**dan's math@home - problem of the week - archives****Problem Archives**page 8**with Answers and Winners.**For Problems Only, click here.**1-10****. 11-20 . 21-30 . 31-40 . 41-50 . 51-60 . 61-70 . 71-80 . 81-90 . 91-100****101-110****. 111-120 . 121-130 . 131-140 . 141-150 . 151+ .**prob index71- Reciprocal Sums 72- Valentines 4 All! 73 - Divisible Dates? 74 - Four Centuries ! 75 - Work/Study Pts. 76 - The Triple Kiss! 77- Polynomial Road 78-RunForYourLife! 79-Open & Shut Case 80-Try Our Products - Problem #71 - Posted Friday, February 4, 2000
- Reciprocal Sums (back to top)
- 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).- You don't always get the best answer by using the biggest possible fraction that still fits.
- 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
- 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)

**WINNERS - Problem 71 .**(back to top)**.**leader board**Kevin Williams**.**Total 16 fractions, denom. sum 1611****Charles Stanton**.**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**.**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)
- Problem #72 - Posted Saturday, February 12, 2000
(back to top)
**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
**WINNERS - Problem 72 .**(back to top)**.**leader board**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**.**Al B.**. . . . . . . . . . . . . . . 4 pts - Good algebra except B*B-1 means B^2 - 1, not B*(B-1) as you meant.**Colleen Frye**. .**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
- Divisible Dates? (back to top)
- 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.

**WINNERS - Problem 73 . . .**(back to top)**.**leader board**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**.**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**. .**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
- Four Centuries! (back to top)
- 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.

**WINNERS - Problem 74 . . .**(back to top)**.**leader board . . . .**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
- Work/Study Points (back to top)
- 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?
- Explain your strategy carefully, it's the essay question for your scholarship!

**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**.

**WINNERS - Problem 75 . . .**(back to top)**.**leader board . . . .- . . .. Please don't send attachments, unless they're GIF files.
- These answers were correct at x = 15, y = 20, P = 125:
**Phil Sayre**. .**Kevin Williams**. . . . . 6 pts - It turned out that x and y were integers, but it's not guaranteed.**Patrik Petersson**.**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.

**WINNERS - Problem 76 . . .**(back to top)**.**leader board . . . .**Kevin Williams**. . . . . 9 pts - You were the winner but some steps and accuracy missing**Phil Sayre**. . .**Allen Druze**. . . . .**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
- Pacific Polynomial Road (back to top)
- 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:
**L**os Angeles (3,4),**N**ewport Beach (5,1),**P**asadena (4,5),**S**anta 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
**C**hatsworth 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)

**WINNERS - Problem 77 . . .**(back to top)**.**leader board . . . .**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**. . . . .**Patrik Petersson**.**Phil Sayre**. . .**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.

- Problem #78 - Posted Wednesday, April 5, 2000
- Run For Your Life! (back to top)

- 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.

**WINNERS - Problem 78 . . .**(back to top)**.**leader board . . . .**Arthur Morris**. . . . . . 10 pts - Great simple way to look at potentially complex question.**Allen Druze**. . . . .**Phil Sayre**. . .**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**.**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
- Open and Shut Case! (back to top)
- 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?
**WINNERS - Problem 79 . . .**(back to top)**.**leader board . . . . all answers were correct this time!**Sue B.**. . . . . . . . . . . . .10 pts - Factors activating lockers - fascinating!**Arthur Morris**. . . . . . 7 pts - Right about the perfect squares. What about cubes?**Allen Druze**. . . . .**Andrew Bajunemi**. . . 5 pts - Another victory for the Microsoft Excel juggernaut!**Patrik Petersson**.**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**. . .- Problem #80 - Posted Sunday, April 30, 2000
- Try Our Products! (back to top)
**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****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.

**WINNERS - Problem 80 . . .**(back to top)**.**leader board . . . .**Arthur Morris**. . . . . 10 pts - Nice answer and "suspicion of e". Keep it up!**Phil Sayre**. . .**Allen Druze**. . . . .**Chen Sin Tak**. . . .- (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 **- 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+ **Browse the complete problem list, check out the weekly leader board,****or go back and work on this week's problem!**(back to top) **[ home | info | meet dan | ask dan | matica | lessons | dvc ]**This site maintained by B & L Web Design, a division of B & L Math Enterprises.