq. Default (new) . . . . . .2 pts - 'Given' name? Montta . . . . . . . . . . . 2 pts - You're on! Yvette Bitar . . . . . 2 pts - Good logic. Nick Perri. (new) . . . 2 pts - P ranger. Chris North . . . . . 2 pts - Deduc. good! 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
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."

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 . . . . . . . . . . . . . 4 pts - Good answer, hope you enjoy the 'spotlight'!
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 . . . . . . . . . . . . . . 3 pts - Nice approach: 1/3 + 1/4 + 1/5 + 1/7 + 1/9 = 1 and 141/3780.
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 . . . . . . . . . . 2 pts - Right answer but there were indeed 180 birds left, not 179.
Tim Nelson. . . . . . . . . 2 pts - Programming doesn't take much more time than doing it by hand!
Sudipta Das (new) . . . . 2 pts - Thanks for entering and explaining yourself so well!

Problem #100 - Posted Saturday, November 11, 2000
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.

Ludwig Deruyck . . . 10 pts - Right, it's recursive; 1/2/3/5/8/13/21/34/55/89 adds to 233 - 2.
Tim Nelson. . . . . . . . . 7 pts - Why think it out when you can get a C compiler to do it 4U!
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 . . . . . . . . 3 pts - Thanks for taking the trouble to make a tree diagram in text!
Montta. . . . . . . . . . . . . . 3 pts - Well-explained and mathematically correct, just not first in!
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 . . . . . . . . 2 pts - Two tries ; the average worked out to be pretty close!
Amy Maris. . . . . . . . . 2 pts - Same as Dan; the right Fibonaccis but need to total em up.
John De Laurentis . . 2 pts - Good work but your 8, 13, 22, threw the rest off a bit.
Filipe Gonçalves. . . . 2 pts - You did a nice table of Fibon's but didn't addemallup...
Marowen Ng . . . . . . . 1 pt - Not sure if the factorials work out, should add not multiply.
Phil Sayre . . . . . . . . . . 1 pt - Light on steps, but made up for it by being just a bit late! ;-}

 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 dansmath.com - Dan the Man Bach - 3*23*29 A.D.
Problem Archives Index

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