dan's math@home - problem of the week - archives
Problem Archives page 15

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

142 - - Strike A Chord
143- Easy as A-B-C-D
144 - - - Click to 222 !
145 - Up The Volume!
146 - Maxes & Means
147 - Circle the Roots!
148 - - Weird Bases !!
149 - Indep'd'ce Events
150 Hpy Bday Hermen

 Problem #141 - Posted Monday, February 18, 2002 Where's The #%@ Road ? (back to top) After parachuting into the woods at night, you land right next to a sign that says : "Road: One Mile." But it has fallen over, and there's no way to know what direction the road is. Describe the shortest path you might take that will guarantee reaching (touching) the road. (You can't see the road until you reach it!) What's the maximum distance you might walk on your path? ROAD : 1 mi

Solution: This is a great problem, and deceptively complicated. Here are some ideas:
 Dan's Note: As Arthur and Jack pointed out, it makes a difference if the road is not known to be straight; I'm responsible for the implication that it was a straight road. I apologize for the imprecision of my statement; with this I agree the road might come to a point 1 mile from the victim (you), and 1 + 2pi is the best you can hope for. See scoring changes below, but Lisa is still this week's winner!
Several of you sent in a reasonable (but not shortest) path: Walk 1 mile in any direction, then go in a circle
of radius 1 centered at your original location; max distance needed would be L = 1 + 2 pi ~=~ 7.283 mi .
(see fig. 1) I tried this problem last year and thought I had solved it. I approached it in two ways

Dan's way #1: Try going in a regular polygon with base 1 mile away from the center; you can traverse n-1
sides of a regular n-gon of side S making (n - 1)S, plus one hypotenuse length Sqrt(1 + (S/2)^2).
We then get S in terms of n, and minimize this quantity as a function of n: Evidently from a picture (not
pictured) we have (S/2)/1 = tan((2pi/n)/2) ==> S = 2 tan(pi / n). And use 1 + tan^2 = sec^2 as well...
Total path is L = 2(n - 1) tan(pi/n) + Sqrt(1 + (tan(pi / n))^2) = 2(n - 1) tan(pi/n) + sec(pi / n) (see fig. 2)
Here we have some (n, L) pairs: {3, 8.9282}, {4, 7.41421}, {5, 7.04841}, {6, 6.9282}, {7, 6.88881},
{8, 6.88138}, {9, 6.8877}, {10, 6.90002}, {11, 6.91475}, {12, 6.93016} (from Mathematica) ;
Shortest path is an OCTAGON! (I drew a heptagon for aesthetic reasons; slightly longer.

Dan's way #2: Instead search in a 'Pacman' shape, a circular arc with a piece missing, with center 1 mile
from the center of the missing segment. The radius R is more than a mile but the angle is less than 360 deg.
say 2 pi - 2T. From another picture we see cos T = 1/R, so R = sec T. We will then travel a maximum of
L = R + R(2 pi - 2T) = R(1 + 2 pi - 2T) = (sec T)(1 + 2 pi - 2T) . This seems to min out at T = 0.290 rad,
giving L = 6.99528; not as good as the polygonal approach. So I thought the answer was 6.88138 miles .

But along came the awesome web research powers of Lisa Schechner, and this time she has found the
optimum solution, as part of Isbell's class of "escaping from a region" series, this problem is "escaping
from half-plane with known distance to edge." Go AB, then BC, then in an arc CD, then 1 last mile DE.

New entrant Ed suggested: "Climb a tree, wait till morning light, then walk 1 mile. Dist ~ 5300 feet."

Here are pictures of each (not the tree); I'm way too busy to be drawing this, but it was fun! - Dan
 The most popular answer, but not the shortest path. Going from A to B is 1, etc. Here ABC = 1 + 2 pi ~ 7.283. For an n-sided polygon the angle ø = pi / n; here n = 7.  The 'radius' AB is sec(pi/n) Min path ABC is 6.881 (n = 8). The radius is sec ø , Path ABC has length (sec ø)(1 + 2pi - 2ø) Min 6.995 at ø = 0.29 rad Radius 1, AC = 1, CD = 7pi/6 Pythm : AB = (2/3)sqrt[3] BC = (1/3)sqrt[3]. ABCDE = 1 + sqrt[3] + 7pi/6 ~ 6.397.

Judges scoring decision: 2 extra pts to Arthur for correctly raising the issue of the straight road, and a bonus
point to each person whose new score on this problem is in Blue. See the explanation and apology above.
Lisa Schechner . . . . . 9 pts - Great find ; I had to figure out the path and method so I charged 1 pt.
Jack Dostal . . . . . . . . 6 pts - Good job finding the regular octagon; would have been winner...
Arthur Morris . . . . . 5 pts - First submission of the popular "1 + 2 pi" answer. No GPS!
Drew . . . . . . . . . . . . . 4 pts - Said "If your eyesight is so bad, walk 1 mile to the eyeglass store"
Quasi-C . . . . . . . . . . . 4 pts - The 1 + 2 pi was not so bad, but I didn't understand the "is it nec." argt.
Steve Lawrie . . . . . . . 4 pts - Right; make the segment tangent to the circle. 1 + 2 pi can be shortened.
Hermen Jacobs . . . . . 4 pts - Yes, in practice, it will be difficult to avoid all obstacles in the dark!
Phil Sayre . . . . . . . . . 3 pts - 1+2pi shorter than 8 sides of octagon but not 7. Bonus pt for polygon.
Ed Wern (new) . . . . . . 3 pts - Welcome to the contest ; good idea figuring 'average trip length'.
Tim Poe . . . . . . . . . . . 3 pts - Just like these other 1 + 2 pi but with a resubmission discount!
Jon Stearn (new) . . . . 3 pts - Hello! Yes, tying a 1-mile rope might be problematic !
Chris Jones (new) . . . 3 pts - Welcome! I like the idea of "1 + 2 pi - (1 / infinity)", thanks!
Zahi Yeitelman . . . . . 2 pts - Good solid answer, just not the shortest path. Thanks for entering!
Matt Situ . . . . . . . . . . 2 pts - Join the 7.28 club! Thanks for .bmp attachment; those I can handle.
Robert Hussey . . . . . 1 pt - Welcome back! Method is ok; circumf is 2pi not pi... oops!
Basem Chaikhouni. . 1 pt - Aha, you again! Be sure to add one radius onto the 6.28 mi (not ^2)
Dodopotpie (new). . . . 1 pt - I like the polygon idea, but I think 4 sides gives 6+sqrt[2] not 9?
Hyunshin Cho (new) . 1 pt - Good idea, walking sqrt[2] mi in 4 direc's, but I get 7*sqrt[2] not 5.

 Problem #142 - Posted Wednesday, March 13, 2002 Strike a Chord ! (back to top) The circle at right has some interesting properties: 1) The chords are at right angles (perpendicular), 2) Red chord is one inch longer than green chord, 3) The values of a, b, c, d are given by clues below. What is the area of the circle, in square inches? Clues for a, b, c, d : a d = b c (true in any chord situation) Length of a = average of b and d ; (a+b+1)(a+b-1) = bad

Solution: This problem woke up the long-dormant former all-time leader Andy Murdock ...
First let's find a, b, c, d: I can do this by hand, I promise, but I used Mathematica:
In: Solve[{a d == b c, a == (b + d)/2 , b + c == a + d + 1, (a + b + 1)(a + b - 1) == b a d}, {a, b, c, d}]
Out: {{c -> 6, d -> 4, a -> 3, b -> 2}, and a bunch of horrible stuff} So (a, b, c, d} = (3, 2, 6, 4}.

As for myself, I then decided to try a coordinates approach; the chords are perpendicular,
so have the origin at the intersection of the chords, then the four points on the circle are:
(0, a), (-b, 0), (0, -d), and (c, 0). Using (x - h)^2 + (y - k)^2 = r^2, we should get 4 equations
in the 3 unknowns h, k, and r. (Hopefully redundant and not inconsistent!)
Using a, b, c, d values just computed, plug in (x, y) = (0, 3), (-2, 0), (0, -4), (6, 0) and solve:
In: Solve[{ (0 - h)^2 + (3 - k)^2 == r^2, (-2 - h)^2 + (0 - k)^2 == r^2,
(0 - h)^2 + (-4 - k)^2 == r^2, (6 - h)^2 + (0 - k)^2 == r^2 }, {h, k, r}]
Out: {{r -> -(Sqrt[65]/2), h -> 2, k -> -(1/2)}, {r -> Sqrt[65]/2, h -> 2, k -> -(1/2)}}
So the center of the circle is at (2, -1/2) and the radius is Sqrt[65]/2 , giving
our circle an area of pi * ( Sqrt[65]/2 )^2 = (65 pi / 4) sq.in. ~~ 51.05 sq.in.

Other approaches included these:
Andy M. set things up so that r was the hypotenuse of either a 2 by 3.5 right triangle, or a
4 by 0.5 triangle, either of which gives r = (1/2)sqrt[65]. Then A = pi r^2 = 16.25 pi.
Steve L. played around to find a,b,c,d while Ludwig used Solve in Matmca like I did.
Ludwig also pulled out some secret Belgian formula r = (a b c) / (4 A) for the radius.
And Hermen from nearby Holland knew about a^2 + b^2 + c^2 + d^2 = 4 r^2. Hmm!
Phil (using Python) found a second set of answers (that I overlooked) fitting the four eqns:
(a, b, c, d} = (0.798, 0.369, 2.657, 1.227} giving r = 1.527 and A = 7.333 Whattaya know!
I did want to see the exact answer ; 16.25 pi was ok but just 51.0508 got a 1-pt inexactness penalty

Steve Lawrie . . . . . . . 10 pts - Good explanation leading to x^2 + 7x + 12 = 16 ; diam = sqrt[65]
Ludwig Deruyck . . . . 6 pts - Nice use of Mathematica; did it give you the other set of solutions?
Arthur Morris . . . . . . 5 pts - Gee, my very existence was a hint (not to use approx decimals!) ;-}
Nikita Kuznetsov . . . 4 pts - Glad algebra is easy 4U, but do show steps. I'll put harder probs in.
Anirban Bhattacharyya. . 3 pts - Nice diagram on your attachmt but 3.14 was a 1-pt roundoff deduction
Phil Sayre . . . . . . . . . 3 pts - I added a pt for your alert inclusion of the second set of solutions
Andy Murdock . . . . . 2 pts - Thanks for waking up from a long hi8us, can we expect more later?
Dangkhoa Pham (new) 2 pts - I was able to read your attachmt but your values were off by a bit.
Kathy Serbin . . . . . . . 1 pt - I got to read some formulas but no images loaded; no numeric values.
Hermen Jacobs . . . . . 1 pt - Good answer, I'll give you a contest pt because problem was still up

Problem #143 - Posted Sunday, March 24, 2002
Kudos to Phil for retitling it as "The ABC's of D" ; I'll take that to mean "what makes Dan tick".
A is the number of real solutions to the equation (x^2 -- 9x + 19)^(2x^3 -- x^2 -- 10x) = 1
B and C are such that the x solving the equation 3^(2 -- x) = 2^(3x -- 1) is log(base B) of C
D is the number of ways A+B+C can be written as a sum of increasing positive integers
What is the value of A + B * C ^ D ?

Solution: From a variety of sources, here goes: A : From Steve L...
If a^b = 1, then either (1) a must be 1, (2) b must be 0, or (3) a is -1 AND b is a multiple of 2.
This is because if a^b = 1, then a = the 'b' root of 1. Except when b = 0, the only numbers that can be
a root of 1 are 1 and -1. So, we'll start with the 1st case: (1) a must be 1. That means:
x^2 - 9x + 19 = 1 ; x^2 - 9x + 18 = 0 ; (x - 6)(x - 3) = 0 ; x = 6 or x = 3. That is two real roots so far.
(2) b must be 0. That means that: 2x^3 - x^2 - 10x = 0 ; x(2x^2 - x - 10) = 0 ;
x(2x - 5)(x + 2) = 0 ; x = 0 or x = 5/2 or x = -2 . We get 19^0, (11/4)^0, and 41^0, all of which equal 1.
That's 5 real roots so far. (3) a is -1 AND b is a multiple of 2. Start with the a = -1 : x^2 - 9x + 19 = -1
x^2 - 9x + 20 = 0 ; (x - 4)(x - 5) = 0 ; x = 4 or x = 5 . Now, we see what b would be in each case:
x = 4 --> b = 2(4)^3 - 4^2 - 10(4) = 72, (-1)^72 ; x = 5 --> b = 2(5)^3 - 5^2 - 10(5) = 175, (-1)^175.
So x = 5 does not work, but x = 4 does. So the real roots to the equation are -2, 0, 5/2, 3, 4, 6.
That's a total of 6 real solutions, so
A = 6.
Now from Tim Poe : B and C are such that the x solving the equation 3^(2 -- x) = 2^(3x -- 1)
is log(base B) of C ; 3^(2-x) = 2^(3x-1) ; (3^2)/(3^x) = (2^3x)/2 . (I added a few parentheses - Dan)
9/(3^x) = 8^x/2 ; 18 = (3^x)*(8^x) =(3*8)^x = 24^x
18=24^x ; log(base 24) 18 = x ; so
B = 24 and C = 18 .
Part D (The idea is from Anirban, who had A=5): D is the number of ways A+B+C can be written as the sum of
increasing positive integers. Now A+B+C = 6+24+18=48. Considering the least of the positive integers as 1,2,3,4........15 , the number of cases is 22+21+19+17+16+14+13+11+10+8+7+5+4+2+1 = D =170
.
But this only counts the ways of writing 48 as sum of three increasing integers. According to Tim Poe's second messsage:
and a visual basic program, from 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 11 , . . . , to 23 + 24 ; there are 2589 ways that 47 can be written as a sum of increasing positive integers. Wonder how many for 48? Anirban states that the number of ways is
the coeff of x^47 in the expansion of (1 + x)(1 + x^2) . . . (1 + x^47) which is 2590 ; one more. This would include 0+0+47
which forces you to accept a sequence of one number {47} as 'increasing'. So I used Mathematica to find the
coeff of x^48 in (1+x)(1+x^2) . . . (1+ x^48) , from (1+2+...+8+12) to (48) , to be D = 2910 ;
this then leads to our answer of 6 + 24 * 18 ^ 2910 ~=~ 1.6718645622517823469493584 * 10^3654.
(As Nick and others of you put it, such a number (3655 digits) is 'mind-bogglingly big.' I agree, as the number
of particles in the universe has been roughly estimated at 10^130. - Dan)

Steve Lawrie . . . . . . . 8 pts - Nice job finding A=6 and B&C but 'incr' is not nec 'consec'.
Anirban Bhattacharyya. . 5 pts - Good prelim answer exc. A=5 and then I liked the 'power series'.
Nikita Kuznetsov . . . 4 pts - Yes, A=6 (wasn't easy for some); 'incr' not 'consec'. You're right abt. #138.
Tim Poe . . . . . . . . . . . 4 pts - 5+24*18^2589 is pretty close; extra point awarded after consultation
Phil Sayre . . . . . . . . . 4 pts - You got an extra pt by executive decision ; 5 + 24 * 18 ^ 2590.
Hermen Jacobs . . . . . 3 pts - I originally figured out just the 3-term sums too, but we need all of em.
Omar Gymboski . . . . 3 pts - Got B, C but A=6 and D is very large, not 1 or 8; but good try.
Arthur Morris . . . . . . 2 pts - Hope you're doing ok there; the B+C=1 constraint was interesting
Joe Alvord (new) . . . . . 2 pts - Welcome! A=5 was popular; yes; D=large not nec consec.
Nick McGrath (new) . . 2 pts - Thanks for the positive feedback; happy to boggle your mind!

 Problem #144 - Posted Monday, April 8, 2002 Click to 222 ! (not factorial, not 'Room 222') (back to top) You start at 0, so does k . Your goal is to get to 222 (two hundred twenty-two). There are three buttons you can click ; one will increase the value of k by 1, another decreases k by 1, the third will ADD k to your total. (For example the sequence [incr][ADD][incr][incr][ADD] gives 1+3=4 in 5 clicks.) How can you reach your goal in the fewest number of clicks? Ties with same number of clicks will be won by fewest [decr] then fewest [incr]. [incr k]   [decr k]   [ADD k]   Click any button!

Returning contestant Jon Stearn sent the first best solution:
INC * 12 ; Add * 1 ; INC * 2 ; Add * 15 . Total of 30 clicks with no decrements.
Explanation : the sqrt of 222 is 14.something.... -> the quickest route is going to be to
add up a bunch of 14's or 15's. I chose 14's for above, but you can also do it w/ 15's
in 30 clicks. effectively 14 * 16 is 224 (2 greater than the desired total) thus the 12
put in first. 17*13 = 221, so using 17s or 13s would take 31 clicks.
As you move further from the sqrt, the # of clicks increases.

Nick McGrath gave the second winning solution with 30 clicks and adds:
Give an even number of clicks, 2N say,the maximum total we can achieve is N^2 which we can get by
N*[incr],N*[ADD] . For an odd number of clicks, 2N+1 say, the largest total is N(N+1) by either
For the current problem, we can establish that we cannot make 222 in less than 30 clicks because the
max total for 29 clicks is 14*15 = 210. The maximum we can make with 30 clicks is 15^2 = 225.
Any total between 211 and 225 takes a min. of 30 clicks, but each can be done in exactly 30 clicks.

Jon Stearn . . . . . . . . . . . 10 pts - Welcome back! I liked your answer enough to quote it; thanks!
Nick McGrath . . . . . . . . 7 pts - Excellent second solution; no "point charge" (as they say in physics)
Jack Dostal . . . . . . . . . . . 6 pts - Nice solution, close behind Nick in time, so the extra point
Steve Lawrie . . . . . . . . . . 5 pts - Good thinking; 14^2 < 222 < 15^2 is the key to the 30 clix
Abbas Mehrabian (new) . 4 pts - Welcome; nice use of calculus on function y = 1 + x + 222/x.
Joe Alvord . . . . . . . . . . . 3 pts - Well put: add as much as possible ; this means near sqr root.
Ritwik Chaudhuri (new) 3 pts - Right: 222 = 10*9+11*12 for 31 clix; 15*2+16*12 for 30.
Phil Sayre . . . . . . . . . . . . 3 pts - That's it: 12*1 + 15*14 also has 14 [incr] and 30 clicks.
Arthur Morris . . . . . . . . 2 pts - 31 clicks is next best; you were earliest (with this ans at least)
Quasi-C . . . . . . . . . . . . . 2 pts - You seemed to be adding [tot] and not [k]. You weren't alone...
Nikita Kuznetsov . . . . . 2 pts - 12i , 12a , 1i , 6a is 31 clicks; not far from optimal.
Tim Poe . . . . . . . . . . . . . 2 pts - Got the right idea on the 2nd try ; woulda been 3 pts if 1st try
Derek Chin . . . . . . . . . . 2 pts - 222 = 6*37 ; 6+37 = 43 clicks is inefficient; yes; 31 is a lot closer.
Lisa Schechner . . . . . . . 2 pts - Better late than never! Might increase to 3 pts if you re-send answer.
Hermen Jacobs . . . . . . . 1 pt - Correct not to use [decr] at all, but 43 clicks can be beat by mixing
Anirban Bhattacharyya. . . . 1 pt - If you have to alternate [incr] with [add] , then 38 clix is good!
Ludwig Deruyck . . . . . . 1 pt - No explicit list of how Matica gave 32 clix; you can e-me the code
Matt Situ . . . . . . . . . . . . 1 pt - A few days after the others; 32 clicks is pretty good...
Drew . . . . . . . . . . . . . . . . 1 pt - Another answer got in before the answer went up - 32 clicks

Problem #145 - Posted Tuesday, April 30, 2002
Mia needed to calculate the volume of a rectangular room. She multiplied the length and
the width correctly but the width had been incorrectly written down, it was one-third
larger than it should have been. To make up for this, she lowered the height by one-third,
then multiplied it on. She figured this was okay since the width was equal to the height.
She then found her volume was off by 20 cubic meters. Why was Mia wrong, and what
was the actual volume? Please explain steps & reasoning...

Solution: Steve L. had a clear path to this answer: "Mia's approach doesn't work because it is not a simple
matter of taking 1/3 from one side to compensate for an extra 1/3 elsewhere. ... to get a width that is 1/3
larger, we actually would have to multiply the width by 4/3. Thus, to fix the error, we would have to
multiply the height by the reciprocal, 3/4, thus taking away a quarter, not a third. Look what happens
when we do it like Mia instead: actual length = L ; actual width = W ; actual height = H
inaccurately measured width = W + (1/3)W = (4/3)W ; adjusted height = H - (1/3)H = (2/3)H
Thus, the actual volume would be: Va = LWH, but the measured volume would be: Vm =
L*(4/3)W*(2/3)H = (8/9)LWH = (8/9)Va . Thus, Mia's meas. vol would only be 8/9 of the actual vol.
Since we already know that Mia's measurement is off by 20 cubic meters, then the actual volume (Va)
minus the measured volume (Vm) equals 20 (measured vol must be lower since it is 8/9 of actual vol)
Va - Vm = 20 ; from above: Vm = (8/9)Va ; Va - (8/9)Va = 20 ; (1/9)Va = 20 ; Va =Volume = 180."
----------
Hermen Jacobs included that and another interpretation: since width = ht, are they both mult by 4/3?
"w=h; the volume is l x w x = l x w^2 ; first answer of Mia was l x 4/3 x w x 4/3 x w = 16/9 x l x w^2
she corrects h= 2/3 x 4/3 x w = 8/9 x w ; Her second answer is l x 8/9 x w x 8/9 x w = 64/81 x l x w^2
The difference is 80/81 x l x w^2 = 80 m3 ; V = lwh = 81 m3. But I believe my first answer."
So do I.
----------
From: Drew ; A third way of interpreting! "one third larger" might mean an extra 1/3 of a meter...
"The true area of the room is LWH ; her new area is L(W-1/3)(H+1/3) or LWH-1/3H-1/9+1/3W
If these were equal then LWH = LWH-1/3H-1/9+1/3W ; 1/9+1/3H=1/3W ; 1+3H=3W ; or,
W must be one third greater than H. I don't yet see a way to find the area of the room, it is possible?"

Steve Lawrie . . . . . . . . . 10 pts - Good answer; you're this week's quotable and rising up the charts!
Anirban Bhattacharyya. . . . 7 pts - Through the power of algebra we make the difficult seem simple!
Hermen Jacobs . . . . . . . 5 pts - Nice - I like both ans. but in the second is h = w or = 8/9 w ?
Arthur Morris . . . . . . . . 4 pts - Right, it didn't matter that w = h; in fact the incorrect ones weren't.
Quasi-C . . . . . . . . . . . . . 4 pts - Noticed inf many real (x, x, 180/x^2) but (6, 6, 5) integer sol'n
Tim Poe . . . . . . . . . . . . . 3 pts - Yes, correct multiplier s.b. 3/4 not 2/3 to cancel out the 4/3.
Omar Gymboski . . . . . . 3 pts - Ok, same suggestion as Tim for Mia's improvement.
Jack Dostal . . . . . . . . . . 3 pts - Clear notation; Vr and Vm for real world and Mia's World.
Jon Stearn . . . . . . . . . . . 3 pts - Right you are; I hope Mia can take all this criticism!
Nikita Kuznetsov . . . . . 3 pts - The factors (1 + 1/3)(1 - 1/3) will never = 1, but 180 is right.
Lisa Schechner . . . . . . . 2 pts - The first of the "very close" solutions; s.b.V/9 = 20 not V/8
Joe Alvord . . . . . . . . . . . 2 pts - Another early answer; got 180 but 8/3 not 8/9, not sure about
Dangkhoa Pham . . . . . . 2 pts - Good equation 4W/3 L 2/3 H = 8V/9 ; that's why Mia was wrong.
Nick McGrath . . . . . . . . 2 pts - Some steps skipped but better'n what Mia did!
Drew . . . . . . . . . . . . . . . . 2 pts - Excellent point (as noted above); "larger" is ambiguous, eh?
Zahi Yeitelman . . . . . . . 2 pts - Right, product of base will always be 8/9 x^2, so V = 180
Matt Situ . . . . . . . . . . . . 2 pts - Good; "this is a trick that works w/add'n and not multipl'c'n."
Phil Sayre . . . . . . . . . . . 2 pts - Ok , the diff was 1/9 of V and was also 20 ; that's it!
Myung Hwa Yang . . . . 1 pt - That's close; 8/9 V = 20 does give answer of V = 22.5.
Richard Hillman (new) . 1 pt - I didn't see this until after #146, but good answer anyway!

Problem #146 - Posted Wednesday, May 15, 2002
I have a list of positive integers; not necessarily distinct. The number 68 appears in the list.
The arithmetic mean (average) of the list is 56; but if the 68 is removed, the average drops
to 55. What is the largest number that could have appeared in my list? Explain steps, reasoning.

Solution: From our Netherlands correspondent Hermen J. Jacobs, who keeps it short and sweet:
There are n positive numbers: the number 68 and (n-1) numbers with the sum x,
Then 68 + x = n * 56 and x = (n-1) * 55 ; thus
(insert algebra here) n=13 and x = 660.
The sum of the 12 positive numbers are 660, eleven are 1 or more . The
largest is 660-11 = 649 or less.
Also look for new scoring for Tim & Phil on Problem 143 due to multiple interpretations of "increasing" (e.g. one number)

Steve Lawrie . . . . . . . . . 10 pts - First again, your average isn't dropping like in this problem
Arthur Morris . . . . . . . . 7 pts - Yes, the average drops one point so 68 - 55 = 13 numbers.
Anirban Bhattacharyya. . . . 5 pts - Your name might be long but your solution was concise ;-)
Dangkhoa Pham . . . . . . 4 pts - Right; n = 13 and s = 728 so cn = 649.
Drew . . . . . . . . . . . . . . . . 4 pts - I like your strong opening: "649, and here's why."
Quasi-C . . . . . . . . . . . . . 4 pts - Law of avgs: (x1+...+xn)/n=56 and (x1+...+xn-1)/(n-1) = 55
Tim Poe . . . . . . . . . . . . . 3 pts - Search in Viz Basic finds the solution; does it prove it unique?
Ludwig Deruyck . . . . . 3 pts - Good set of equations: T/N = 56 and (T-68)/(N-1) = 55.
Hermen Jacobs . . . . . . . 3 pts - You got quoted this week. Keep on entering, Hermen!
Lisa Schechner . . . . . . . 3 pts - That's it; you let the variables take out the guesswork
Joe Alvord . . . . . . . . . . 3 pts - That's right: 1, 1, 1, 1, 1, 1, ... , 1, 68, 649 will do it!
Phil Sayre . . . . . . . . . . . 3 pts - You mean you solved a pair of equations without Python?
Jack Dostal . . . . . . . . . . 3 pts - The idea of sacrifice; want one biggest, make others small
Jon Stearn . . . . . . . . . . . 2 pts - That makes it seem easy; s/n = 56, (s-68)/(n-1) = 55.
Nikita Kuznetsov . . . . . 2 pts - Ok it was an easy-ish problem, and thanks for the joke.
Nick McGrath . . . . . . . . 2 pts - Good system of equations makes it systematic!
Omar Gymboski . . . . . . 2 pts - I'm glad this solution is yours and not a corporate product.
Zahi Yeitelman . . . . . . . 2 pts - That's it, the sum affects the average and vice versa.
Marcel Birthelmer (new) 2 pts - Welcome, and be sure to put problem number in subject!
Abbas Mehrabian . . . . . 2 pts - Glad to have you back; and 660 - 11*1 = 649 is good step.
Musthaffa Kannipoyil (new) 2pts Welcome to my contest, good solution and notation.
Ed Wern . . . . . . . . . . . . . 2 pts - You got it; N, S, 13, 660, gets you 649 in a nutshell
Matt Situ . . . . . . . . . . . . 2 pts - Jump on the bandwagon, license plate #MAX649.
Richard Hillman . . . . . 1 pt - Hi (again)! 649 was proper one; 0 isn't positive so 660 too big.

Problem #147 - Posted Tuesday, June 4, 2002
#1. A semicircular arched doorway is 6 ft high, 1 ft from the edge.
What's the height of the arch 5 ft from the edge (above the floor)?
#2. The square of the diagonal of square ABCD is 8. Point P on side BC makes the ratio
of PC to PB equal to 3. If a circle passes through A, P, and the center of the square,
how far is the center of the circle from the point D?
#3. How much is the quantity : sin[arcsec(17/8) - arctan(-2/3)] ?
Give each answer in simplest form (A *B ) / C , with B square-free, and find the product
of these three answers, also written in this form. Explain steps; resubmissions carry a 1-pt penalty.

Solution: Art allowed himself square-free rationals which wasn't my intention but not really prohibited. Here's Phil:

Part 1. The equation for the doorway, origin at center, is R^2=x^2+y^2, of course; at x=R-1 (one foot from the edge),
y=6, the eqation yields R=18.5 ft. Hence at x=R-5, y=sqrt[R^2-x^2]=sqrt(160)= 4*sqrt(10).

Part 2. Using analytic geometry, let center of square be at origin (point O). If length of side is 2k, then coordinates of
vertices are: A:(-k,k); B:(k,-k); C:(k,-k); D:(-k,k). The length of the diagonal AC is sqrt[2*(2k)^2]=k*sqrt(8). Given
that AC^2=8, k=1. The point P has coords (k,s)=(1,s). The ratio PC/PB=(k+s)/(k-s)=3 or s=k/2=1/2. P is (1,1/2).
Suppose the center of the circle is point Q with coordinates (x0,y0).
The circle is described by the three equations: R^2=QA^2=(x0+1)^2+((y0-1)^2 = x0^2+y0^2+2x0-2y0+2 ;
R^2=QO^2= x0^2+y0^2 ; R^2=QP^2=(x0-1)^2+(y0-1/2)^2=x0^2+y0^2-2x0-y0+5/4
From the first two of these we obtain y0=x0+1. From the second two, y0=-2x0+5/4.
Equating these, we get x0=1/12, hence y0=13/12. The distance QD is sqrt[(x0+1)^2+(y0+1)^2]=[sqrt(794)]/12

Part 3.
Let angle a=arcsec(17/8); then sec (a)=17/8 or cos (a) = 8/17. First note -arctan(-2/3)=+arctan(2/3). Let angle
b = arctan(2/3); then tan (b)=2/3. We substitute into the formula sin(a+b)=sin(a) cos(b) + cos(a) sin(b) the corresp
expressions from the right triangles defined by cos(a)=8/17, namely sin(a)=[sqrt(17^2-^2)]/17=15/17, and tan (b)=2/3,
sin(b)=2/sqrt(13); cos(b)=3/sqrt(13). Hyp is sqrt(2^2+3^2). We get sin(a+b)=61/[17*sqrt(13)]= 61*sqrt(13)/221.

Final result: product of above: 4sqrt(10)*[sqrt(794)/12]*[61sqrt(13)/221] = 122*sqrt(25805)/663 ~ 29.56 ft^2.

Tim Poe . . . . . . . . . . . . . 9 pts - Got it right on 2nd e-mail, before any other correct answers!
Phil Sayre . . . . . . . . . . . 7 pts - Couldn't have explained it better myself, so I didn't!
(see above)
Steve Lawrie . . . . . . . . . 5 pts - Nice work, and a set of attached jpeg pictures no less!
Arthur Morris. . . . . . . . 3 pts - Mostly ok but 61/(17sqrt13) wasn't kosher
and ans a bit off
Anirban Bhattacharyya . . . 3 pts - Nice clear expl; all parts ok but product has a 122 not just 2
Nick McGrath. . . . . . . . 3 pts - You got sqrt[397]/3 for part 2; and were not alone with that.
Lisa Schechner . . . . . . . 3 pts - Mostly ok but for sqrt[397]/3
by thinking square side = 4sqrt2
Ludwig Deruyck . . . . . 2 pts - Part 1 was sqrt 160 not sqrt 105 (you used 5 not 6) but good try
Hermen Jacobs . . . . . . . 2 pts - Part 2 close with (sqrt 394)/2 ; other parts ok. HB80 again!
Quasi-C . . . . . . . . . . . . . 2 pts - 2nd and 3rd answers ok; first needed some simul-solving.
Joe Alvord . . . . . . . . . . . 2 pts - Yeah, I guess 4 sqrt 10 ft is a tall door. Good on part 3 too.
Jon Stearn . . . . . . . . . . . 1 pt - I like your use of Pythm on part 2; ans a bit off there
Nikita Kuznetsov . . . . . 1 pt - Saw your attachmt; sqrts were hard to read (131/2); good try!

Problem #148 - Posted Saturday, June 22, 2002
Our base 10 system has columns for 1's, 10's, 100's places, etc.;
digits can be 0, 1, ..., 9 in each. What if we use other systems?
A. Factorial Base: Columns 1, 2, 6, 24, etc; digits in k! col. can be 0, 1, ..., k.
B. Fibonacci Base: Columns 1, 2, 3, 5, 8, etc. Digits can only be 0 or 1 each.
C. Square Base: Cols 1, 4, 9, 16, etc. Digits 0-3 in 1's, 0-2 in 4's, 0-1 others.
(1) Express one million (uniquely) in factorial base.
(2) Express one hundred in as many ways as possible for the other two systems.

Solution: Here's our Alaskan entrant Joe A. with part A: "One million in factorial base is 266,251,220
10! is more than a million and 9! is less, so the number has 9 digits. Take as many 9!'s from a million
as you can, record that number in the 9!'s place, the number that is left is made up from the rest of the
digits. Repeat with each factorial 8!, 7!, 6!, etc. until the leftover number is zero.
2 * 9! = 2 * 362,880 = 725,760
6 * 8! = 6 * 40,320 = 241,920
6 * 7! = 7 * 5,040 = 30,240
2 * 6! = 2 * 720 = 1,440
5 * 5! = 5 * 120 = 600
1 * 4! = 1 * 24 = 24
2 * 3! = 2 * 6 = 12
2 * 2! = 2 * 2 = 4
0 * 1! = 1 * 0 = 0

For parts b) and c) we turn to 2002 high school grad and imminent Cornell freshman Nikita K.
b) Fibonacci Base : 9 ways - "there's a nice pattern here!!! ­ the first five columns just repeat themselves every three rows"
 1 2 3 5 8 13 21 34 55 89 Number 0 0 1 0 1 0 0 0 0 1 1000010100 1 1 0 0 1 0 0 0 0 1 1000010011 1 1 1 1 0 0 0 0 0 1 1000001111 0 0 1 0 1 0 0 1 1 0 110010100 1 1 0 0 1 0 0 1 1 0 110010011 1 1 1 1 0 0 0 1 1 0 110001111 0 0 1 0 1 1 1 0 1 0 101110100 1 1 0 0 1 1 1 0 1 0 101110011 1 1 1 1 0 1 1 0 1 0 101101111

c) Square Base : . . . there are 10 ways in square base. You can also start with the largest value column. -- Dan
 1 4 9 16 25 36 49 64 81 100 Number y o u c a n 1 1000000000 0 100001003 f i l l i n 0 100000122 0 10100000 t h e s e 0 10010102 0 10010023 c o l u m n s 0 10001123 * 0 1100112 y o u r s e l f ! 0 1011101 0 1011022

* most often overlooked way

Phil Sayre . . . . . . . . . . . 9 pts - First correct answer list but with no details on part A I hadda docku
Nikita Kuznetsov . . . . . 7 pts - Congrats on HS grad and Cornell entrance. Carl Sagan lives!
Steve Lawrie . . . . . . . . . 5 pts - You showed surprise that my factorial base was unique!
Joe Alvord. . . . . . . . . . . 4 pts - Good answer style from the southern part
of our northern state!
Nick McGrath. . . . . . . . 3 pts - Woulda been 5 or 4 pts but base numerals not given ...
Tim Poe . . . . . . . . . . . . . 3 pts - I had you rescored to 2 pts
but your later followup was great!
Ed Wern . . . . . . . . . . . . 3 pts - You're welcome; glad you're enjoying these problems!
Dangkhoa Pham. . . . . . 3 pts - Glad you figured out the system so well
; left off final 0 in A
Arthur Morris. . . . . . . . 2 pts - Prompt response as usual this time short & off on part C answers
Anirban Bhattacharyya . . . 2 pts - Rewrite numerals in part A, a couple more each in parts B and C
Ludwig Deruyck . . . . . 2 pts - You did find most of them and all yours were ok as far as I can tell
Jon Stearn . . . . . . . . . . . 2 pts - You got almost all of em but missed 10001123 at Base Squaresville.
Quasi-C . . . . . . . . . . . . . 2 pts - Missed same one; you do need 2*2^2 otherwise no 8 possible
Hermen Jacobs . . . . . . . 2 pts - Good on A; got all but 2 each on B and C - can u find which?
Jack Dostal . . . . . . . . . . 2 pts - Good job on A, all 9 on B, a few short on the squares.
Paul Botham (new). . . . . 2 pts - Welcome! Later ans, correct but no expl'n or steps on B and C
Drew. . . . . . . . . . . . . . . . 1 pt - Part A seemed good; didn't hear back about b and c. How was 'camp'?

Problem #149 - Posted Thursday, July 4, 2002
Three sisters : Venus , Serena , and the well-hidden little Cathy,
play one tennis opponent per week. The chances of each sister
beating any unrelated player are 7/10, 8/10, and 9/10, respectively.
The probabilities of any sister beating another are given in the
table at right (for example Venus beats Cathy 7/10 of the time).
Each week, one pair of sisters plays each other (at random) ;
the third plays an outsider. In a 50-match season, which
player is likely to win the most matches, and how many
matches would each one expect to win, on the average?
 \ V S C V .0 .6 .7 S .4 .0 .9 C .3 .1 .0

Solution: Here's a thorough solution from brand new contestant AZ Runner!
"Each week a sister has a 2/3 chance of playing against a sister. Thus a 1/3 (1- 2/3) chance of playing against an
unrelated opponent. Since it is random which other sister is played each week, the 2/3 chance of playing either
sister becomes a 1/3 chance of playing against a specific sister. Therefore, the probability for any sister playing
an opponent is: 1/3 opponent unrelated, 1/3 opponent is sister one, 1/3 opponent is sister two.
Using the above information and the information provided in the question we can calculate the average number
of matches each sister is likely to win if 50 matches are played during the year, one a week.
For Venus: 50* [1/3(7/10) + 1/3(6/10) + 1/3(7/10)] = 33 1/3 matches Venus will win on average.
For Serena: 50*[1/3(8/10) + 1/3(4/10) + 1/3(9/10)] = 35 matches Serena will win on average.
For Cathy: 50*[1/3(9/10) + 1/3(3/10) + 1/3(1/10)] = 21 2/3 matches Cathy will win on average.
In a 50-match season, Serena is likely to win the most matches.

Steve Lawrie . . . . . . . . .10 pts - Yes, each plays 50 matches, not 50 total for all players.
Jack Dostal . . . . .
. . . . . 7 pts - I like it: "This explains why Cathy is heard of so seldom!"
AZ Runner (new) . . . . . . 5 pts - Welcome to my contest; keep on entering for further fame!
Jon Stearn . . . . . . . . . . . 4 pts - Yep, hard for anyone to win a partial match. Watch for roundoff!
Nick McGrath. . . . . . . . 4 pts - Is X the outsider or a fourth sister? I hear there are older W. kids!
Phil Sayre . . . . . . . . . . . 4 pts - You move up a step with complete answer and others' roundoffs.
Ludwig Deruyck . . . . . . 3 pts - I think it's possible for "beating" not to be a transitive relation.
Hermen Jacobs . . . . . . . 3 pts - The cases are equally likely so you're ok.
Lots of Dutch players!
Paul Botham . . . . . . . . . 3 pts - I like your complete table of probab's. The .txt attachmt was ok.
Omar Gymboski . . . . . . 2 pts - Welcome back! Careful about 'assuming'; equal probs was given
Ed Wern . . . . . . . . . . . . 2 pts - VSC wereVanilla (not Sky), Strawberry (not Darryl), & Choc.
Tim Poe . . . . . . . . . . . . . 2 pts - Nice method to give a 'typical 3-week period' and 'extrapolate'.
Quasi-C . . . . . . . . . . . . . 2 pts - Ok but Vanessa & Cindy don't play tennis,
verified by A.Andersen.
Joe Alvord. . . . . . . . . . . 2 pts - Right, 50-match season means each, as in baseball's 162 games.
Lisa Schechner. . . . . . . . 2 pts - Good answer; I'd have liked to see the (2.1 / 3)(50)
calc'd out.
Nikita Kuznetsov . . . . . 1 pt - Turned out you did (just) get it in on time; I was slow about #150!
Abbas Mehrabian . . . . . 1 pt - Arithmetic ok, but you claimed Cathy was winner, meant Serena.

Problem #150 - Posted Sunday, July 28, 2002
Hermen knows how old he is turning this birthday; you don't. He is as many years old as
the largest number of divisors of any integer N less than or equal to 20,000 (twenty thousand).
How old is Hermen turning, and what's the smallest such N? (Bonus point for giving the three
next-smallest N's with the same no. of divisors.)

Solution: Many of you used computer programs (mentioned in purple) to check or find the solution(s),
ranging from the complex (C++, Python, Delphi) to the simple and/or elegant (MATLAB, BASIC, Mathematica).
. . . Art's code: (he figured these candidates would be multiples of 2 and of 3 hence counted by 6's) : %prob150.m ;
for n=[0:6:24000]; d=0; for k=[1:sqrt(n)]; if mod(n,k)==0; d=d+1; end; end; if d>39; fprintf(1,'%6i%6i\n',n,2*l); end; end .
Kudos to Joe from Alaska for discovering the connection between # of divisors and prime fac exponents.
If we search all N up to 20000 the largest number of divisors we find is 80 for N = 15120 smallest.
If we prime-factorize N = p^a * q^b * r^c * ... then N has d(N) = (a+1)(b+1)(c+1)... divisors
(think of columns and choices; see my Number Theory math lessons for details...)
Then we look for others with 80 divisors and find 18480, 21840, and 22680 are next.
80 = 2*2*2*2*5 so we group 5*4*2*2 finding expons 3, 2, 1, 1 so N = 2^3*3^2*5*7 = 15120;
if we group 5*2*2*2*2 it's N = 2^4*3*5*7*11 = 18480. I call these 'ordered' and 'saturated'
meaning (as Quasi-C pointed out), the small primes want larger exponents, and p=2, q=3, r=5, etc.
But then 21840 = 2^4*3*5*7*13 is not saturated, and 22680 = 2^3*3^4*5*7 is not ordered.
Some other interpretations were that 10080 has 72 divisors, or that 20001, 20002, 20006 have 8 divisors.

* indicates bonus point included ... Thanks for including computer code if used.
Steve Lawrie . . . . . .*11 pts - (Delphi) Got em all, though checking up to n/2 is kinda inefficient.
Arthur Morris. . . . . . 7 pts -
(MATLAB) Very elegant short program; overlooked 21,840.
Joe Alvord. . . . . . . . . 5 pts - Glad that your brain keeps expanding! Some smaller n's work too.
Tim Poe . . . . . . . . . .
*5 pts - (VBasic) Thanks for the lists of divisors & all N<50000 with d(N)>=80.
Quasi-C . . . . . . . . . . *4 pts - For optimal N I agree with ordered and saturated, but not record-tiers.
Jon Stearn . . . . . . . . *4 pts - (C++) Yes, brute force works but go only to sqrt[n] for efficiency.
Jack Dostal . . . . . . . .*4 pts - (C/C++) Thanks for the interactive program; I read it passively.
Nick McGrath. . . . . .*4 pts - Thanks for noticing the pr. fac. method is explained in my site!
Zi Jheng
(new) . . . . . . 3 pts - (Python) Hello; happy to have you, glad you like my site! Good answer.
Hermen Jacobs . . . . . 3 pts - (BASIC) You're welcome ;-} you knowing your age was a birthday present!
Ludwig Deruyck . . . 3 pts - Kim & Justine out early in US Open but Mathematica kept you in game!
AZ Runner . . . . . . . *3 pts - Welcome back! Thanks 4 including examples, reasoning, next 3.
Abbas Mehrabian . .*3 pts - Good observ. that N>10000 for if not, 2N has more divisors.
Nikita Kuznetsov . . . 2 pts - Think you misread 20000 as 2000; that took 40 years off his age!
Anirban Bhattacharyya 2 pts - Good list ; 10080, 12600, 13860, 17640 all have 72 divisors, yes.
Lisa Schechner@. . . . 2 pts - I agree with 15120 and 18480; not clear why the 5-prime limit?
Paul Botham . . . . . . . 2 pts - Yes, 'is' and 'is turning' same age; 15120 ok
Drew . . . . . . . . . . . . . 1 pt - Ok, 2*3*5*7*11=2310 so put extra factors, can beat 72 divs though.
Uwe Buenting . . . . . . 1 pt - Good to have you back; 15120 included among others but not selected
@ Dan's Note: Lisa S. has also been awarded 2 pts for Prob. 144, "Click to 222," for a lost answer.

 THANKS to all of you who have entered, or even just clicked and looked. My site is now in its fifth season - OVER 29,000 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 - 2*7*11*13 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+

or go back and work on this week's problem!