dan's math@home - problem of the week - archives
Problem Archives page 6
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 index
 
51- The Age-Old Lie
52 -- Determine - It !
53 - Chickens'n'Eggs
54-The Three Boards
55 - The Millionaire!
56- Subtrac.Diamond
57 - Close to the Line
58 - - Y 2 Digits - - ?
59- - - Bicycle Shorts
60-Odd Combinations
 
 
Problem #51 - Posted Wednesday, July 14, 1999
The Age-Old Lie (back to top)
A reporter on New Year's Eve 1993 wanted to know, from Pat and Chris, how old they were,
but felt (correctly, it turns out) that one would lie. So the reporter asked them both, "Write down
your age now, your age at the end of next year, add these together, then multiply the result
by 5," quickly followed by: "now add the last digit of the year you were born." They had no
time to fake that last digit; Pat answered 281, while Chris announced 229. Who was lying, and
what were their real ages at the time?

Solution: Let p be Pat's age, c be Chris' age, dp and dc the last digits of their birth years. Ok, follow along:
 
Then Pat figures (p + (p+1)) * 5 + dp = 281 , or 10 p + 5 + dp = 281 , or 10 p + dp = 276.
Since 0 <= dp <= 9 , we have p = 27 and dp = 6. Pat being 27 would have been born in 1993 - 27 = 1966,
which does end in a 6, so Pat was 27 and told the truth.
 
Chris went (c + (c+1)) * 5 + dc = 229 , or 10 c + 5 + dc = 229 , or 10 c + dc = 224. Thus c = 22 and dc = 4.
Chris claimed to be 22, but then 1993 - 22 = 1971 would be the birth year, and that doesn't end in a 4.
So Chris was lying, but since dc = 4 , then Chris was born in 1984, 1974, 1964, . . . we don't know!
Chris was really 9, 19, 29, 39, . . . ?
 
Some people claim to be younger than they are, but then some 19-year-olds might wannabe 22, eh?

WINNERS - Problem 51 . . . (back to top) . . . leader board
Andy Murdock. . . 10 pts - Good reasoning except for calling them both "he."
Tim Nelson. . . . . . . 7 pts - You makin funna them funny <= signs?
Cheryl Lim Xuanzi . .5 pts - You have a good clear style. You a teacher?
Beth Wilson . . . . . . 3 pts - Just got it in under the wire! I like your two series approach.
Colleen Frye. . . . . . 2 pts - Good work except for being sure how old Chris was.
Jeff O'Connell . . . . 1 pt - It's a conspiracy! You get a conspiracy point.
 
 
 
Problem #52 - Posted Saturday, July 24, 1999
Determine-It! (back to top)
Find the determinant of the 2 x 2 matrix at the right, where the values are:
A is the absolute value of the difference of two primes, such that the sum of the cubes of the two primes is 6244; B is the largest even number that can't be written as the sum of two odd composite numbers; C is the smallest integer satisfying: C is divisible by 11, C+1 is divisible by 12, C+2 is a multiple of 13, and C is at least 14; and D is the largest odd divisor of 10! (10 factorial).Explain your answers as clearly as you can!

A
B
C
D

Solution: Ok here's Andy's version, which pretty much coincides with mine:
A=6, because 17^3+11^3=6244, and 17-11=6.
B=38 which is the largest even number that can't be written as the sum of
two odd composite numbers.
C=1727. If C is divisible by 11, C+1 by 12, and C+2 by 13, and since 11,
12, and 13 are consecutive, C must be 11 more than a common multiple of the
three numbers, and 1727 is the smallest that works.
D=14175, because starting at 0, 256 is the first number that goes into 10!
an odd number of times, that number being 14175 (which is also 10!/2^8).
So the determinant becomes (6 x 14175) - (38 x 1727) = 19424.

Dan's comments: The determinant of the A B / C D matrix is AD - BC.
Also, p^3 + q^3 = (p + q)(p^2 + pq + q^2) = 6244 = 4 x 7 x 223 , so you suspect that p + q = 28.
B is kinda fun; see if you can list the small odd composites and what they do and don't add up to.

WINNERS - Problem 52 . . . (and maybe more to come!) (back to top) . . . leader board
Andy Murdock. . . . . 10 pts - Four for four -- you were 'determined' to git it.
Rebecca Wooten . . . . 5 pts - Good try and welcome to my contest! A-ok! (and C and D too.)
 
 
Problem #53 - Posted Tuesday, August 3, 1999
Chickens 'n' Eggs (back to top)
I reckon a chicken and a half can lay an egg and a half in a day and a half. Reckon these:
a) How many eggs can ten chickens lay in ten days?
b) How many days does it take ten chickens to lay ten eggs?
c) How many chickens does it take to lay ten eggs in ten days?

Solution: Andy M. decided that one chicken lays eggs at 2/3 per day, so:
(a): 10 ch x 10 days x 2/3 egg/ch/day = 66 2/3 eggs,
(b): Twice as many chickens would lay twice as many eggs in the same amount of time,
so ya gotta reckon 10 chickens lay 10 eggs in 1 1/2 days,
(c): 1 1/2 chickens can lay an egg per day so can lay 10 eggs in 10 days.

WINNERS - Problem 53 . . . (back to top) . . . leader board
Andy Murdock. . . 10 pts - If a student and a half takes a class and a half with a brain and a half...
Beth Wilson . . . . . . 7 pts - Nice way of explaining in terms of rates per ten chickens or days.
 
 
Problem #54 - Posted Saturday, August 14, 1999
The Three Boards (back to top)
Craig was remodeling a house and found three perfectly square boards (not pictured).
He noticed that one of them was five square feet bigger than the smallest, and was
the same amount smaller than the biggest. To his surprise, each of the three boards
measured a whole number of inches on a side! How big were the three boards?

Solution: Look for three squares in arithmetic progression, with a common difference
of 720, because 5 sq ft = 5 x 12" x 12" = 720 sq in .
Kenneth suggested a nice idea of four 180 sq in planks which could be 4 x 45 or 5 x 36,
then look for basically n^2, (n+8)^2 = n^2 + 720, and (n + 18)^2 = n^2 + 1440.
Rebecca wrote a program to find three such numbers that were squares.
The 4 x 45 hinted at a 41 x 41 square inside the planks making a 49 x 49 square;
the correct answer is the boards were 31 in., 41 in., and 49 in. on a side,
making areas of 31^2 = 961 , 41^2 = 1681 = 961 + 720 , and 49^2 = 2401 = 1681 + 720.

WINNERS - Problem 54 . . . (back to top) . . . leader board
Kenneth Potstada. . . 10 pts - Nice geometric intuition; welcome to the contest!
Rebecca Wooten . . . . 8 pts - Hey, a programmer too! I docked you for saying "31 sq in."
Chris Block. . . . . . . . . 2 pts - Good try; one pair was 5 sq ft but the other was 7 sq ft.
 
 
Problem #55 - Posted Friday, August 27, 1999
The Millionaire! (Last problem of the 1998-99 contest) (back to top)
Eddie, the eccentric rich guy, wants to give away one MILLION dollars.
Eddie has two quirks: (1) he gives each person either $1, $7, $49, or some power
of seven, and (2) he won't give more than six people the same size gift.
How many people received money, and exactly how did he distribute his fortune?

Solution: Many of you noticed there was a base 7 thing going on here:
powers of 7 are 1, 7, 49, 343, 2401, 16807, 117649, and 823543 up to a million.
So since Eddie can't give away 7 of the 117649, he gives as follows:
Starts with: $1000000 Total Remaining
1 person receives $823543 $ 823543 176457
1 person receives $117649 $ 117649 58808
3 people receive $16807 each $ 50421 8387
3 people receive $2401 each $ 7203 1184
3 people receive $343 each $ 1029 155
3 people receive $49 each $ 147 8
1 person receives $7 $ 7 1
1 person receives $1 $ 1 0
 
Therefore 1+1+3+3+3+3+1+1 = 16 people received money.

WINNERS - Problem 55 . . . (back to top) . . . leader board
Rebecca Wooten . . . .10 pts - Good approach, using septal numbers (is that the word?).
Bernie Leitner . . . . . . . 7 pts - You not only got the numbers right, but the main recipient's name!
Maria Pereira . . . . . . . . 5 pts - Welcome to the contest; good answer, keep checking in!
Chris Pentacoff . . . . . . 4 pts - If you'd totaled up the people, you'da been second. Good though.
Joon Sup . . . . . . . . . . . . 3 pts - You answered just in time! Again, don't forget to do all parts!
Pooboy . . . . . . . . . . . . . . 1 pt - The amounts had to be powers of 7, as above. Oh, poo!
 
Problem #56 - Posted Sunday, September 5, 1999
Subtraction Diamond (First problem of the 1999-2000 contest) (back to top)
Put four different whole numbers from 1 to 49, in the corners, then subtract the smaller from the larger and put the answers in between. Keep doing this until all four numbers become equal. How many steps can you make it last? The demo is "(14, 30, 18, 37) lasts 3 steps." The winner is the contestant with the first correct longest-lasting entry.
click here
for picture

Solution: I'm not sure if there's a dependable strategy for this one, but Rebecca has analyzed this problem completely.
She not only wins this first problem of my third season, but she now has e-mailed me the complete stats of how many combos there are that last a certain number of steps. I might post that if anyone asks. And I thought the best was 8 steps! (not factorial.)

WINNERS - Problem 56 . . . (back to top) . . . leader board
Rebecca Wooten . . .10 pts - {1, 8, 21, 45} - - 12 steps (56 w/ 12 steps, lots w/ 11 steps)
Kelly K. . . . . . . . . . . . 7 pts - {29, 37, 43, 11} - 8 steps
Chris Pentacoff. . . . . 5 pts - {49, 24, 7, 1}- - - 7 steps
Colleen Frye . . 4, no, 5 pts - {8, 30, 44, 49}- - 6, no, 7 steps (go49ers! Updated score!)
Joon Sup. . . . . . . . . . . 3 pts - {49, 7, 16, 25}- - 6 steps (not 8; sorry)
Nicole Hetzer. . . . . . . 3 pts - {1, 43, 29, 17}- - 6 steps (answer just in time)
Nikki Carr . . . . . . . . . 2 pts - {49, 9, 1, 4} - - - 5 steps
Triono Dawis. . . . . . . 2 pts - {40, 19, 35, 28} - 5 steps
 
Problem #57 - Posted Tuesday, September 14, 1999
Close to the Line (back to top)
The picture at the right shows a line of slope -/2 (square root of 2), through the origin, that passes "close" to the point (2,3). This means that the slope 3/2 is close to -/2. We define "m/n is really close to -/2" to mean that its distance from -/2 is less than 1/n^2. (In our example, |3/2 - -/2| := 1.500-1.414 := 0.086 < 1/4, so 3/2 is "really close.") What are all fractions m/n, with 0<m+n<75, that are really close to -/2? (m and n have to be integers.) The winner is the first entry with the largest difference of (correct fractions) -- (incorrect fractions).
click here
for picture

Solution: Matt R. wrote a PERL program; I created a grid of dots and plotted the line y = -/2 * x
in Mathematica (Mma) and clicked on the dots that looked close to the line (hence the name).
I had Mma round off all the click locations to the nearest integer, and then made the following table.
If the error is to be less than 1/a^2, then d = | error | * a^2 should be less than 1.
 a b b/a decimal c= b/a - -/2 d=| c | * a^2
1 1 1/1 1.0000 -0.41420 0.414
2 3 3/2 1.5000 0.08579 0.343
3 4 4/3 1.3333 -0.08088 0.728
4 6 6/4 1.5000  0.08579 1.37
5 7 7/5 1.4000 -0.01421 0.355
7 10 10/7 1.4286  0.01436 0.704
8 11 11/8 1.3750 -0.03921 2.51
9 13 13/9 1.4444  0.03023 2.45
10 14 14/10 1.4000 -0.01421 1.42
11 16 16/11 1.4545  0.04033 4.88
12 17 17/12 1.4167  0.00245 0.353
13 18 18/13 1.3846 -0.02960 5.00
14 20 20/14 1.4286 0.01436 2.81
. . . (partial list) . . .
That's pretty cool, but I forgot to click on (2, 1)! The answer 2/1 has a d-value of 0.586.
It just didn't look close. Two more answers are 24/17 (d = 0.71) and 41/29 (d = 0.35).
Final list of answers: 1/1, 2/1, 3/2, 4/3, 7/5, 10/7, 17/12, 24/17, 41/29.
 
(Dan's Note: The red fractions are the best approx. to -/2, and also satisfy "Pell's Equation,"
b^2 - 2 a^2 = +1 or -1. This is part of the subject of "continued fractions" in Number Theory.)

WINNERS - Problem 57 . . . (back to top) . . . leader board
Matt Rattigan. . . . . . 10 pts - Good programming skills, that's not cheating !
Patrik Petersson. . . . 7 pts - Welcome to my contest world, Patrik !
Chris Pentacoff . . . . . 4 pts - Close, but no 7/5, and you meant 2/1 not 1/2 ?
Warren Peace. . . . . . . 3 pts - All the fractions you had were pretty close to -/2 !
Rebecca Wooten . . . . 2 pts - Good list but not the < 1/n^2 criterion I meant.
 
Problem #58 - Posted Friday, September 24, 1999
Y 2 Digits? (back to top)
Why not? Figure out all of these whole numbers that you can; each one has two digits.
A's double exceeds its half by 99. B is twice the product of its digits. C is thrice the sum of its digits. Half of D exceeds its third by the sum of its digits. E is increased by 20% if its digits are reversed. F can be squared by sandwiching in two more digits between its original two. G differs from its reverse by twice the product of its digits. The product of H's digits is twice their sum. And turn I upside down and you'll increase it by 12.
The winner will be the first entry with the largest difference of (correct answers) -- (incorrect answers) that's sent in before the deadline.

Solution: In all the problems I'll call the number 10 a + b , where a = 10's digit, b = 1's digit.
A: 2A = A/2 + 99 ; 3A/2 = 99 ; A = 66 .
B: 10 a + b = 2 (a b) ; b = 2 a b - 10 a = 2 a (b - 5) ; b = 6 --> a = 3 ; B = 36
C: 10 a + b = 3 (a + b) ; 7 a = 2 b ; a = 2, b = 7; C = 27
D: D/2 = D/3 + (a + b) ; 1/6 (10 a + b) = a + b ; 4 a = 5 b ; a = 5, b = 4 ; D = 54
E: 10 b + a = 1.2 (10 a + b) ; 50 b + 5 a = 60 a + 6 b ; 55 a = 44 b ; 5 a = 4 b ; E = 45
F: looking at the squares, 95^2 = 9025 and 96^2 = 9216 , so F = 95 or 96. Try (10 a + b)^2 . . .
G: (10 a + b) - (10 b + a) = 2 a b ; 9 a - 2 a b = 9 b ; a = 9 b / (9 - b) ; b = 3 --> a = 9 ; G = 93 or 39
H: a b = 2 (a + b) ; a b - 2 a = 2 b ; a = 2 b / (b - 2) ; H = 36 or 44 or 63
I: Upside downers: 0 --> 0 ; 1 --> 1 ; 6 --> 9 ; 8 --> 8 ; 9 --> 6. I = 86 --> 98 , up 12.

WINNERS - Problem 58 . . . (back to top) . . . leader board
Patrik Petersson. . . . 10 pts - Thanks, I overlooked G = 39 because of my own equation!
Miao . . . =^o^= . . . . . . 7 pts - Welcome to the contest; all correct! Would you like some milk?
Maria Pereira. . . . . . . . 6 pts - "I" admit, there's more than one way to turn a number upside down.
Sue B. . . . . . . . . . . . . . . 5 pts - Ten 4 ten! Not the first, but the sweetest! (honey joke)
Aaron Mikuni . . . . . . . 4 pts - All correct, and just under the wire, too.
Rebecca Wooten . . . . 4 pts - Good question on I, and your E was bakcwrasd.
Danielle Janney . . . . . 4 pts - Thanks for entering! After A, it was all perfect.
Rajwant Singh. . . . . . 3 pts - Mostly ok but your B, D, and G need checking.
Shinsuke Kuzuya. . . . 2 pts - Some ok, some didn't all work, like D.
Daniel Cho. . . . . . . . . . 1 pt - Thanks for showing steps. It helps you think em out!
 
 
Problem #59 - Posted Monday, October 4, 1999
Bicycle Shorts (back to top)
'Bikeman' wants to tour five cities in one day: Anamoose, Brindle, Catfish, Danville, & Easton; located at (1,0), (2,3), (-2,4), (5,-1), and (-3,-2), respectively. He wants to see all cities (not nec. in that order) and travels from town to town in straight lines. What is the shortest total distance possible, where should he start, and what is the sequence of cities he visits? The winner is the first entry with the shortest distance submitted before the deadline. Exact answer or at least to four decimal places.

 Solution: Figure out all the straight line distances by the
Pythag Thm and write to 5 places:
AB = -/10 :=: 3.16228 ; AC = 5.00000 ; AD = -/17 :=: 4.12311 ;
AE = -/20 :=: 4.47214 ; BC = -/17 :=:4.12311 ; BD = 5.00000 ;
BE = -/50 :=: 7.07107 ; CD = -/74 :=: 8.60322 ;
CE = -/37 :=: 6.08276 ; DE = -/65 :=: 8.06226 .
Draw a graph ; a good path seems to be E-A-D-B-C :=: 17.7183 ,
but E-C-B-A-D :=: 17.4912 is the shortest.
The exact shortest distance is ECBAD = -/10 + 2-/17 + -/37.
Any path can be run backwards; same distance!
click here
for picture

WINNERS - Problem 59 . Thanks for all your entries; this is my largest group ever! (back to top) . leader board
Patrik Petersson. . . . 10 pts- (first with ECBAD and correct distance)
Danielle Janney . . . . . 7 pts - (second entry with ECBAD & correct dist)
Miao . . . =^o^= . . . . . . 5 pts - (third with ECBAD & correct dist)
Yuri Sumnicht . . . . . . 4 pts - (last perfect answer)
Rajwant Singh . . . . . . 3 pts - (ECBAD ok but miscalc dist)
Chris Pentacoff . . . . . 3 pts - (likewise but good expl.)
Shinsuke Kuzuya. . . . 3 pts - (First EADBC, the next best path)
Maria Pereira . . . . . . . 2 pts - (Next with EADBC)
Dave Seaman . . . . . . . 2 pts - (Third with CBDAE)
Sue B. . . . . . . . . . . . . . .2 pts - (Fourth along that road. Thanks for returning to the hive!)
Daniel Cho . . . . . . . . . 2 pts - (ECBAD ok but no dist given)
Gohan . . . . . . . . . . . . . . 2 pts - (EADBC but only 2 decimal places. Picky!)
Nicole Hetzer . . . . . . . 1 pt - (last with ECBAD but dist miscalc)
Sarah Mahdavi . . . . . . 1 pt - (EDABC pretty good but 19.47 miles)
 
 
Problem #60 - Posted Wednesday, October 13, 1999
Odd Combinations! (back to top)
You have one each of the numbers 1, 3, 5, 7, 9. Using just basic operations
+, -, *, /, and ( ), 'stitch together' as many answers from 1 to 10 as you can.
(Example: 1 - (5 - 3) + (9 + 7) = 15 , but that's not between 1 and 10.) The winner will be
the first entry with the highest score, figured by (correct answers) - (incorrect answers).

Solution: I'm rewarding multiple answers but what I wanted most was all answers 1-10.
Danielle got all 10 in first, while some of you got 25, 38, and even 86 answers! -^miao^-
Here are the last answers sent in, all 1-10 once, by SueB: 1 = ( 5 + 7 ) - ( 9 - 1) - 3 ;
2 = 3 * ( 5 + 1 ) - ( 7 + 9 ) ; 3 = ( 9 + 5 ) - ( 1 + 3 + 7 ) ; 4 = ( 5 + 9 ) - ( 7 + 3 ) * 1 ;
5 = ( 9 / 3 ) + 7 - ( 5 / 1 ) ; 6 = ( 5 + 7 ) - ( 9 * 1 ) + 3 ; 7 = ( 7 * 9 ) / ( 5 + 3 + 1 ) ;
8 = ( 5 * 7 ) - ( 9 * 3 * 1 ) ; 9 = ( 5 * 7 ) - ( 9 * 3 ) + 1 ; 10 = ( 5 + 7 ) - ( 9 / 3 ) + 1 .

WINNERS - Problem 60 . Thanks again for all your entries! (back to top) . leader board
Danielle Janney . . . 10 pts - First with all 10 answers; keep it up!
Miao . . . =^o^= . . . . . 8 pts - Bonus point for the 76 extra answers!
Connie C . . . . . . . . . . 5 pts - Welcome to my contest; tell your kids!
Sue B. . . . . . . . . . . . . . 4 pts - Just buzzed in under the wire!
Joon Sup . . . . . . . . . . 3 pts - (more points if you'd evaluated them!)
Camilo Jimenez. . . . 2 pts - You got most of the target numbers - good!
Rebecca Wooten . . . 2 pts - You only get to use each number once, but I liked your recursions.
Randy Toy. . . . . . . . . 1 pt - Keep entering - you were very close on some.
Gohan . . . . . . . . . . . . .1 pt - If you worked em out, send em in!
Charles Stanton . . . . 1 pt - Hey, another programmer! Make it pick one of each number.
Dave Seaman. . . . . . . 1 pt - You got a partial result, but here's a full point.
 
 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
 
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 . 121-130 . 131-140 . 141-150 . 151+
 
 
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