dan's math@home - problem of the week - archives
Problem Archives page 4

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

31 - Paint My House!
32 - Squares in a Box
33 - Disk- o -Mania !
34 Party like it's 1999
35 - Fifth Half Asleep
36 -Under Salary Cap
38 - Same Birthday ?
39 -- Holy Triangle !
40 -- More on Twins

Problem #31 - Posted Thursday, December 3, 1998
Aaron, Brenda, Charlie, and Danita are house painters. The team of Aaron, Brenda, and
Charlie can paint my house in 3 days; if Aaron and Danita work as a pair it takes 6 days;
the pair of Charlie and Danita takes 8 days, and the team of Brenda, Charlie, and Danita
can finish in 5 days. How long would it take the whole 4-person crew to paint my house?
(Answer to the nearest minute if necessary)

Solution: Let's let A = days Aaron would take to paint the house himself, B = days for Brenda alone, etc.
Then Aaron can do 1/A of the house in a day. If Aaron, Brenda, and Charlie take 3 days, then the fractions
for one day give : (1/A) + (1/B) + (1/C) = 1/3 ; letting small letters a, b, c, d replace 1/A, 1/B, 1/C, 1/D :
we get the linear system {a + b + c = 1/3 , a + d = 1/6 ; c + d = 1/8 ; b + c + d = 1/5}.
The solution is : a = 3/20 , b = 3/40 , c = 13/120 , d =1/60 ; so that all four together would paint
a + b + c + d = 7/20 of the house in a day; it would take 20 / 7 days which is 2 and 6/7 days.
Ask Beth and she'll tell you this is 2 days, 20 hours, 34 minutes and change.

Beth Wilson . . . . . 10 pts - Nice solution and good expl of time conversion
Andy Murdock. . . 6 pts - I like your (linear) system; how bout hrs & mins?

Problem #32 - Posted Tuesday, December 15, 1998
Suppose you have one square tile of each of these side lengths: 1, 4, 7, 8, 9, 10, 14, 15, and 18 feet.
How can you fit them together into a rectangle? (With no cutting or overlapping, of course.)
To answer, give the coordinates of the lower left corner of each square, starting with the square at (0,0).
 Solution: With the ares hint, we add up 1^2 + 4^2 + 7^2 + . . . + 18^2 = 1056 = 2^5 * 3 * 11. So the rectangle has to be 32 x 33 because it has to be at least 18 feet wide. Then the 18 and 15 must be the long side and the 18 and 14 make the short side. Also, as Andy noted, to fit a 1 x 1 you could put it in a "whorl" (or is it whirl) of 3 consec. lengths; 7-8-9 or 8-9-10. Shuffle them around and get these coords, a la Beth Wilson: Square of area 18 is at (0,14); 15 at (18,17); 14 at (0,0); 10 at (14,0); 9 at (24,0); 8 at (25,9); 7 at (18,10); 4 at (14,10); and 1 at (24,9). (Dan's note: See also Problem #123 -- Tour de Ants!)

Beth Wilson . . . . . 10 pts - Timely solution and correct coordinates. Yay!
Andy Murdock. . . 5 pts - I think I can read them but I wanted coordinates...
Greg Hess . . . . . . . 5 pts - Welcome back; good answer! (Show steps pls)

 Problem #33 - Posted Sunday, December 27, 1998 Disk-O-Mania! (back to top) a) How many circular disks of diameter 1 inch can you fit (without overlapping) inside an 8 by 8 square ? b) What is the smallest n for which you can fit more than n^2 disks of diameter 1 inside an n x n square ? Explain fully why your answers work. click here for picture

Solution: a) The answer is: more than 64. Place 8 disks along the bottom, then a second
row of 7 can be tucked into the hollows as shown. The new top is only (-/3)/2 above
the old top, by using the Pythm on the equilateral triangle formed by three centers.
So we need to see if 9 rows will fit: 1 + 8 (-/3)/2 = 1 + 4 -/3 :=: 7.928 < 8 ; yes!
We get 8 + 7 + . . . + 8 = 5(8) + 4(7) = 68 disks!
. . . . . . b) Solving the inequality 1 + n (-/3)/2 < n gives n > 2 / (2 - -/3) :=: 7.46 ,
meaning that our solution n = 8 is smallest ; 68 > 8^2 = 64.
(Special Offer: What is a function f (n) for the largest number of disks in an n x n square?
For ex. f(3) = 9 , f(7) = 49 , f(8) = 68. One contest point for the first decent answer before Feb.1!)

Andy Murdock. . . 9 pts - Good solution; one inequality was backwds...
Beth Wilson . . . . . 3 pts - Close but check number of disks in each row
Greg Hess . . . . . . . 1 pt - Just for asking. Did I help clear it up for you?

Problem #34 - Posted Thursday, January 7, 1999
a) The number 1999: is it prime or composite? (Prove it's prime or find its prime factorization.)
b) Is 1999 the sum of two perfect squares? Three? How many does it take, what are they?
c) What's the smallest number of perfect cubes that add up to 1999 and what are they?

Solution: a) 1999 is prime: Try dividing all primes from p = 2 ... to ... p = 43 into 1999;
none go in evenly. Trying p >= 47 is not necessary because 47 * 47 = 2209 so if 1999 isn't
prime then it has a smaller factor. Trying non-prime divisors is never necessary.
b) Beth Wilson has found 8 ways to write 1999 as the sum of 4 squares.
This is the most you'll find, I trust her method. Can you web-hitters find them all?
Two are 1999 = 43^2 + 11^2 + 5^2 + 2^2 and 1999 = 39^2 + 21^2 + 6^2 + 1^2
Dan's note: ( 1 ) It has been proven that every natural number n can be written as the sum of at most 4 squares.
. . . . . . . .( 2 ) Those primes p that can be written as the sum of two squares, p = a^2 + b^2 , are the ones that _factor_
if you allow complex integers: p = (a + b i)(a - b i) . (These are exactly the primes that are 1 more than a multiple of 4.)
The other half of the primes, the ones three more than a mult of 4, don't factor over the complex numbers, and also
cannot be written as the sum of two squares. (Those a + b i with integer a and b are called GAUSSIAN INTEGERS.)
c) Beth and Andy both found 1999 = 10^3 + 9^3 + 6^3 + 3^3 + 3^3 , and there's another one too;
so 1999 is the sum of 5 perfect cubes. I think there are no numbers that require more than 9 cubes...?

Beth Wilson . . . . . 10 pts - That's the way to explain it! You're a good programmer.
Andy Murdock . . . 7 pts - I think using the same cube twice is completely overhanded!
Greg Hess. . . . . . . . 1 pt - What was your original answer on this one?

Problem #35 - Posted Monday, January 18, 1999
In a 3:00 presentation Mr. Forth gave to less than 104 fifth-graders, there were three boys for every five girls.
At 3:04 some were asleep and some were awake; by 3:05 one-fifth of those awake fell asleep, and one-fourth
of those that had been asleep woke up. There were then as many kids awake as asleep.
How many fifth-graders were there; how many boys and how many girls?

Solution: (by new entrant Charleen Yen)
If a = people awake before; s = people asleep before; g = girls; b = boys;
a + s < 104 ; (4/5)a +(1/4)s = (1/5)a + (3/4)s ; simplify, you got s = (6/5)a
solve for "a" from the two equations, you got a < 47.27
a = 40 ...(select a number less than 47.27); s = 40 (6/5) = 48
Now b : g = 3 : 5 ; so 3 g = 5 b, g = (5/3) b ; g + b = 40+48 ; (5/3)b + b = 88 ;
b = 33 ; g = 55 . . . Ans. Total of 88 fifth graders, 33 boys and 55 girls.

I got LOTS of responses to this question! Is this going to be my life now?
Beth Wilson . . . . . 10 pts - Once again, a clear and correct approach
Andy Murdock . . . 6 pts - Your logic is impeccable, captain, although a bit awrbtstruse.
Charleen Yen. . . . . 5 pts - Nice answer; you're the featured solution!
Jeff O'Connell. . . . 3 pts - Former student makes good (teacher). Show your steps dude!
Tim Nelson . . . . . . 3 pts - You can have all the time you need. Hurry up!
Lisa Cassidy . . . . . 2 pts - Hello there; thanks for entering. Be sure to explain your answer.
Huan-yu Chung . .. 2 pts - Good solution - nice use of common multiple of 8 and 11.
Yu Chen Wang . . . 2 pts - Yu too! Now do your calculus homework. ;-}
Some answers that weren't quite right, but you still get a point this time! (listed in order received)
Erik Thompson . . . 1 pt - Your answer was short but first.
Greg Hess . . . . . . . . 1 pt - And another point for the previous Problem 34.
Sarah DeSanctis. . . 1 pt - Nice to have you back in my on-line (dysfunctional) family!
Matt Whitlow . . . . . 1 pt - Too many kids - if 10 were sick you'd be right.
Daniel Bonilla . . . . .1 pt - You only need 8 kids to be sick.
Erfan Gunawan . . . 1 pt - Good try but your algebra misled you
Mike Birnbaum . . . 1 pt - Two suggested answers, both close.

Problem #36 - Posted Thursday, January 28, 1999
An NBA basketball team has 12 players and the total team salary must be sixty million dollars. If players can earn only 2, 4, 6, or 8 million, in how many "ways" can the 60 million be paid out? List all the ways! (Give the lists of possible salaries, from largest to smallest. For example, one "way" would be 60 = 8+8+8+8+8+8+ 2+2+2+2+2+2.)

Solution: I count 25 different salary distributions. Here they are, in order of # of 8's (Wilson/Bach method)
 8+8+8+8+8+8+2+2+2+2+2+2 8+8+8+8+8+6+4+2+2+2+2+2 8+8+8+8+8+4+4+4+2+2+2+2 8+8+6+6+6+6+6+6+2+2+2+2 8+8+6+6+6+6+6+4+4+2+2+2 8+8+6+6+6+6+4+4+4+4+2+2 8+8+6+6+6+4+4+4+4+4+4+2 8+8+6+6+4+4+4+4+4+4+4+4 8+8+8+8+6+6+6+2+2+2+2+2 8+8+8+8+6+6+4+4+2+2+2+2 8+8+8+8+6+4+4+4+4+2+2+2 8+8+8+8+4+4+4+4+4+4+2+2 8+6+6+6+6+6+6+6+4+2+2+2 8+6+6+6+6+6+6+4+4+4+2+2 8+6+6+6+6+6+4+4+4+4+4+2 8+6+6+6+6+4+4+4+4+4+4+4 8+8+8+6+6+6+6+4+2+2+2+2 8+8+8+6+6+6+4+4+4+2+2+2 8+8+8+6+6+4+4+4+4+4+2+2 8+8+8+6+4+4+4+4+4+4+4+2 8+8+8+4+4+4+4+4+4+4+4+4 6+6+6+6+6+6+6+6+6+2+2+2 6+6+6+6+6+6+6+6+4+4+2+2 6+6+6+6+6+6+6+4+4+4+4+2 6+6+6+6+6+6+4+4+4+4+4+4

Jeff O'Connell solved a sys of 2 eqns in the 4 unknowns a=#2's, b=#4's, c=#6's, d=#8's. I like it!

Again, I got LOTS of responses! Sorry about the delay- my furnace broke and I been COLD!
Jeff O'Connell. . . . 9 pts - Nice method ; officially I wanted the actual sums!
Huan-yu Chung . .. 7 pts - Your first and third e-mails were correct!
Beth Wilson . . . . . .5 pts - Great answer, just not first in my mailbox!
Some of you didn't find all 25, but you still get points! (listed roughly in order received)
Jen-Kai Pan . . . . . . 3 pts - You got all but one, and you were the first response
Greg Hess. . . . . . . . 2 pts - That's most of them. Try to be systematic
Yu Chen Wang . . . 2 pts - You got all but one too!
Andy Murdock. . . . 1 pt - There were smore ways to spend the 60 mil.
Jason McCargar. . . 1 pt - Nice notation 8*3p + 4*9p etc.
Kiyomi Sakata. . . . 1 pt - Thanks for trying; you got most of them.
Matt Whitlow. . . . . 1 pt - Do I have to make a comment for everyone?
Colleen Frye. . . . . . 1 pt - No actual salaries, but three intersting methods!
Utku Aksu. . . . . . . . 1 pt - Keep on entering my humble contest, everyone!

Problem #37 - Posted Saturday, February 6, 1999
A former student of mine mistakenly canceled the 6's and got lucky:
16 - 1
64 - 4 . . . (This says 16/64 = 1/4.)
Find three other two-digit examples like this and at least two three-digit examples of lucky
canceling without using the digit 0. The numbers must have different digits; i.e. 33/33 = 3/3 doesn't count.

Solution: There are three other proper fractions that work:
19/95 = 1/5 , 49/98 = 4/8 , 26/65 = 2/5 . (I like that last one.)
Andy Murdock noticed you can use reciprocals: 64/16 = 4/1 , etc.; not illegal!
Of the three-digit ones, here are a few:
119/595 , 116/464 , 492/984 , 493/986 , 494/988 .
Check back for others or keep suggesting them! ~ ;-}

Andy Murdock. . 10 pts - I wasn't thinking improperly, Dan said fractionally.
Tim Nelson. . . . . . 5 pts - Your answers were legal at the time!
Dave B. . . . . . . . . . 4 pts - Good answers; but I wanted three 2-digit ones.

Problem #38 - Posted Monday, February 15, 1999
a) Choosing four people at random, what is the probability that none of them have their birthdays on the same day of the week?
b) What is the smallest number of (random) people such that there's a better-than-50% probability of at least two of them having the same birthday? (such as Feb. 14 ; not nec. the same year!)

Solution: a) Many of you got this part: The prob. the second person has a different day than the first is 6/7; the prob. the third is diff than the first 2 is 5/7; the prob. the 4th is diff is 4/7; so Prob(all 4 diff) = (1)(6/7)(5/7)(4/7) = 120/343 :=: 0.349854; about 35%.

b) Using the same reasoning, to find the prob. thaty all b'days are diff't, multiply all fractions (365/365)(364/365)(363/365) . . . then see how far to go until the product is less than 1/2.
With n people you mult down to (365 - n + 1) / 365 ; notice that (365/365)(364/365) . . . (344/365) :=: 0.5243 while (365/365)(364/365) . . . (343/365) :=: 0.4927. ; this is for n = 23 people.

The prob that some 2 people of the 23 will have the same b'day is about 50.73%. 30 people: 70.63%; 40: 89.12%; 50: 97.04%. (Good question: what if you allow Feb. 29 in 1/4 as many cases as a usual b'days ? . . .)

Beth Wilson. . . . . . . . 10 pts - Not the first answer, but the best and correctest.
Colleen Frye . . . . . . . . 5 pts - Dangerous rounding in a), evidence destroyed in b)
Yu Chen Wang . . . . . .4 pts - Interesting odds, but your source rounded off the 23.
Huan-Yu/Richy Chang. 3 pts - You were just barely off, another vote for 24.
Zheng Shuuwei. . . . . . 1 pt - It's not as many people as most people think!
Matt Whitlow . . . . . . . 1 pt - You may be surprised it's only 23 people too!
Rebecca Frye . . . . . . . . 1 pt - I heard you helped in an important way!

Problem #39 - Posted Thursday, February 25, 1999
 Holy Triangle! (back to top) If I rearrange the four pieces on the left (base = 5, ht = 13) to make the picture on the right (same base, same ht), there's one empty square! What the hole is going on here? Explain how the same pieces seem to occupy different areas.

Solution: (from new entrant Jeffrey Hunt):
The trick here is that your holy triangle is not a triangle at all. It is some type of four sided
polygon instead. Here is how my logic goes:
First of all I summed the areas of the pieces in the left triangle and came up with an area of 32.
I did the same with the triangle on the left; 32. From this I deduced that the trick wasn't in the shapes
but in the triangular shape overall. What helped me deduce this was that the area of the triangle on
the right and the left one too, should have been 32.5, but the sums of the pieces were 32 and 33,
counting the hole. This led me to check the slopes of the hypotenuese to see if this was a triangle
after all. Bingo, if it was a triangle then the two smaller triangles would have a slope equal to the
one of the hypoteneuse. Not so. (5/2) does not equal (8/3) does not equal (13/5). The result is that
the shape on the left is not the shape on the right, so they don't have the same area.

(Dan's note: The areas of the four pieces are 12, 8, 7, and 5, which adds up to 32. The shapes plus
the hole would make 33. A 5 by 13 right triangle would have area 32.5 , but that's not what we have!)

Huan-Yu/Richy Chang. 10 pts - Good answer to a fun problem. Thanks! ;-}
Andy Murdock. . . . . . . 6 pts - You got somethin' against trick questions?
Yu Chen Wang. . . . . . . 5 pts - Excellent answer, nicely explained.
Jen-Kai Pan. . . . . . . . . . 4 pts - Those angles aren't the same; look again.
Joao Sousa. . . . . . . . . . . 3 pts - Welcome back; yes, it's not a triangle!
Jessica Bubb. . . . . . . . . 3 pts - Right; the slopes are different; hole appears.
Jeffrey Hunt . . . . . . . . . 3 pts - Welcome to the contest; keep it up!
Colleen Frye. . . . . . . . . 1 pt - Now tell your family & friends how it works!
Matt Whitlow. . . . . . . . 1 pt - Keep entering, for fun and profit!
Dave B. . . . . . . . . . . . . . 1 pt - The triangles do all have integer 'legs'...

Problem #40 - Posted Sunday, March 7, 1999
"Twin primes" are prime numbers that are 2 apart (such as 17 and 19).
Consider the set of all pairs of twin primes, including (17, 19).
a) If we pick a pair of twin primes, each less than 100, what is the probability that their sum is divisible by 12?
b) If we pick a pair of twin primes between 10 and 1000, then what's the prob. their sum is div. by 12?

Solution: (Kudos to Richy, Beth, and Jeff H. for their numerically or logically complete answers.)
a) The pairs of twin primes under 100 (not counting the "Siamese twins" 2 and 3) are:
(3,5), (5,7), (11,13), (17,19), (29,31), (41,43), (59,61), (71,73).
7 of 8 of these add up to multiple of 12; in other words their average is a mult. of 6.
So the prob. a randomly picked pair sums to 12n is 7/8 (or .875 or 87.5%).

b) All odd primes are either 1, 3, or 5 more than a mult of 6 : 6n + 1, 6n + 3, or 6n + 5 (let's say 6n - 1).
But 3 goes into 6n + 3 , so a pair of TWIN primes has to add up to (6n - 1) + (6n + 1) = 12 n . Aha!
So the only twin primes that DON'T add up to a multiple of 12 are 3 and 5 . (There are 33 twin prime pairs
between 10 and 1000.) So the probability a pair of twin primes above 10 adds to a mult. of 12 is 1 (100%).

[ Dan's note: It is well-known that there are an infinite number of primes, and that the infinite sum of
(1/p , over all the primes p) is a divergent series. But it is still not known whether the number of twin
primes is infinite, (it's unsolved, a big mystery, up for grabs,....) What they have done is proved that the
sum of (1/p + 1/(p+2) , over all twin prime pairs p, p+2) is a CONVERGENT series, so that doesn't prove
whether or not it's an infinite sum. (If it diverged it would have to be infinite!) ]

Beth Wilson. . . . . . . . . 10 pts - Nice method- I hadn't seen it! (sum is a mult of 4 and a mult of 3.)
Huan-Yu/Richy Chang. . 7 pts - Nice complete list. Didja write a program?
Jeffrey Hunt . . . . . . . . . 5 pts - You were very persistent and covered all the bases!
Dave B. . . . . . . . . . . . . . 4 pts - If you had found and checked them all, you woulda been first!
Colleen Frye. . . . . . . . . 4 pts - Ditto for you. (The fact that 12 has a lot of factors attracts primes nearby!)
Andy Murdock. . . . . . . 3 pts - You got somethin' against trick questions?
Yu Chen Wang. . . . . . . 3 pts - Also needed to check all or prove it. (The average does help.)
Kristy Poindexter. . . . . 2 pts - Thanks for entering! 7 was right but 28 was a bit short... ;)

 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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