dan's math@home - problem of the week - archives
Problem Archives page 12

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

111 -- Wedding Cake
112 - Hands On Time
113 - The Two Cubes
114 - Price of Stamps
115 - Thinking Caps!
116 -- Hoop It Up ! !
117 - Kissing Circles
118 -Cubes & Squares
119 Better than Perfect
120 -- Rotating Tires!

Problem #111 - Posted Tuesday, February 20, 2001
"Sorry, I'm all out of two-tiered cakes," said the pastry chef. "What about
this square one; could you make two layers of it?" "Sure, I could make two
square tiers-- the bottom with a side that's twice as long as that of the top.
And to avoid crumbling, I'll use the fewest number of straight cuts possible."
What is that smallest number of cuts, and exactly how is it done?

Solution: It was understood that you should use all the cake. Otherwise you could just cut a 1" square and
a 2" square and stack 'em. And some of you decided to cut horizontally to create more layers, then making and
stacking quarters of layers, but there was still some waste.

The key is to measure areas first. The square can be any size, so assume it's a 10" by 10" square and ignore the height.
The total area is 100 in2, so the layers are x by x and 2x by 2x , and: (x)^2 + (2x)^2 = 100 ; so x = sqrt(20) = 2 sqrt(5).
The layer sides will be 2 sqrt(5) and 4 sqrt(5) .
 Arthur found an elegant solution with four cuts: make four right triangles with hypotenuse 10 in and legs 2 sqrt(5) and 4 sqrt(5) ; fit the four together to make the lower tier, and the little square in the middle makes the upper square.   (Dan's note: The optimal solution is with three cuts, gotten by joining corners with opposite midpoints, as shown.)
Tim, Lisa, Tat, and Kristen suggested two cuts: cutting into 4 pieces along the 2/3 , 1/3 lines, making two squares,
one 1/3 by 1/3 , the other 2/3 by 2/3 , and discarding two rectangles, each 2/3 by 1/3. (Dan's note: Place the rectangles
side by side on top of the big square, then top it with the small square; it has the shape of a two-tiered cake, but it's really three, isn't it.)

Arthur Morris . . . . . .8 pts - Nice symmetry there, and the triangles fit! But 3 or 2 cuts suffice!
Ludwig Deruyck . . . 5 pts - Cut into layers, quarter twice and stack - nice idea but wasted cake.
Tim Boomer (new) . . . 4 pts - First with the idea of 2/3, 1/3 ratio - but wasted the rectangles.
Lisa Schechner. . . . . 3 pts - Two tries, one like Ludwig, one like Tim. Re-use the rectangles!
Tat Tsui . . . . . . . . . . . 3 pts - Two layers, three cuts, seven quarters - but 1 1/2 qtrs wasted.
Kristen Hehr. . . . . . . 3 pts - Welcomoe back! Baker can make 2 cuts but should reuse the rectangles
Phil Sayre . . . . . . . . . 3 pts - Your 5 cuts and "fudging" with the icing did involve 4.47 ~ sqrt[20]
Richard Clarke. . . . . 2 pts - Nice to hear from you! After you cut it in quarters, what do you do exactly?

Problem #112 - Posted Friday, March 2, 2001
"What time is it, Rory?" asked Cory one lazy day. "When I checked my watch this morning, the
hour hand was where the minute hand is now, and the minute hand was one minute before where
the hour hand now sits. I notice both hands are now at exact minute divisions."
What is the time now? When did Rory check this morning?

Solution: From Sudipta Das (Calcutta , India): Suppose the time now is x minutes past y. Then the minute hand
coincides with the x th minute mark and the hour hand coincides with the (5y + x/12) th minute mark.
As (5y + x/12) is an integer , x must be a multiple of 12 .Also, as 0<=x<60, x is either 0 or 12 or 24 or 36 or 48.
Again , suppose the earlier time was z minutes past k. Then , the minute hand was coinciding with the z th minute
mark, and the hour hand was coinciding with the (5k+ z/12)th minute mark.This means that z was a multiple of 12.
Now, z = 5y + x/12 - 1 and x = 5k + z/12 . . . z/12 can take values 0,1,2,3 or 4.
So, x is any one of (5k) , (5k+1) , (5k+2) , (5k+3) or (5k+4). Equating these with the values x can take, we get
x k z 5y-1
0 0 0 0
12 2 24 23
24 4 48 46
36 7 12 9
48 9 36 32

Of the above cases , y is an integer only when 5 y - 1 = 9 . . . So, x = 36 , k = 7 , z = 12 , y = 2 . . . Thus , the
time now is 2:36 p.m. and Rory checked the time earlier in the morning at 7:12 a.m.

Sudipta Das . . . . . . . 10 pts - Great solution/proof. Thanks for submitting it; you're web-published!
Arthur Morris . . . . . .7 pts - You got up before West Coast time but not before Calcutta! Good method.
Ludwig Deruyck . . . 5 pts - You EXCEL at adapting clunky software to do cool math problems!
Lisa Schechner . . . . 4 pts - Good to cut down to V = multiples of 12, then V = Y - 1 solves it!
Al B . . . . . . . . . . . . . . 3 pts - 36 min past is a mult of 12 and looks like 7; good!

 Problem #113 - Posted Sunday, March 11, 2001 The Two Cubes (back to top) Woody had two solid cubes; one was 10 cm on a side and the other was slightly larger. He showed me how he'd cut a square tunnel completely through one cube(which remained intact) so the other could pass through it. The amazing thing was that it was the smaller cube that had the hole in it !! How was this possible, and what is the biggest theoretical side of the larger cube? (Hint: Look at the smaller cube from one corner.) graphic by dan bach

Solution: This is one of my all-time favorite problems. Thanks to those of you who tried it, or even thought it was impossible! -Dan
From a corner, the small cube will look like a hexagon. The 10 cm side is now "foreshortened" by the angle of view...

New contestant Tim Poe attached a nice picture and solved the problem this way:
 "If you slice horizontally through the midpoint, you'll cut the corner off of six sides from midpoint on one edge through the midpoint of an adjacent edge. The cross section forms a hexagon with each edge measuring (10/2)sqrt(2) cm = 5sqrt(2) cm. However, that leaves material hanging "over the edges" of the hexagon. What would have been the corners of the cross-section described above are the midpoints of the sides of a larger hexagon created by the overhanging points. If the distance from the center to the midpoint of the side is h = 5sqrt(2) cm, then the sides of the hexagon will be h^2 = s^2 - (s/2)^2 = s^2 - (s^2)/4 = (3s^2)/4; h = 5sqrt(2) = s*sqrt(3)/2; s = 10sqrt(2)/sqrt(3).   "If the square is aligned so that all four corners lie on the edges of the hexagon, then the line from the center to the corner of the square (see portion of figure in the black outline) can be described as a line from the lower-left corner of an equilateral triangle with sides 10sqrt(2)/sqrt(3), rising at 45 degrees to intersect the upper-right edge of the triangle. This line bisects the triangle into two parts; the lower part has angles A = 45 and C = 60; therefore, angle B = 75. Side b (opposite angle B) is 10sqrt(2)/sqrt(3) cm long.   "We're looking for the length of side c (opposite angle C). By the law of sines, c/sinC=b/sinB=a/sinA=2R, so c/sin60=10sqrt(2)/sqrt(3)sin75; c=10sqrt(2)sin60/sqrt(3)sin75 (approx. 7.3205). The diagonal would be twice that distance, or 20sqrt(2)sin60/sqrt(3)sin75 (approx. 14.6410). The side of a square with that diagonal is diagonal/sqrt(2) = (20sqrt(2)sin60)/(sqrt(2)sqrt(3)sin75) = 20sin60/(sqrt(3)sin75) (approx 10.3528cm)."
Dan's note: I have to admit I thought if you stand a cube with corners up and down, so from the front it looks like a
regular hexagon, that the top corner would be directly over the bottom, and that you had a vertical hexagon distance of
10-/3 but it's not true.That assumption led to a 10.86 cm cube being able to pass through. In fact the largest square hole
happens when the hexagon is not regular, but the max seems to be closer to 10.3 cm. I had to resort to my 3D program
I used for the illustration. I think Tim's solution can be slightly improved by tilting the cube forward to show more top.
Tim Poe (new) . . . . . . 10 pts - Welcome to the contest! Nice picture and very detailed solution.
Arthur Morris . . . . . .7 pts - Great answer, very clearly explained. (The Art Morris express rolls on!)
Lisa Schechner . . . . . 5 pts - Similar result; thanks! (Some steps glossed over so no bonus pt...)
Slanky . . . . . . . . . . . . 3 pts - Good try but answer was too wide to leave more than 2 opposite points.
Sue B . . . . . . . . . . . . . 3 pts - Thanks for entering, I'm not sure where your 5 by 10 -> 11.18 lives.
Tat Tsui. . . . . . . . . . . 2 pts - I agree, the problem seems totally impossible but it turns out ok!

Problem #114 - Posted Wednesday, April 4, 2001
Philo just spent a total of about 32 dollars on two types of stamps.
In the first batch, he bought as many stamps as the average cent value of each stamp.
For the second batch the average price per stamp was five times the number of stamps.
If he spent a total of \$3.19 less on the second set than on the first, what was the value
(and the number bought) of each stamp, and what was the exact total spent?

Solution: From: Sudipta Das ( Calcutta , India!)
Suppose the average price of the stamps in the first batch was x cents and that in the second batch was y cents.
Then the total amount ( in cents ) spent = x^2 + (y^2)/5 . . . Again, x^2 -(y^2)/5 = 319 ....(1)
If the total amount actually spent was (3200 + k) cents, then x ^2 + (y^2)/5 = 3200 + k ....(2)
Solving for x from equations (1) and (2) , we get x = sqrt [(3200 + k + 319)/2] = sqrt [(3519+k)/2]
The minimum value of k for which x is an integer is k = 9 , when x = 42. Putting x = 42 in (2) we get y = 85.
So, the value of each stamp of the
first batch was 42 cents and the number of stamps bought were 42.
The vaue of each stamp of the second batch was 85 cents and 17 stamps were bought in this batch.
The actual
total spent was \$32.09.
Dan's note: The next x = 43 gives \$33.79 and y = Sqrt[306], no good. There are other solutions but none are near \$32.
Tim Poe noted that if Batch 1 had 30 @ 8c and 12 @ 127c , and Batch 2 had 6 @ 8c and 11 @ 127 c it would work.

Tim Poe . . . . . . . . . . 10 pts - Excel finds x = 17; visual basic finds many others (one above); good!
Arthur Morris . . . . . .7 pts - Used the 'average' to your advantage. Yes there are other sol'ns.
Sudipta Das . . . . . . . 6 pts - Nice proof ; bonus point for being number-theoretic and quotable.
Sue B . . . . . . . . . . . . . 5 pts - Good idea, rounding to integers; batch 1 ~ \$17.595 ~ 17.64 = n^2.
Lisa Schechner . . . . . 4 pts - "Solved it by pull-push (#@^&*)" Is that swearing or computer code?
Phil Sayre . . . . . . . . . 3 pts - Welcome back; you got the answer right, even if I can't tell how!
Tat Tsui. . . . . . . . . . . 2 pts - The averages mean you can assume each batch is a fixed price.
Mercedes Foster (new) 2 pts - Good! - \$32.04 is closer to \$32 but the 12 at \$1.20 is 10x not 5x.

Problem #115 - Posted Thursday, April 19, 2001
Three subjects of an experiment are to be let into a room. They each recieve, at random,
either a red or blue cap. After entering the room each person must simultaneously either
guess their own cap color or say nothing; they can see the other two caps but not their own.
If anybody guesses right and nobody guesses wrong, the three will split six million dollars.
But if there is no guess or a wrong guess, they get nothing. They cannot communicate in
any way once they enter the room but can agree on a strategy in advance.
What strategy will maximize their chance of winning?

Solution: This problem has been making the rounds lately-- I didn't make it up. See the New York Times
Science section from Tuesday, April 17 for an excellent discussion (I knew Prof. Berlekamp at Berkeley)
If we call the subjects A, B, and C, one strategy is for A to guess "red" and for B and C to say nothing.
This has a winning percentage of 50%. But that's not the best. Here's a strategy with a 75% win rate:
If a subject sees two different color hats, they say nothing (pass). If they see the same color, guess the
opposite color. There are eight possibilities, listed below. Letter is hat color; italics for those who guess.
 R R R (lose) R B R (win) B B B (lose) B R B (win) R R B (win) R B B (win) B B R (win) B R R (win)
As you can see, there are 6 wins and just two losses; as Arthur and Lisa noted (avid NY Times readers),
it's good to 'concentrate the mistakes' in a small number of places.

Sudipta had a great answer, but I judged it to violate the 'no communication in any way' rule:
Suppose they plan that if any of them sees one blue and one red hat , he will remain quiet .
They also agree that they will mentally count 10 first . If one of the guys sees 2 hats of the same color ,
he will slowly say: "The color of my hat is ... ( the color other than that he will be seeing)."
However , if after counting 10 , all of them start saying " The color of my hat ...", they will immediately
end the line with " ... is ...( the color of the hat each is seeing)." This wins 100% of the time.

Arthur Morris . . . . . 10 pts - Thanks - your 'table' answer was clearer than the one on the Times!
Lisa Schechner . . . . . 7 pts - Good concise answer - I read the Science Times every Tuesday too.
Tim Poe . . . . . . . . . . . 5 pts - Your second answer was on the money (but no longer as early)!
Sudipta Das . . . . . . . 5 pts - This would have won (see above) except for the 'communication'.
Bruce . . . . . . . . . . . . . 4 pts - Your strategy wins (3/4) * 50% + (1/4) * 100% = 62.5% of the time.
Jon Wilson (new) . . . . 2 pts - Good answer, 75%, and just under the deadline. Smart family! %;-}
Alan Druze . . . . . . . . 2 pts - Welcome back! Your strategy works out to a 50% win rate.

Problem #116 - Posted Sunday, April 29, 2001
In the eternal battle to excel, two basketball players (we'll call them
Chris and Pat, to be gender-neutral) compared their stats after a tough game.
"I had a better game," said Chris. "I scored more points than you did tonight."
"No, I had the better game," countered Pat. "I had a better percentage on my two-point
shots, a better percentage on my three-point shots, and I took more total shots than you."
"That's bizarre," said Chris. "Maybe we figured this wrong; we'd better check our math."
(For you rabid or at least avid basketball fans, we are assuming neither player attempted any free throws.)
Was this really possible? Give an example, or else prove that it's impossible.
The winner was the first entrant with a solution with the lowest total attempts by both players.

Solution: This IS possible, although it sounds like a contradiction...
Here's hall-of-famer Beth Wilson returning with this week's featured answer:
* If you have to try a shot to receive a percentage, the solution is as follows:
Chris shot 1 2-pointer and missed; three 3-pointers and made two. Pat shot four 2-pointers and made 1;
Chris made 67% of three-pointers; Pat made 100%. Chris makes 4 shots, Pat makes 2.
* If you got 0% for making no shots: Chris makes no 2-point attempts, and makes 2 out 3 3-pointers.
Pat makes 1 out of 3 2-pointers and 1 out of 1 3-pointers.
* If you get 100% for making no shots: Chris makes 0 out of 1 2-pointers and 1 out of 2 3-pointers.
Pat makes 1 out of 4 2-pointers and 0 out of 0 3-pointers.
Dan's Note : I only accepted the first kind , and the fewest total attempts (9) was the tie-breaker.
 Player 2-pts / att / pct 3 pts / att / pct shots points Chris 0 - 1 (0%) 2 / 3 (66.7%) 4 6 Pat 1 - 4 (25%) 1 - 1 (100%) 5 5

Tim Poe . . . . . . . . . . . 10 pts - Covering your bases, the legal answer was the acceptable one!
Sudipta Das . . . . . . . . 8 pts - Your bases too! Bonus pt. for A > 1.5 B , 1.5D > C for pcts.
Beth Wilson . . . . . . . . 6 pts - Welcome back! You're very quotable this week.
Arthur Morris . . . . . . 5 pts - Good; your Chris and Pat took a few more shots than the low-tal of 9.
Tat Tsui . . . . . . . . . . . 3 pts - That was a small total (7 attempts) but 0-0 doesn't have a percent!
Michael Browell (new) 2 pts - I agree with your intuition that it seems impossible. But numbers don't lie!
Manuel Fukea (new) . .1 pt - Very astute to notice whoever gets the most points gets the glory!

 Problem #117 - Posted Thursday, May 10, 2001 The Kissing Circles (back to top) Inside the (orange) unit circle we fit two blue circles of radius 1/2. a) The yellow circle is tangent to the two blue circles and the inner edge of the orange circle. What is its radius, a? b*) The green circle is tangent to one of the blue circles, the yellow circle, and the orange circle. What's its radius, b? c*) If the orange circle is centered at (0, 0) what are the coords of the centers of the four inner circles? (The blue ones are easy.) Solution: Allen Druze was the last entrant so I'll make him semi-famous:   Let diameter COD run through the centers of the two blue circles.Since CD = 2 ,CO = 1 and OD = 1. The radius of both blue circles equal 1/2, coordinates of (-1/2,0) and (1/2,0). Draw a perpendicular line from the top of circle a to center point 0. Let the radius of circle a = a and the distance from circle a to center 0 be x. Then 2a + x = 1 ( note: by connecting the centers of the circles I can prove congruency by s.a.s. hence center a lies on the line that is perpendicular to COD.). Connect the center of circle a to the first blue circle(left side) and you will have a right triangle that will yield (1/2)^2 + (a+x)^2 = (a + 1/2)^2, since x = 1 - 2a , by substituting for x, the first equation will yield a = 1/3, or coordinates (0, 2/3) . Draw a line tangent thru pt C and construct a semi circle equal to circle a that is tangent to both circle b as well as one of the blue circles. Let b equal the radius of the green circle, therefore a + 2b + a = 1 , since a = 1/3 , b = 1/6, hence circle b has coordinates (-1/2, 2/3). Check out my new page on this circle problem!

Arthur Morris . . . . . .10 pts - Good use of symmetry and Pythagoras, esp. the 3-4-5.
Sudipta Das . . . . . . . . 7 pts - Nice development of equations and logic, a class act.
Lisa Schechner. . . . . . 5 pts - You got these circles - can you go between a, b, and 1/2?
Slanky . . . . . . . . . . . . 4 pts - Welcome back, and nice use of color coordinated text!
Andy Hsieh (new). . . . 4 pts - A bonus point for the thoroughness of your solution!
Phil Sayre . . . . . . . . . 3 pts - Good setup - managed to factor Trig into the equation!
Beth Wilson. . . . . . . . 2 pts - Good try and explanation; some radii off by factor of 1/2
Tat Tsui . . . . . . . . . . . 2 pts - You also got 1/6 and 1/12 - are you guys growing ?
Hermen Jacobs (new) . 2 pts - Welcome to you in Holland! Correct answers; please give steps.
Tim Poe . . . . . . . . . . . 2 pts - Blue, aqua, and yellow pinned down, green one 'b' floatin'!
Alan Druze . new score 2 pts - The answer was a bit late but - now the full 2 pts!

Problem #118 - Posted Sunday, May 20, 2001
Here are three nice puzzles about perfect squares and cubes.

a) The sum of the cubes of the digits of 407 is :4^3 + 0^3 + 7^3
= 64 + 0 + 343 = 407 itself. What's the smallest number, greater
than 1, not containing any zeroes, with this cubical property?

b) Find two whole numbers > 1 such that the difference of their
cubes is a square, and the difference of their squares is a cube.

c) What cube has volume numerically equal to its surface area?
Can a cube have the cube of its area equal the square of its volume?

Solution: Many of you used spreadsheets or programs - go geeks!
a) 1^3 + 5^3 + 3^3 = 1 + 125 + 27 = 153 , also 371 works.
b) Lisa and others pointed out I didn't require the numbers to be different, so for example:
2^3 - 2^3 = 0^2 and 2^2 - 2^2 = 0^3. But what I had in mind (please read it) was the answer
10 and 6 : 10^3 - 6^3 = 1000 - 216 = 784 = 28^2 and 10^2 - 6^2 = 100 - 36 = 64 = 4^3.
c) side = x ; x^3 = 6 x^2 ==> x = 6 units; vol = area = 216. c+) NO WAY because
(6 x^2)^3 = (x^3)^2 ==> 216 x^6 = x^6 ==> either x = 0 or 216 = 1 , both not so good.

Arthur Morris. . . . . . . . . . . 10 pts - Used EXCEL as more than a shaver, yes 216 =/= 1.
Sudipta Das . . . . . . . . . . . . . 7 pts - Yes, all 4 are 153, 370, 371, 407 ; eliminate zeroes.
Lisa Schechner . . . . . . . . . . . 5 pts - A compromise between extra credit and partial credit.
Tim Poe . . . . . . . . . . . . . . . . . 4 pts - If initial search comes up empty, turn on the lights!
Beth Wilson. . . . . . . . . . . . . . 4 pts - Correct and within minutes of Tim, hence extra point!
Nikita Kuznetsov . . . . . . . . . 3 pts - Yes, 153; 10 & 6 or a & a ; A = v --> E = 6. horoxo.
Sue B . . . . . . . . . . . . . . . . . . . 3 pts - Nice steps in a) and correct on b^3 = b^3 = 0^2...
Hermen Jacobs . . . . . . . . . . . 3 pts - Got your answer to b) on second try, thanks!
Slanky . . . . . . . . . . . . . . . . . . 2 pts - Brute force and algebra win out over other methods...
Hareenda Yalamanchili (new) 2 pts - Good setup - managed to factor Trig into the equation!

Problem #119 - Posted Wednesday, May 30, 2001
A "perfect number" is the sum of its proper divisors, like 1 + 2 + 3 = 6.
The proper divisors of a "doubly perfect number" add up to twice the number, and so on.
a) What are the next two perfect numbers, after 6? b) What is the sum of the reciprocals
of the proper divisors of a perfect number? c) Find the first doubly perfect number, and
figure out the sum of reciprocals of its proper divisors. d) Find the second doubly perfect
or the first triply perfect number. (One bonus point for both!)

Solution: a) The first two perfect numbers are widely known : 6, 28, . . . .It's not hard to prove
that all even perfect nos n look like n = 2^(p-1) * (2^p - 1) ; where p and 2^p - 1 are prime.
(Dan's note: Nobody has found any odd perfect nos, nor proven they don't exist, but many bounds & cond'ns exist.)
So p = 2 gives n = 2^1 (2^2 - 1) = 2 * 3 = 6 ; p = 3 gives n = 2^2 (2^3 - 1) = 4 * 7 = 28 ;
and p = 5 gives n = 2^4 (2^5 - 1) = 16 * 31 = 496 ; the next one for p = 7 is n = 64 * 127 = 8128.
(Sudipta sent in: 1^3 + 3^3 = 28 ; 1^3 + 3^3 + 5^3 + 7^3 = 496 ; 1^3 + 3^3 + . . . + 15^3 = 8128. Cool!)
b) Let's find the sum of the recips of all the divisors of, say, 28:
1/1 + 1/2 + 1/4 + 1/7 + 1/14 + 1/28 = (28 + 14 + 7 + 4 + 2 + 1)/28 = 56/28 = 2.
This works identically with any perfect number; thia means leaving off the recip of n
we get 2 - 1/n for the answer. (I think the first question using all divisors is better.)
So 13/6 , 57/28 , 991/496 , etc. are the answers for the first three perfect nos.
c) The first doubly perfect is 120 = 2^3 * 3 * 5 ; sum of divs = 15 * 4 * 6 = 360 = 3 * 120.
The sum of the recip's of all these (16) divs is (120 + 60 + . . . + 2 + 1) / 120 = 360 / 120 = 3,
(my intended question), so without the 120 it's 359/120 = 2 119/120 .
d) The next doubly perfect is 672 = 2^5 * 3 * 7 , because sum of div = 63 * 4 * 8 = 3 * 672.
There are no other doubly perfects under 100,000. As for triply perfects, I let Mathematica
tell me the first two are 30240 = 2^5 3^3 5 7 and 32760 = 2^3 3^2 5 7 11.
122,522,400 is the first number whose sum of div is over five times the number itself, but
I hear from a website given to me by Phil, that the first quadruply perfect is about half a billion.
http://www.uni-bielefeld.de/~achim/mpn.html

Programs used: It's all in what you're used to!
 Visual BASIC from Tim P. Sub Prob119() Documents(1).Activate ActiveDocument.Select Selection.Collapse direction:=wdCollapseEnd num = 2 While num < 1000 If num Mod 1000 = 0 Then Selection.TypeText ("...passing " & num & vbCrLf) End If divsum = 0 For div = 1 To num - 1 If num / (num \ div) = div Then divsum = divsum + div End If Next If divsum = num Then Selection.TypeText (num & vbTab & divsum & vbTab & "**" & vbCrLf& vbTab & "(") For div = 1 To num - 1 If num / (num \ div) = div Then Selection.TypeText (div & " + ") End If Next Selection.TypeText (")" & vbCrLf) Else Selection.TypeText (num & vbTab & divsum & vbCrLf) End If num = num + 1 Wend Selection.TypeText ("End of run") End Sub BASIC from Hermen J. 10 cls: N=6:K=2 OR K=3 20 FOR I=2 TO N/2 30 IF N/I=INT(N/I) THEN S=S+I 40 NEXT I 50 IF S+1 = K*N THEN PRINT N 60 S=0: N=N+1:GOTO 20     (code not submitted: MATLAB from Art Morris Mathematica from Dan Bach)
Tim Poe. . . . . . . . . . . . . . . . . 11 pts - Bonus pt for 672 and 30240, plus all divisors and code!
Hermen Jacobs . . . . . . . . . . . 7 pts - I love the simplicity of your program; N/2 beats num - 1
Art Morris. . . . . . . . . . . . . . . 5 pts - Matlab strikes again. Steps glossed but 672 & 30240 both.
Sudipta Das . . . . . . . . . . . . . 4 pts - Good, 120, 672, 523776 doubly; 30240, 32760, 2178540 triply!
Sue B . . . . . . . . . . . . . . . . . . . 4 pts - 'Nearly twice' is a good description of answer to part b!
Lisa Schechner . . . . . . . . . . . 3 pts - Said Descartes found 120 and 30240 back in 1638, thanks!
Phil Sayre . . . . . . . . . . . . . . . 3 pts - Recycled a program used on Problem #103 - Weird numbers!
Slanky . . . . . . . . . . . . . . . . . . 3 pts - It pays to have a helpful office mate; two brains are better'n, uh...
Nikita Kuznetsov. . . . . . . . . 3 pts - Good job and thanks for the General Formula k - 1/n for k-tuply.
Ludwig Deruyck . . . . . . . . . 3 pts -Welcome back - - Extra point for lots of extra answers!
Allen Druze . . . . . . . . . . . . . 2 pts - Thanks for 2^(p-1)(2^p - 1); just short of 2 on b), see above for d).

Problem #120 - Posted Thursday, June 21, 2001
Tires placed on the rear of your car will wear out after 21000 miles, while tires on the front
of your car will last for 29000 miles.
Suppose you have a new car and five identical new tires (four installed and one spare).
a) What is the maximum distance you can drive, assuming you can easily change the tires
any time you want? b) Describe a rotation schedule that allows you to drive this distance.

Solution: A good way to look at this one is to use "rates" of wear in 'tires per mile.'
a) Let X = 21,000 and Y = 29,000. Assuming that the the rate of tread wear is a linear function of miles
driven, then 2/X + 2/Y = (total tread wear per mile). So, the maximum number of miles that can be
driven is 5 / (2/X + 2/Y) = 5 XY/ (2X + 2Y) = 30,450. Answer: 30,450 miles .
b)
One strategy to achieve this maximum mileage is to rotate: S -> LR ->LF -> RF -> RR -> S
every
30450/5 = 6090 miles.

Tim Poe. . . . . . . . . . . . . . . . . 10 pts - That's the right idea, define the amount of treadwear per mile!
Lisa Schechner . . . . . . . . . . . 7 pts - Good: distance = D = (2/29000 + 2/21000)/5 = 30450.
Phil Sayre . . . . . . . . . . . . . . . 5 pts - I received your answer in 2006 but it still hadn't 'worn out .'
Arthur Morris. . . . . . . . . . . . 5 pts - Answer of 30450 was good but rotation not 'spelled out.'
Hermen Jacobs . . . . . . . . . . . 4 pts - Nice rotation schedule, but 690.000 or 690,000 is too far.
Sue B . . . . . . . . . . . . . . . . . . . 3 pts - Roundoff error cost you 252 miles. Beyond the AAA tow limit!
Ludwig Deruyck . . . . . . . . . 2 pts - Method works for 29000, 'sharing the spare' on the rear only.
Allen Druze . . . . . . . . . . . . . 2 pts - Your averaging method would be 5/4(25000) = 31250?

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