dan's math@home - problem of the week - archives
 
 
Problem Archives page 5
with Answers and Winners. For Problems Only, click here.
 
Probs & answers . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
Probs & answers . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Problems only . . . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Probs & answers . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+
Problems only . . . . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+
 
41- Car AND Train!
42 - Co-Perfect Nos.
43 -"Divisorly" Nos.
44 - The Lying Truth
45 - Not Sold at 7-11
46- Round the Corner
47 - Roll That Prime!
48 - Cut The Cheese!
49 - Not So Peachy!
50 - Bouncing Ball...
 
 
Problem #41 - Posted Tuesday, March 16, 1999
Car AND Train!! (back to top)
Every day, Ellie takes the commuter train and arrives at the station 8:30 AM, where she's
immediately picked up by a car and driven to work.
One day she takes the early train, arrives at the station at 7:00 AM, and begins to walk
towards work. The car picks her up along the way and she gets to work 10 minutes
earlier than usual. When did Ellie meet the car on this day? (Explain fully.)
Solution: If the car drives 10 minutes less than it usually does, then it drives 5 minutes less time in each
direction, so Ellie gets in the car 5 min. early, or 8:25 AM.
WINNERS - Problem 41 . . . (back to top) . . . leader board
Andy Murdock. . . . . . . 10 pts - It's true, Ellie was early, not surly.
Jeffrey Hunt. . . . . . . . . . 6 pts - Your logic is good; the total was 10 min saved.
Jessica Bubb. . . . . . . . . . 3 pts - Picking her up at 8:15 would make her 30 min early.
Huan-yu (Richy) Chung. 1 pt - There was more information in there than you might think!
 
 
Problem #42 - Posted Sunday, March 28, 1999
Co-Perfect Numbers (back to top)
A perfect number is one whose positive proper divisors add up to the
number itself. (For example: 6 = 1 + 2 + 3 ; 28 = 1 + 2 + 4 + 7 + 14 ; the next is 496.)
Can you find a pair of numbers, so that the proper divisors of each
number add up to the other one? (Give a complete proof.)
Solution: What do all the divisors of a number n add up to (incl. n)? Call the sum s(n).
You can see if each number goes into n, then keep a running total.
 
Or for more efficiency you can look at the prime factorization of n. If n is a power of a prime
(like n = 9 = 3^2) the only divisors are 1, 3, 9. Then s(9) = 1 + 3 + 9 = 13. Also s(8) = 1 + 2 + 4 + 8 = 15.
If n has more than one prime in it then the s's multiply: s(72) = 1+2+3+4+6+8+9+12+18+24+36+72 = 195,
while s(8) * s(9) = 15 * 13 = 195. The proper divisors would add up to 195 - 72 = 123.
Playing with prime factors might lead you to the pair 220 = 2^2 * 5 * 11 and 284 = 2^2 * 71 ; both s(n) = 504.
 
s(220) = s(4) s(5) s(11) = (1+2+4)(1+5)(1+11) = 7 * 6 * 12 = 504 ; 504 - 220 = 284.
s(284) = s(4) s(71) = (1+2+4)(1+71) = 7 * 72 = 504 , and again 504 - 284 = 220.
 
A familiar way to say it is : sprop(220) = 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284 ,
while on the other hand . . . sprop(284) = 1 + 2 + 4 + 71 + 142 = 220 . Whaddaya know!
 
Notice that the number of divisors d(n) is the product of how many divisors each prime power has:
d(72) = d(8) d(9) = d(2^3) d(3^2) = 4 * 3 = 12 ; d(220) = d(4) d(5) d(11) = 3 * 2 * 2 = 12 , etc.
 
The numbers 220 and 284 are the first "amicable pair" - I called them co-perfect so you couldn't just do
an internet search on perfect and amicable numbers! Beth Wilson also found the next few friendly pairs.
WINNERS - Problem 42 . . . (back to top) . . . leader board
Jeffrey Hunt. . . . . . . . . . 10 pts - Good answer; explain what path took you to the nos.
Wilson Matheson. . . . . . .7 pts - Welcome to our contest! Can you enter again?
Beth Wilson . . . . . . . . . . . 6 pts - Should be 5 for 3rd ans but you wrote & explained a program!
Jen-Kai Pan. . . . . . . . . . . . 2 pts - 1, 2, 4, 7, 14, 28 aren't the only divisors of 56, etc.
AD/Marsha Reichle. . . . . 2 pts - 1 + 7 = 8, 1 + 2 + 4 = 7. Good try, but 7 isn't proper div of 7.
 
 
Problem #43 - Posted Friday, April 9, 1999
Divisorly Numbers (back to top)
A 'divisor' of a natural number is one that goes in with no remainder (18 has 6 divisors: 1, 2, 3, 6, 9, 18.)
a) What's the smallest number with at least 14 divisors?
b) What's the smallest number with exactly 14 divisors?
c) Find the smallest number with exactly 100 divisors.
(Hint? See the solution for last week's problem in the archives.)
Solution: The more factors of small primes in a number, the more divisors it has. So we look at things like
12, 24, 60, etc., with powers of 2, 3, and 5. Also it's most efficient if there are more factors of 2 than of 3, etc.
Let d(n) = number of divisors of n; for ex. d(18) = 6. d(12) = 6, d(24) = 8, d(36) = 9, d(48) = 10,
d(60) = 12, d(96) = 12, d(120) = 16, d(144) = 15 (count 'em up, or use prime factors; see my Number Theory lessons)
a) The smallest no. with at least 14 divisors is 120, which has 16: {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120}
b) The smallest number with exactly 14 divisors is 192 : {1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192}
(14 = 7 * 2 ; the number is 2^6 * 3^1 = 64 * 3 = 192. (Note 2^13 = 8192 and 2^1 * 3^6 = 1458 also have 14 divisors.)
c) Now how can n have 100 divisors? 100 = 2^2 * 5^2 , so write 100 as a product of factors, then use
exponents that are 1 less (see my Number Theory lessons). For example:
100 = 10 * 10 , n = 2^9 * 3^9 = 10,077,696 (not so small), and d(n) = 100. Some other ways:
100 = 10 * 5 * 2 , n = 2^9 * 3^4 * 5^1 = 207,360 (better),
100 = 5 * 5 * 4 , n = 2^4 * 3^4 * 5^3 = 162,000 (still better),
100 = 5 * 5 * 2 * 2 , n = 2^4 * 3^4 * 5^1 * 7^1 = 45,360 (best with exactly 100 divisors.)
WINNERS - Problem 43 . . . (back to top) . . . leader board
James Layland. . . . . . . . . 10 pts - Good job, how'd you find em? Welcome to the contest.
Tim Nelson. . . . . . . . . . . . .7 pts - Writing a program isn't cheating, but is it an efficient one?
 
 
Problem #44 - Posted Saturday, April 24, 1999
The Lying Truth (back to top)
You missed math class yesterday and you want to know if there's a test today.
There are two students outside the classroom; one always lies and one always tells
the truth, but you don't know which is which. What single yes-or-no question can
you ask (one of them) so that no matter who answers, you'll know if there's a test?
Solution: I got several interesting responses to this, some work, some don't, some I made up.
Which questions work? (Which ones allow you to tell if there's a test, not tell if they're a liar!)
1. Are you waiting for today's test?
2. If I were to ask you if there was a test today, what would you say?
2 1/2. If I were to ask you if there was a test today, would you say 'NO'?
3. Have you studied hard enough for the math test today?
4. What would the other one say about [whether there's a] test?

 

 
The lying truth is that these all work, except for #1 and #3. Well ok, that's only half of them.
I prefer #2, because with #4 you have to assume each one knows what the other one is, but
not so with #2. Like I always say, it takes a truthful person to admit to being a liar!
WINNERS - Problem 44 . . . (back to top) . . . leader board
Tim Nelson . . . . . . . . . . 10 pts - The original classic problem is no longer politically correct.
Huan-yu (Richy) Chung. . 7 pts - I like your question (2 1/2), it made me think twice!
Colleen Frye . . . . . . . . . . 4 pts - Your question tried to answer too much...
Yu Chen Wang . . . . . . . . 2 pts - Even if there's a test, people might not study for it.
Jen-Kai Pan . . . . . . . . . . . 1 pt - These students wouldn't give a satisfactory answer...
 
 
Problem #45 - Posted Tuesday, May 4, 1999
Not Sold at 7-11! (back to top)
The local convenience store sells a gallon of milk for $7 and a box of cereal for $11. If you bought 3 milks and 2 cereals you'd spend $43, but there's no way to spend exactly, say, $20. Starting at $1, how many whole numbers (such as $20) can't equal the "total" of some milks and cereals, and what are they? (It's ok to buy all milk or all cereal. Hint: After a certain point, all totals are possible!)
Solution: I'm pretty sure there are 30 unattainable numbers, Beth's list is 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 13, 15, 16, 17, 19, 20, 23, 24, 26, 27, 30, 31, 34, 37, 38, 41, 45, 48, 52, 59. Tim Nelson claims that (7-1)(11-1) = 60 has someting to do with this. Richy Chung used a nice argument with numbers arranged in rows of 11 and then 7.
Dan's note: If you got to exchange food for cash you can make any amount from $1 on; too bad 7-11 doesn't buy food!
WINNERS - Problem 45 . . . (back to top) . . . leader board
Colleen Frye . . . . . . 10 pts - Sure cereal is expensive; (cost) = k * (convenience).
Richy Chung. . . . . . . 7 pts - Right; I liked your visual 'table' approach!
Tim Nelson. . . . . . . . 6 pts - You were first with the right list but not the total number.
Beth Wilson . . . . . . . 4 pts - Same here, but I like the way you say how you solved it!
 
Problem #46 - Posted Saturday, May 15, 1999

Round the Corner! (back to top)

a) There's a big circle (7 feet across) painted in the corner of a room, and a little circle crammed in that just touches the big circle. What's the diameter of the little circle?

b) Do the same problem using spheres: a big ball (of diam 7 ft) just touches the three walls, and a little ball touches three walls and the big ball. What's the diameter of the little ball?

Give answers involving fractions or radicals, not decimal approximations.

Solution:
a) Instead of similar triangles, let R be the radius of the large circle, r be the (unknown) radius of the small circle, and x be the distance from the corner to the lower edge of the small circle. The distances from the corner to the centers of the circles are R-/2 and r-/2, so we have two equations:
R + 2 r + x = R-/2 , and r + x = r -/2 ; subtracting to cancel x's we get:
R + r = R-/2 -- r -/2 , or r = R (-/2 - 1) / (-/2 + 1). The diameters are in the same proportion, so
d = D (-/2 - 1) / (-/2 + 1) = 7 (-/2 -- 1) / (-/2 + 1) = 7 (3 -- 2-/2) ft (approx. 1.201 ft.)
 
b) In the 3-dimensional case, the R, r, and x can be defined similarly, but the distance to the corner is the radius times -/3 , so in the same way we get
d = D (-/3 - 1) / (-/3 + 1) = 7 (-/3 -- 1) / (-/3 + 1) = 7 (2 -- -/3) ft (approx. 1.876 ft.)
 
(Dan's note: Following our running 7-11 theme, the small circle has radius about 7" while the small ball has radius about 11". ;-})
WINNERS - Problem 46 . . . (back to top) . . . leader board
Richy Chung . . . . 7 pts - Good but I wanted diam not radius, problem with 3D ans.
Andy Murdock . . 5 pts - Gotcha, finally, but the 3D case is different, honest!
Tim Nelson. . . . . . 4 pts - One mystery quadratic that works, where from? No 3D...
(Richy - Sorry about unreliable DVC e-mail!)
 
 
Problem #47 - Posted Monday, May 24, 1999

Roll That Prime! (back to top)

Three standard fair dice are thrown. What's the probability that the total of the three is a prime number?

Solution: There are 6 x 6 x 6 = 216 ways to throw the dice. The possible prime totals are: 3, 5, 7, 11, 13, 17.

The number of ways to get each of these is: 3 (1 way), 5 (6 ways), 7 (15), 11 (27), 13 (21), 17 (3). For example for 7 you could get: 1, 1, 5 (3 ways), 1, 2, 4 (6 ways), 1, 3, 3 (3 ways), 2, 2, 3 (3 ways); 15 ways in all.

The total number of ways to roll a prime total: 1 + 6 + 15 + 27 + 21 + 3 = 73 ; the probability of a prime roll is 73 / 216 (about 33.8%).

WINNERS - Problem 47 . . . (back to top) . . . leader board
Beth Wilson . . . . . 8 pts - You got all the parts right but missed a 3 adding it up!
Colleen Frye. . . . . 6 pts - Part right, but diff dice are always diff; 3 ways to roll 2-2-1, etc.
Tim Nelson. . . . . . 4 pts - Not sure how you got your answer if you hold on to the 'details.'
Andy Murdock. . . 3 pts - There're 6ix ways of rolling, say, a 2-4-5, etc. Close, but cigar!
 
 
Problem #48 - Posted Saturday, June 5, 1999
Cut The Cheese! (back to top)
Take a big cube of cheese (12" x 12" x 12"), and cut through it with a knife. Only cut through the corners and midpoints shown, and look at the cross sections you get. a) What are all regular polygons you can get, and what are their areas and perimeters? b) Allowing non-regular (but planar) polygons, which has the most area, and what is its area? c) Which planar polygon has the most perimeter, and what is that longest perimeter? d) Which polygon has the largest area- to- perimeter ratio, and what is that ratio?
 
Solution: by Rebecca Wooten
A REGULAR
Equilateral PHE, PH'E', EFQ, E'F'Q, FGR, F'G'R, GHS, G'H'S
Cutting off the corners (small) : A = (1/2)*6rt2*3rt6~31.2 . . . P = 3*6rt2~25.5 . . . R = 1.22
Equilateral AB'D', A'BD, BC'A', B'CA, CD'B', C'DB, DA'C', D'AC
Cutting off the corners (large) : A = (1/2)*12-/2*6-/6~124.7 . . . P = 3*12-/2~50.9 . . . R = 2.45
Squares EGG'E', HFF'H', PQRS,
Cutting block in half (not on the diagonal): A = 12*12 = 144 sq. in. . . . P = 4*12 = 48 in. . . . R = 3
Hexagons EHSG'F'Q, FEPH'G'R, GFQE'H'S, HGRF'E'P
Cutting many midpoints : A = (1/2)*6-/2*3-/6~187.1 . . . P = 6*6-/2~50.9 . . . R = 3.68
 
B NON-REGULAR
Type 1 PBD, PB'D', QAC, QA'C', RBD, RB'D', SAC, SA'C'
Isosceles (Small) : A = (1/2)*12-/2*6~50.9 . . . P = 12-/2+2*6-/3~37.8 . . . R = 1.34
Type 2 AH'E', BE'F', CF'G', DG'H', A'HE, B'EF, C'FG, D'GH
Isosceles (Large) : A = (1/2)*6*6-/2~25.5 . . . P = 2*6-/3+6-/2~29.3 . . . R = .87
Type 3 EHH'E, EFF'E', HGG'H', GFF'G', PHFQ, PH'F'Q, SHER, SH'E'R,
SGEP, SG'E'P, RGEQ, RG'E'Q
Rectangle (Small) : A = 12 * 6-/2 ~ 101.8 . . . P = 12-/2 + 24 ~ 41.0 . . . R = 2.48
Type 4 AB'C'D, BCD'A', CDA'B', DAB'C', AA'C'C, BB'D'D
Rectangle (Large) Cutting block in half along diagonal
. . . A = 12 -/2 *12 ~ 203.6 . . . P = 24 -/2 + 24 ~ 57.9 . . . R = 3.52
 
C Largest Area: Type 4 (big rectangle) Largest Perimeter: Type 4 also
 
D Largest Ratio: Hexagon R = 3.68
WINNER - Problem 48 . . . (back to top) . . . leader board
Rebecca Wooten . 10 pts - Great effort - and I forgot about the isosceleses! <| :-}

 
Problem #49 - Posted Friday, June 18, 1999
Not So Peachy! (back to top)
As a fuzzy fruit farmer, I know how many peaches people produce. For fun I figured that last year my peach trees averaged as many peaches per tree as there were trees. But this year I reckoned the average was only 97 ppt (peaches per tree); that's 3500 fewer total peaches with the same number of trees. How many peach trees do I farm? Explain your solution!

 
Solution: Let's say there are n trees ("there are n trees!"); last year there were n(n) = n^2 peaches.
This year there were 97(n) peaches, 3500 fewer than last year's n^2, so n^2 - 3500 = 97 n .
Solving n^2 - 97 n - 3500 = 0 ; factor and get (n - 125) (n + 28) = 0 ; n = 125 or n = - 28 ; if you want help,
see my factoring lessons, or use the quadratic formula. I'd say in this case throw out the negative answer, so
that n = 125 peach trees. Checking; 125^2 = 15625 ; 97(125) = 12125; 15625 - 12125 = 3500. Check!
WINNERS - Problem 49 . . . (back to top) . . . leader board
Cheryl Lim Xuanzi . 10 pts - Welcome to the contest! Good work, sorry about delay.
Beth Wilson . . . . . . . . 6 pts - Nice answer as usual; I'd like to see a couple more steps
Tim Nelson. . . . . . . . . 4 pts - Equation was ok but you grew 25 more trtees while solving
 
 
Problem #50 - Posted Tuesday, June 29, 1999
The Bouncing Ball . . . (back to top)
A basketball is dropped straight down from the rim height of 10 feet. Each bounce, it
rises to the same percentage (p%) of its previous height. The ball finally comes to rest
after traveling a total of 70 (vertical) feet. What is the mystery "rebound" percentage p?
Solution: The total distance involves the sum of a geometric series with ratio p% (or p/100, which we'll call r):
The ball bounces twice at each height except for the first one, so add, double, and subtract 10:
Total dist = 2 (10 + 10 r + 10 r^2 + . . .) - 10 = 2 (10 / (1 - r) ) - 10 = 20 / (1 - r) - 10 = 70 ;
the formula S = a / (1 - r) is well-known; our solution is 20 / (1 - r) = 80 ; 1 / (1 - r) = 4 ; 1 - r = 1/4 ; r = 3/4 ;
we discover our mystery rebound percentage: p = 100 (3/4) = 75%.
Checking; 2 (10 + 7.5 + 5.625 + 4.21875 + . . .) - 10 = uh, yep, 70 ! (not factorial!). Ok, don't check it that way.
WINNERS - Problem 50 . . . (back to top) . . . leader board
Tim Nelson. . . . . . 10 pts - Thanks for the update, I mean proof, of formula S = a / (1 - r) .
Beth Wilson . . . . . 7 pts - Just got it in under the wire! I like your two series approach.
Andy Murdock. . . 3 pts - You found the height (not sure of what), not the percentage.
 
 THANKS to all of you who have entered, or even just clicked and looked.
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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 . 91-100
Problems only . . . 1-10 . 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100
Probs & answers . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Problems only . . . 101-110 . 111-120 . 121-130 . 131-140 . 141-150 . 151-160 . 161-170 . 171-180
Probs & answers . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+
Problems only . . . . 181-190 . 191-200 . 201-210 . 211-220 . 221-230 . 231+
 
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