dan's math@home - problem of the week - archives
Problem Archives page 8

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

71- Reciprocal Sums
72- Valentines 4 All!
73 - Divisible Dates?
74 - Four Centuries !
75 - Work/Study Pts.
76 - The Triple Kiss!
78-RunForYourLife!
79-Open & Shut Case
80-Try Our Products

Problem #71 - Posted Friday, February 4, 2000
If n is a natural number, the reciprocal of n, 1/n, is called a 'unit fraction.'
Historically, numbers were expressed as sums of unit fractions, like 4/7 = 1/2 + 1/14 or 1 = 1/2 + 1/3 + 1/6.
Write each of these as sums of distinct unit fractions (with denoms at least 2; read rules below):
a) 3/23 . . . b) 14/15 . . . c) 7/11 . . . d) 11/7
RULES: The winner will be the one using the smallest total number of fractions. In case of a tie, the winner is the
one with the smallest sum of all denominators. You can use the same fraction for more than one of the numbers.

Solution: The first one was part of an early problem, #3 - Unit Fractions (see link below).
Here are the best answers I received for each part. Remember, only fractions with 1 on top count.
a) 3/23 = 1/8 + 1/84 (almost all of you)
b) 14/15 = 1/2 + 1/3 + 1/10 (most of you)
c) 7/11 = 1/2 + 1/11 + 1/22 (from Al B. and Andy H.; most had 1/2 + 1/8 + 1/88)
d) 11/7 = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/14 + 1/28 + 1/70 (by Patrik)
Other interesting ones:
3/23 = 1/10 + 1/46 + 1/115 ; 11/7 = 1/2 + 1/3 + 1/4 + 1/5 + 1/10 + 1/14 + 1/21 + 1/28 + 1/30 (by Po-Nien)
11/7 = 1/2 + 1/3 + 1/4 + 1/6 + 1/7 + 1/9 + 1/15 + 1/1260 (Kevin)
11/7 = 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/9 + 1/97 + 1/122220 (Charles; Sue B. close)

Kevin Williams. . . . . . 10 pts - Total 16 fractions, denom. sum 1611
Charles Stanton . . . . . . 7 pts - Welcome back! 16 fractions, d. sum 122,646
Patrik Petersson . . . . . 5 pts - 17 fracs. The 1/3 + 1/6 was 1/2, or you woulda won!
Po-Nien Chen . . . . . . . 4 pts - 18 fracs, smallest denom total on 11/7. Welcome 2 the contest!
Sue B. . . . . . . . . . . . . . . 2 pts - 15 fracs but your 11/7 was off by a few millionths
Al B. . . . . . . . . . . . . . . . 3 pts - 12 fracs; great start on a,b,c but repeated 1/2 on d. BONUS PT INCLUDED!
Melina Hidayat . . . . . . 1 pt - Good entry on a); some fracs had numerators over 1.
Andy Harmon . . . . . . . 1 pt - Best answer on c, slightly off on d.

Extra Point Winner:
Al B. sure enjoys a challenge, and was the first to send in better answers to part d:)
11/7 = 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/12 + 1/14 + 1/24 (d-sum 73)
11/7 = 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/10 + 1/14 + 1/40 (d-sum 87)

 Valentines 4 All!   In Miss Lizzy's . . . . .4th grade class, all the kids gave each of the other kids a valentine. The girls received a total of 798 valentines and the boys got 684. How many kids were in Miss Lizzy's class, how many girls and how many boys? Explain your reasoning carefully.

Solution: (By several people)
Let's say there were n kids; they each sent n-1 valentines, so there were a total of
n*(n-1) = 798 + 684 = 1482 ; n^2 - n - 1482 = 0 ; use quadratic formula
n = (+1 +- Sqrt(1+4*1482))/2 = (1 +- Sqrt(5929))/2 ; n = 39 or -38.
These are kids, not anti-kids, so n > 0 , making n = 39.
Each girl got 38 val's, so there were 798/38 = 21 girls;
Each boy got 38 val's, so there were 684/38 = 18 boys;

There were 21 girls and 18 boys, making 39 kids in Lizzy's class. Go figure!

Two of you complained that the problems aren't hard enough. This one might not be that hard, but at least you
have to think, and that's all I'm here for. You keep thinking about stuff; it means you're still alive. -- Dan

Sue B. . . . . . . . . . . . . . . 10 pts - You got it & explained it, but decimals are still dangerous.
Patrik Petersson . . . . . 7 pts - Good use of a system of two quadratic equations in two variables!
Kevin Williams. . . . . . 5 pts - You had a good equation but I'd like more expl'ns.
Al B. . . . . . . . . . . . . . . . 4 pts - Good algebra except B*B-1 means B^2 - 1, not B*(B-1) as you meant.
Colleen Frye. . . . . . . . . 4 pts - Welcome back, Colleen... give yourself a valentine! :)
Dave Seaman. . . . . . . . 3 pts - Dave proves he loves algebra with this trio of variables.
Sarah DeSanctis . . . . . 2 pts - The return of an original contestant! Is Dan's website still up? Hi Sarah!
Phil Sayre. . . . . . . . . . . 2 pts - Good answer, but the early worms were taken (pardon my metaphor)
Po-Nien Chen . . . . . . . 2 pts - You got it. And all the kids did too!
Cheryl Fredricks . . . . 2 pts - Good explanation, and you get 2 pts on Problem 70 also!
Esther Li. . . . . . . . . . . . 2 pts - The e-mail bouncing took 10 days to get me your answer!
Andrew Bajunemi . . . 1 pt - Right answer ; more points next time with early entry & more steps
Shinsuke Kuzuya. . . . 1 pt - Reasoning was good but no total and some steps skipped
Sarah Mahdavi. . . . . . 1 pt - Good try, but the kids gave out lots of valentines, not 1 each.

Problem #73 - Posted Sunday, February 20, 2000
The posting date (Feb 20) of this problem can be written as 2|20|2000.
Notice the 2 goes into the 20 (without remainder) and the 20 goes into 2000.
a) What are all the days this year that have this "divisible date" (dd) property, how many
dd's are there, and what percentage of the days of 2000 do they amount to (nearest 1/10 %)?
b) In France they would name Feb 20th as 20|2|2000, not a "dd." But October 5 would be a French dd.
How many French dd's are there this year? c) How many days are either dd, French dd, or both?

Solution: Phil Sayre and others) had a nice way of organizing these:
a) The possible divisible days are 1, 2, 4, 5, 8, 10, 16, 20, and 25. For each one, the months that go into it are:
1|1, 1,2|2, 1,2,4|4, 1,5|5, 1,2,4,8|8, 1,2,5,10|10, 1,2,4,8|16, 1,2,4,5,10|20, and 1,5|25. A regular list is:
1/1, 1/2, 1/4, 1/5, 1/8, 1/10, 1/16, 1/20, 1/25 (9 in Jan.); 2/2, 2/4, 2/8, 2/10, 2/16, 2/20 (6 in Feb.);
4/4, 4/8, 4/16, 4/20 (4 in Apr.); 5/5, 5/10, 5/20, 5/25 (4 in May); 8/8, 8/16 (2 in Aug.); 10/10, 10/20 (2 in Oct.).
This is a total of 27 dd's; representing 27/366 = 7.4% (approx.) of the Y2K days.
b) The French dd's are found by listing the divisible months: 1, 2, 4, 5, 8, 10; then days:
1/1 (1 in Jan.); 1/2, 2/2 (2 in Feb.); 1/4, 2/4, 4/4 (3 in Apr.); 1/5, 5/5 (2 in May); 1/8, 2/8, 4/8, 8/8 (4 in Aug.);
1/10, 2/10, 5/10, 10/10 (4 in Oct.), for a total of 16 French dd's (or Swedish, as Patrik points out). (about 4.4%.)
c) There are 6 days that are both: 1/1, 2/2, 4/4, 5/5, 8/8, and 10/10. So the days that are some sort of dd will number:
27 + 16 - 6 = 37 days that are either dd, Frdd, or both.

Al B . . . . . . . . . . . . . . 10 pts - First answer without a boo-boo; show all steps (27+16-6)
Tom Falbo . . . . . . . . . 7 pts - Welcome Tom! Nice explanation, including prime fact'z'n of 2000.
Andrew Bajunemi . . 5 pts - Notice any patterns if you also list the French ones?
Those sending mostly (or partially) correct answers were:
Dave Seaman. . . . . . . 3 pts - Good lists; but Feb. 25 wasn't a legal dd, and see c.
Arthur Morris. . . . . . 3 pts - Thanks for entering; good answer except May 25 missing.
Sue B. . . . . . . . . . . . . . 3 pts - Second try was good improvement, dunno how you got c) 38.
Patrik Petersson . . . . 3 pts - Feb 5 is not a dd even in Sweden, sorry! Good answer otherwise
Tim Nelson. . . . . . . . .3 pts - You got it mostly except 'either' not 'both' in c, and 366 days.
Ileen Huh. . . . . . . . . . 3 pts - Woulda been 4 pts but miscalc'n in the percentage.
Phil Sayre . . . . . . . . . 2 pts - Good way to organize dates; some mos don't go into 2000.
Kevin Williams. . . . .2 pts - Right totals, but even easy steps and list need to be shown
Colleen Frye. . . . . . . 2 pts - The French do things their own way; day first.
Tim Gavin. . . . . . . . . 2 pts - Welcome to the contest; remember to show lists and steps
Esther Li. . . . . . . . . . 1 pt - Good try but need 4, 8, 16 in some months.
Judy Kane . . . . . . . . .1 pt - Welcome! That's a good start; there are a few more dates.

Problem #74 - Posted Tuesday, February 29, 2000
The posting date of this problem (Feb 29) is the 'leap day' in the only leap year that ends in 00 for the next 400 years. This is because of the rule of "a leap year every 4 years, except for every 100 years, except for every 400 years."
a) Prove that February 29, 2400 will be on the same day of the week as Feb 29, 2000. (This says the days of the week follow a 400-year cycle, not 2800 years.)
b) How many times, during the 400-year period from Jan 1, 2000 to Dec 31, 2399, is the "first of the month" on a Saturday, as was 1/1/2000? (There are 4800 months.)
c) Which day (or days) of the week have the most "first days of the month," and how many "firsts" are on that day in a four-century period? Answer any or all the parts you can; please include your methods and reasoning.

Solution: These are sketchy for now; I need sleep.
a) There are 97 leap years and 303 regular years, so total number of days in 400 years is:
. . .97*366 + 303*365 = 146097 days = 7 * 20871, so it's a whole number of weeks!
b) According to Sue B, there are 684 Saturdays! (The first two occur in the first 4 months!)
c) 'Sunday' is the overall winner; I was wrong here last week. Here are the 'daily' totals:
. . Sun: 688, Mon: 684; Tue: 687; Wed: 685; Thu: 685; Fri: 687; Sat: 684.
Good job by Kevin also with the huge spreadsheet, the 365 rows instead of 400 leaves me confused.

Sue B. . . . . . . . . . . . . .10 pts - I trust your programming and built-in calendar!
Kevin Williams. . . . . 4 pts - I think your totals are too 'spread' out (variance), great try!
Al B . . . . . . . . . . . . . . . 4 pts - I like your idea of a quasi-repeating sequence,
Shinsuke Kuzuya . . . 3 pts - Good try; I like the inventive chart
Sarah Mahdavi . . . . . 2 pts - You were first with an answer, needed more tallies!
Mary Jane Santos . . . 1 pt - Welcome to the contest! Thanks for your entry.
Michael Williams . . . 1 pt - You helped, so you get a point. You're welcome!

Problem #75 - Posted Wednesday, March 8, 2000
You want to win a scholarship by compiling the most "work/study points."
You get 3 points for each semester-unit of classes you take, and 4 points for
each hour of work (per week) on your job.
. . (1) You have to take at least 7 units,
. . (2) Triple your units, plus your work-hours, is at most 75,
. . (3) The sum of the squares of your units and your work-hours is at most 625.
How should you plan your schedule (x units and y work-hours) to maximize your work/study points?

Solution: This is like a 'linear programming' problem, except the inequalities aren't all linear.
That's the trick, the solution isn't at one of the 'corners' of the region. Let's graph the region:
(1) x >= 7 ; (2) 3x + y <= 75 ; (3) x^2 + y^2 <= 625 ; (4) y >= 0.
See the graph at the right for the corresponding boundary colors. ---> [ graph coming later ]
The light parallel lines represent 'level curves' or places of constant work/study points P = 3x + 4y.
The values of P increase away from the origin, so we want the last line that still touches our region.
This gives a slope of -3/4, meaning it's tangent to the circle at (15, 20) where the radius has a slope
of 4/3; the negative reciprocal of the perpendicular tangent line. Phil saw both these strategies!
So you should take 15 units and work 20 hours per week, for a total of 15(3) + 20(4) = 125 points.

. . .. Please don't send attachments, unless they're GIF files.
These answers were correct at x = 15, y = 20, P = 125:
Phil Sayre . . . . . . . . . 10 pts - Very elegant ant thorough solution, even without the addendum.
Kevin Williams. . . . . 6 pts - It turned out that x and y were integers, but it's not guaranteed.
Patrik Petersson . . . . 6 pts - Nice answer; woulda been more points with more steps explained
Al B . . . . . . . . . . . . . . . 5 pts - Same attachment to integers; I'm taking a 1.5-unit class right now
Sue B. . . . . . . . . . . . . . 4 pts - Nice distinction between 'line points' and 'circle points'
Andrew Bajunemi . . 4 pts - Good method except show you tried all cases
Most of these were x = 20, y = 15, P = 120; 'intersection points only':
Po-Nien Chen . . . . . . 3 pts - Never could open that .bmp, I'll resort to a PC soon
Dave Seaman. . . . . . . 3 pts - Never mind what's 'feasable', this is math!
Shinsuke Kuzuya . . . 2 pts - Good try; careful simplifying algebra
Filipe Gonçalves. . . . 2 pts - Welcome! Your equations were ok but your totals were off

 Problem #76 - Posted Saturday, March 18, 2000 The Triple Kiss (back to top) The term "kissing circles" means mutually tangent. If the circles at the right have radii 3, 4, and 5 cm, what is the area of the (green) shaded region between the three circles? Explain your procedure carefully; give your answer to the nearest thousandth of a sq cm.

Solution: From new contestant Allen Druze: (Dan's comments)
Connect centers to create a 7,8,9 triangle. Using the law of cosines extract the corresponding angles.
(Angles are 48.1897, 58.4119, and 73.398 degrees)
Then use (angle)/ 360 * (area of circle) to calculate each segment.That would be
10.513358 + 8.15583 + 5.7647 = 24.433888. (Use Heron's Formula to find the area of a triangle
with sides 7, 8, and 9) using the sq.root of s(s-a)(s-b)(s-c) to obtain 26.8328 .
Subtract this value from the area of the triangle and the difference is 2.399 cm^2.

Kevin Williams. . . . . 9 pts - You were the winner but some steps and accuracy missing
Phil Sayre. . . . . . . . . . 8 pts - Nice solution, you got Kevin's point. sin(u/2) = \/((s-x)(s-y)/xy)?!
Allen Druze . . . . . . . . 6 pts - Great answer, welcoeme to my contest!
Al B . . . . . . . . . . . . . . . 5 pts - Good steps & method; woulda been more but no calculations shown
Brian Lin. . . . . . . . . . . 4 pts - Nice answer & steps; you did h = 5.963 instead of Heron's formula.
Filipe Gonçalves . . . . 4 pts - Good answer, just rounded to 2.4 ; sorry about the e-mail delay!
Shinsuke Kuzuya . . . 3 pts - Good work but all areas rounded up and made 2.400
Dave Seaman. . . . . . . 3 pts - Your alias threw me; I like the "Law/cos, Law/sin, Subtract."
Sue B. . . . . . . . . . . . . . 3 pts - You did great but for severe rounding lost accuracy to 2.398
Po-Nien Chen . . . . . . 2 pts - I finally read your attachments, but 2 sectors were off
Tim Nelson. . . . . . . . .2 pts - Welcome back! Your book gave you sector arclength not area.

Problem #77 - Posted Monday, March 27, 2000
In Problem 59, we tried to minimize the length of a zig-zag path that went through 5 cities.
Now let's build a smooth curving road through 4 cities:
Los Angeles (3,4), Newport Beach (5,1), Pasadena (4,5), Santa Monica (2,3).
a) What's the smallest degree polynomial y=f(x) that will pass through all four cities? . . .
b) What is the exact equation of this polynomial? (Hint: use fractions not decimals)
c) Would this road go through Chatsworth at (1,6)? (Actual approximate SoCal-ordinates!)

Solution: a) How many points determine a line? Two, of course; and a line is a polynomial of degree 1.
So to fit 4 points you'd need a degree __ curve: That's right, _3_. A cubic. An S-curve.
b) Let y = f(x) = a x^3 + b x^2 + c x + d; notice there are 4 unknown coeff's; a,b,c,d, to fit our four points.
Now plug in the x's and y's given to get four linear equations in a, b, c, and d.
Don't let a little algebra make you run out for a copy of Mathematica; you can do it by hand, or with
matrices and maybe some help from a TI-83, -86, or -89 calculator. The solutions are:
a = -5/6, b = 15/2, c = -62/3, d = 21; f(x) = (-5/6)x^3 + (15/2)x^2 - (62/3)x + 21.
c) Well, once you have the correct f(x), see if f(1) = 6 or not. It's 7, so (1,7). Sorry Chatsworth! (see pic)

Arthur Morris. . . . . . 10 pts - Good presentation of system of equations and steps.
Andrew Bajunemi. . . 8 pts - You were first with the answer, but no equations or steps!
Filipe Gonçalves. . . . 7 pts - Nice proof it can't be quadratic! I got your #76 answer too.
Allen Druze . . . . . . . . 5 pts - Nice sys of eqns and reduction of variables; light on steps.
Patrik Petersson . . . . 5 pts - Thanks for telling me - matrix function! (on TI-calculator?)
Phil Sayre. . . . . . . . . . 4 pts - Explain the Lagrange Interpolation Formula! Ok, I'll look it up.
Kevin Williams. . . . . 3 pts - Your Excel 'trendline' can be a cubic curve? Gates got math!
Po-Nien Chen . . . . . . 3 pts - Your second answer was correct; I graphed both of them!
Dave Seaman. . . . . . . 3 pts - Glad you remembered multiple equations; you don't get forever!
Chen Sin Tak. . . . . . . 3 pts - Good answer and explanation; welcome to my contest!
Sue B. . . . . . . . . . . . . . 2 pts - Yes, there're formulas; good but explain what you used on yr TI-86.
 Here's the cubic road (red) missing (1, 6). Dotted road is degree 4 for Chatsworthers.
One day Dan was out bicycling. After entering a one-lane tunnel and riding one-fourth of the way through
it, he glanced back over his shoulder and saw a truck approaching the tunnel entrance at 80 miles per hour!
Doing the quick mental math, Dan realized that if he accelerated immediately to his top speed, he could just
escape with his life, whichever direction he rode. What is Dan's top biking speed?

Solution: I received mostly algebraic solutions; here's an exemplary one from Phil:

Let d=distance truck is in front of tunnel entrance, L=length of tunnel, x=Dan's speed.
Case 1: Dan turns around and heads for entrance, a distance of L/4.
Dan and truck get to entrance at same time T1=d/80=(L/4)/x.
Case 2: Dan streaks for exit, a distance of 3L/4. Dan and truck get to exit at same time
T2=(L+d)/80=(3L/4)/x.
Solving both equations for x and setting them equal, x=(L/4)*80/d=(3L/4)*80/(L+d)
After simplifying, d=L/2, hence x=40. Answer: Dan's top biking speed is 40 mph.

Arthur sent in this streamlined gem, and (so) it reached me first:
In the time he could get to the start of the tunnel, he could also be half-way down the tunnel.
At the time he is half way down, the train [truck] is at the entrance. He must be travelling at
80/2 = 40 miles per hour to reach the end of the tunnel at the same time as the train[uck] does.

Arthur Morris. . . . . . 10 pts - Great simple way to look at potentially complex question.
Allen Druze . . . . . . . . 7 pts - Good solution; I'd love it if you defined variables carefully
Phil Sayre. . . . . . . . . . 5 pts - You got honorable mention this time - literally!
Dave Seaman. . . . . . . 4 pts - Good to mention assumptions of const truck speed etc.
Daniel Branscombe . 4 pts - Ok to assume L = 4 mi; belated welcome to my contest!
Patrik Petersson . . . . 3 pts - Slight error a/80 = 2L/x should be L/x; I'm not that fast!
Filipe Gonçalves. . . . 2 pts - L/80 doesn't take into account the distance to the tunnel.
Al B . . . . . . . . . . . . . . 2 pts - The truck isn't at the tunnel yet, so Dan isn't 60 mph.
Chen Sin Tak . . . . . . 1 pt - Your quadratic eqn is for free-fall, maybe for a vertical tunnel.
Andrea Archie. . . . . . 1 pt - It turns out the tunnel length isn't important, just proportions.

Problem #79 - Posted Tuesday, April 18, 2000
 At the Metric Academy there are 100 students, and 100 lockers numbered 1-100. One day the lockers are all closed, and student #1 opens all of them. Then student #2 shuts every other locker, starting with #2, then #4, etc. Student #3 changes the state of every third one, starting with #3: opens it if it's shut, shuts it if it's open. Student #4 changes the state of every 4th locker, etc. This continues until the hundredth student changes the state of locker #100. Exactly which lockers are now open, and which are closed, after all the students have open and shut them as described? Show your steps and reasons.

Solution: This is a cool problem with a beautiful solution, as some of you have noticed.
Here's Arthur's: The number of times each door will be changed is the number of integer
divisors of that door number. Only perfect squares have an odd number of divisors.
Thus the doors that are open are 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100.
Thanks, Phil, for calling this problem a "little gem" -- I wish I'd written it!
(Dan's note: The reason perfect squares have an odd number of divisors d is that the
factors pair up: d, n/d except when d^2 = n a perfect square, then d = n/d. See? Si')
For example: 36 = 1 * 36 = 2 * 18 = 3 * 12 = 4 * 9 = 6 * 6. The 6 only counts once.
Filipe noticed 60, 72, 80, 96 have 12 divisors (the most). 1024 is the smallest with exactly 11 divisors; anyone know why?

Sue B. . . . . . . . . . . . . .10 pts - Factors activating lockers - fascinating!
Arthur Morris . . . . . . 7 pts - Right about the perfect squares. What about cubes?
Allen Druze . . . . . . . . 5 pts - Good solution; I'd have given you 6 if you'd included a list.
Andrew Bajunemi. . . 5 pts - Another victory for the Microsoft Excel juggernaut!
Patrik Petersson . . . . 4 pts - Great! Try my number theory beta lesson for more info.
Filipe Gonçalves. . . . 3 pts - I agree, this problem is made for divisors!
Al B . . . . . . . . . . . . . . . 3 pts - Nice to notice the closed lockers are groups of 3, 5, 7, 9, ...
Phil Sayre. . . . . . . . . . 3 pts - Good use of prime factorizations. What kind have most divisors?

Problem #80 - Posted Sunday, April 30, 2000
Answer all of these; they increase in difficulty.
a) What is the maximum possible product of two whole numbers whose sum is 10? What are the numbers?
b) What's the maximum product of any number of whole numbers whose sum is 10? What are they?
c) What's the maximum product if the sum of the set of whole numbers is 20? What are the numbers?
d) What's the maximum product of any number of whole numbers whose sum is 100? What are they?
e) What's the maximum product of any whole number of real numbers whose sum is 100? What are they?

Solution: a) A required problem in every precalc and calculus class: Let the numbers be x and y;
then x + y = 10 ==> y = 10 - x; the maximum of the product P = xy = x(10 - x) = 10x - x^2 occurs
when x = 5, so the numbers are 5 and 5 ; the max product is 5 * 5 = 25.
b) Try different sets that add to 10: Max product with 2 nos:5*5 = 25; try 3 nos: 4 * 4 * 2 = 32 ;
4 * 3 * 3 = 36 ; 5 * 3 * 2 = 30 ; now 4 nos: 3 * 3 * 2 * 2 = 36 ; 4 * 2 * 2 * 2 = 32.
c) Notice that 3 * 3 = 9 while 2 * 2 * 2 = only 8, so use at most 2 twos.
But 4 * 4 * 4 = 64 while 3 * 3 * 3 * 3 = 81, so use at most 2 fours; lots of 3's seem to be best.
For a sum of 20, 4 * 4 * 3 * 3 * 3 * 3 = 1296 ; max is 3 * 3 * 3 * 3 * 3 * 3 * 2 = 1458.
d) To total 100, use 32 3's and two 2's (or one 4): 3^32 * 2^2 = 7.412 * 10^15 approx.
e) This letter is the key: e. The closer all the factors are to e ~ 2.718, the better.
100 / e ~ 36.79 ; getting a cue from part a), make the numbers all equal: 37 factors of 100/37.
(100/36)^36 = 9.400 * 10^15 ; (100/37)^37 = 9.474 * 10^15 ; (100/38)^38 = 9.294 * 10^15.

Arthur Morris . . . . . 10 pts - Nice answer and "suspicion of e". Keep it up!
Phil Sayre. . . . . . . . . . 7 pts - Good job; I like the proof that the max of (C/x)^x is at (C/x) = e.
Allen Druze . . . . . . . . 4 pts - Yes, e is involved, can you show how the answers come from it?
Chen Sin Tak. . . . . . . 3 pts - Good proof on a) and right answer on b). You can c) the rest.
(Dan's note: The number e can be used; 36e + 2.14185 ~ 100 ; e^36 * 2.14185 ~ 9.23*10^15 ; pretty close but no cigar.)

 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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