dan's math@home - problem of the week - archives
Problem Archives page 2
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 index
 
11 -Impossible Twins?
12 -The ABC's of Cars
13 - Spider and the Fly
14:Intersecting Bubbles
15 -The Right Numbers
16 -The Magic Number
17 - How Many Ants ?
18-What Time Was It?
19 - World Cup Soccer
20:The Average Speed?
 
 
Problem #11 - Posted Thursday, March 12, 1998
Impossible Twins? (back to top)
Anna, the older twin, was born three hours before her younger sister Lana. On Lana's 21st birthday,
they both went out to a club with some friends. The bouncer was checking ID's and let Lana go in.
But when the older Anna tried to follow her in, the bouncer said, "I'm sorry, your 21st birthday
isn't for another two days." How is this possible? Explain fully and correctly to win the contest.

Solution: (by Andy and Aaron Murdock)
For this problem, the "impossible twins" aren't impossible just improbable. They were both given birth on a boat (or a plane or helicopter, etc, just anything capable of crossing the international dateline). Anna, the older twin, was born on March 1 of a non-leap-year on the Asian side of the International Dateline, three hours later, after they had crossed to the other side of the Dateline, Lana was born, the date being February 28. 21 years later, which happens to be a leap-year, they go to a night club on the 28th of February, Lana's birthday, yet Anna has to wait two days, the 29th and then the 1st.

WINNERS - Problem 11 . . . (back to top) . . . leader board
Andy Murdock - 8 pts - You go Murdox! (but share the 10 winners points)
Peter Williams - 3 pts - Got the key idea in there.
Aaron Murdock - 2 pts - Yes the Int'l Date Line is involved!
Joe Sousa - 2 pts - I'm not sure what an "elliptical year" is; is it a southern hemisphere thing?
 
 
Problem #12 - Posted Wednesday, March 25, 1998
The ABC's of Cars (back to top)
The student parking lot has 81 cars in it; all Acuras, Beetles, and Camrys. There are half as many Acuras as Beetles, and the number of Camrys is 80% of the number of Acuras and Beetles together. How many of each kind of car is in the parking lot? Show all steps to win contest!

Solution: (by Andy M & George C)
Let A = # of Acuras, B = # of Beetles, and C = # of Camrys.
Then A + B + C = 81 , B = 2 A , C = .8 (A + B).
Subbing in gives A + 2 A + .8 (A + 2 A) = 81 , 5.4 A = 81 ,
A = 15 , B = 2 (15) = 30 , C = .8 (45) = 36
Answer: 15 Acuras, 30 Beetles, and 36 Camrys.
 
We now have respondents from as far away as my old stomping grounds of Columbus, Ohio. Welcome!
(I had two requests to make the problems more challenging. So try this week's problem and advertise it to your smart friends!)

WINNERS - Problem 12 . . . (back to top) . . . leader board
Andy Murdock - 9 pts - Define variables carefully
George Caleodis - 6 pts - Perfect solution but I had to fill in a few steps
Joe Sousa - 3 pts - Some good news and some errors; be careful!
 
 
Problem #13 - Posted Monday, April 6, 1998
The Spider and the Fly (back to top)
A spider, in the top-left-front corner of a 10 x 10 x 10 foot room, sees a big fat fly in the bottom-right-back corner. Describe the shortest path, and the length of the path, that the spider can crawl to get the fly. That's crawl, not jump, fly or spider-web express! Your explanation must be clear. (Not affiliated with the squished fly from Problem #2.) ;-}

Solution: Think of unfolding the room and seeing the ceiling and right wall as a 10 x 20 domino. The shortest path is a straight line; the diagonal of this domino, which is -/(10^2 + 20^2) = -/500 = 10-/5 ft , or about 22.36 feet.
Answer: 10-/5 ft , which is 22.36 ft to nearest hundredth.
Flavio's answer:
 
Note: This is a simplified version of a famous problem, in which the room is 12 x 12 x 30 ft, the spider is 1 ft below the ceiling in the middle of one 12 x 12 ft end wall, and the fly is 1 ft above the floor in the middle of the opposite end wall. Now what's the shortest path? There are different ways the story can "unfold" here!

WINNERS - Problem 13 . . . (back to top) . . . leader board
Andy Murdock - 8 pts - Some logical holes
Flavio Cherarola - 6 pts - Enjoyed your proof! Use 10 - x not 10 + x
Kevin Marshall - 4 pts - Hi Kevin! Your intuition was right.
 
 
Problem #14 - Posted Wednesday, April 15, 1998
The Intersecting Bubbles (back to top)
Two overlapping spherical soap bubbles, whose centers are 50 mm apart, have radii 40 mm
and 30 mm. What is the diameter D of their circle of intersection?

Solution: A cross-section picture will reveal a triangle with sides 30, 40, 50 ; the two radii and the distance
between centers. Since this forms a right triangle, drawing in the radius r of the circle of intersection gives
similar triangles and the proportion r/30 = 40/50 , 50 r = 1200 ; r = 24 ; D = 2 r = 48
Answer: The diameter of the circle of intersection is 48 mm.
Note: This is easier to see with a picture, so it's a challenge to express in words. You can send a picture
with your answer as an e-mail attachment; just keep the file size small please.

WINNERS - Problem 14 . . .(back to top) . . . leader board
Andy Murdock - 4 pts - Good start - try similar triangles
Joe Sousa - 3 pts - Define variables carefully
 
 
Problem #15 - Posted Friday, April 24, 1998
The Right Numbers (back to top)
(a) The area and volume of a certain sphere are both 4-digit integers times Pi. What is the radius?
(b) The integers 1, 3, 8, and N have the property that the product of any two, when added to 1,
gives a perfect square. What is the smallest positive integer value of N?

Solution: Largely by Andy Murdock, smallly by Dan Bach
(a) A = 4 Pi r^2 , and V = (4/3) Pi r^3 ; so 1000 <= 4 r^2 < 10000 and 1000 <= (4/3) r^3 < 10000 ; 250 <= r^2 < 2500 means 15.8 < r < 50 and 750 <= r^3 < 7500 gives 9.08 < r < 19.57, so for both we need r = 16, 17, 18, or 19.
But for (4/3) r^3 to be an integer, r must be a multiple of 3, and so r = 18 units.
(b) 1*N + 1 is to be a perfect square, so N must be a^2 - 1. We try N = 2^2 - 1 = 3 , but 3*3 + 1 = 10 =/= square. If N = 8, 8*8+1 = 65 =/= square. If N = 15, then 1*15 + 1 = 16, 8*15 + 1 = 121, but 3*15 + 1 = 46 =/= square. N = 24 is no good since 24*3 + 1 = 73 ; N = 35 , N = 48 , N = 63 , N = 80, and N = 99 are all no good, but N = 11^2 - 1 = 120 works:
1*120 + 1 = 121 = 11^2 ; 3*120 + 1 = 361 = 19^2 ; 8*120 + 1 = 961 = 31^2. Yay-us.

WINNERS - Problem 15 . . .(back to top) . . . leader board
Andy Murdock - 10 pts - Good job, summarized steps well
Joe Sousa - 4 pts - Didn't mean area = volume; 2nd part ok.
Problem #16 - Posted Tuesday, May 5, 1998
The Magic Number (back to top)
A certain six-digit number is split into two parts; the first three digits and the last three digits
are added (as 3-digit numbers), the resulting sum is squared, and the answer is the original 6-digit
number! What was the number? (There might be more than one answer!)

Solution: (First number submitted by Andy Mudrock):
998 + 001 = 999 and 999^2 = 998,001. Now is 001 a 3-digit number? That's a pretty good answer, but we'd
have to wonder how to mane everything, like three = 00000003 .
The other answer (more legit, possibly) is 494,209: 494 + 209 = 703 and 703^2 = 494,209.
To find something like this you could count backwards and find the first one, and for the second one you
could write a simple program on a TI-8x calculator, or take my MATHEMATICA class at DVC in September!
E-mail me any time if you have a good way to hunt that yields 494,209.

WINNERS - Problem 16 . . . (back to top) . . . leader board
Andy Murdock - 6 pts - Good reasoning to limit sum between 316 and 999.
 
 
Problem #17 - Posted Friday, May 15, 1998
How Many Ants? (back to top)
At least a dozen ants are marching through my kitchen!
If the ants walk in rows of 7, 11, or 13, there are 2 ants left over,
while in rows of 10, there are 6 left over.
What is the smallest number of ants there could be?

Solution: from Andy M.
One thing to notice is that 7 * 11 * 13 = 1001, so the number can be (and must be) 2 more than a multiple of 1001.
This gives 1003, 2004, 3005, 4006. The last one is the first one that's 6 more than a multiple of 10.
Therefore there were 4006 ants (and one uncle, if you count me!).

WINNERS - Problem 17 . . . (back to top) . . . leader board
Andy Murdock - 10 pts - Yes, 1001 = 7 * 11 * 13 ; see note below.
Sarah DeSanctis - 3 pts - slight miscount, but 366 works for 7, 10, and 13.
 
NOTE: A fun trick is to get someone to use a secret 3-digit number; have them multiply it by 7,
then the answer by 11, then that answer by 13. They tell you the result, and you "guess" the original number!
 
 
Problem #18 - Posted Monday, May 25, 1998
What Time Was It? (back to top)
A basketball playoff game started between 3pm and 4pm, and ended between 6pm and 7pm.
The positions of the minute hand and the hour hand were reversed at the end of the game,
compared to the beginning. What was the exact time the game started and ended,
and how long was the game? (Try to give exact times, not rounded to the nearest anything.)

Solution: (by Dan Bach; no contestants invoked the power of algebra!)
 
Ok, we know the game started a little after 3:30, since the minute hand was between 6 and 7. Let's make up a variable: let "t" be the number of minutes after 3:00 ; we figure 30 < t < 35. One minute of time is 360/60 = 6 degrees of angle for the minute hand, and 1/12 of that, 1/2 degree, for the hour hand. So at the start, while the minute hand is at 6 t degrees, the hour hand starts at 3:00 (90 degrees) and after t min is at 90 + t/2 degrees.
Now at the end of the game, a little after 6:15, the minute hand must be at 90 + t/2 degrees, which is
(90 n+ t/2) / 6 = 15 + t/12 min past 6, causing the hour hand to move
(90 + t/2) / 12 = 15/2 + t/24 degrees past 6, which has to match up with the original 6 t degrees:
180 + 15/2 + t/24 = 6 t ; mult by 24 and solving gives
4320 + 90 + t = 144 t ; 4410 = 143 t ;
. . Starting time = t = 4410 / 143 = 30 120/143 min past 3 , or 3:30 120/143 , approx. 3:30.83916;
. . Ending time = 15 + t/12 = 15 + 2205 / 1716 = 16 489/1716 min past 6 , approx 6:16:284965.
. . Time of game = 2 1/2 hrs + 16 489/1716 min - 120/143 min = 2 hrs 45 765/1716 min.
 
Approx values in decimal seconds are :
Start = 3:30:50.35 , End = 6:16:17.10 , Duration = 2:45:26.75.

WINNERS - Problem 18 . . . (back to top) . . . leader board
Andy Murdock - 4 pts - Your predictor / corrector method is very good!
Greg Vitan - 3 pts - pretty close for a "hands-on" approach!
 
 
Problem #19 - Posted Tuesday, July 7, 1998 (back from vacation!)
World Cup Soccer Standings (back to top)
In this year's Coupe du Monde 98, there are 4 teams in each Group, and they each play each of the other 3 teams once. Here are the final "Pts standings" of Groupe C, with the W . L . T . PTS records (a win is 3 pts and a tie is 1 pt):
France. . . . . . . 3 . . 0 . . 0 . . 9
Denmark . . . . 1 . . 1 . . 1 . . 4
South Africa . .0 . . 1 . . 2 . . 2
Saudi Arabia. . 0 . . 2 . . 1 . . 1
How many "Pts standings" are possible, and are any gettable in more than one way? (not counting order or particular teams) This one would be called 9-4-2-1.

Partial Solution: By Andy Murdock -
For the number of total possible combinations, I got 29 -
If no teams tie, then all scores have to be multiples of three with a sum of 18 total:
9-6-3-0, 9-3-3-3, 6-6-6-0, 6-6-3-3
If there is only one tie, the sum is now 17 and each must have 2 occurrences
of 1,4,or 7 (each being a 1+a multiple of 3):
9-6-1-1, 9-4-4-0, 9-4-3-1, 7-7-3-0, 7-6-4-0, 7-6-3-1, 7-4-3-3
With two ties, the sum is now 16, and similar to the last case where there
were 2 cases of one being added to a multiple of three, it happens four times here:
9-4-2-1, 7-5-4-0, 7-5-3-1, 7-6-2-1, 6-5-4-1, 6-4-4-2, 5-4-4-3
With three ties the logic still follows and the sum is 15:
9-2-2-2, 7-5-2-1, 7-4-2-2, 7-4-3-1, 6-5-2-2, 5-5-4-1, 4-4-4-3
With 4 ties, the # of possible outcomes drops sharply: 7-3-2-2, 5-4-3-2
With 5 ties: 5-3-3-2 , . And with all six ties: 3-3-3-3
Making a total of 29 possible outcomes for El Cupo Mundial Roundo Primero.
 
Dan's note: I have 20 confirmed. I have verified the ones in red;
I'm not convinced of the existence of the rest, or if they exist, I havent found them yet.
I also find 6-6-4-1, 6-4-4-3 (one tie), 5-5-3-2, 5-4-4-2, 5-5-5-0 (three ties)

WINNERS - Problem 19 . . . (back to top) . . . leader board
Andy Murdock - 3 pts - You got sum answers but are they possible?
 
 
Problem #20 - Posted Sunday, July 26, 1998
What's The Average Speed? (back to top)
a) Aaron rides his bicycle at 20 kph (kilometers per hour) to his job, and 30 kph back home along the same road. What was his average speed for the round trip?
b) Andy walks 3 mph to his class. How fast does he have to run back home in order to average 6 mph for the round trip? . . . Please show your steps and reasoning.

Solution: HINT : a) not 25; . . b) not 9. . .
You can't just average the speeds, because you spend longer going at the slower rate.
Here's a good solution, submitted by Andy Murdock:
 
a) 24 kph average speed (if it was average velocity it would be entirely different). It
doesn't matter what the distance is as long as it is the same both ways. If the road is
60 km long, it would take him 3 hrs one way 2 hrs the other for a total of 5 hrs over
120 km, or 24 kph.
(Bach's note: Andy picked a convenient number; 60.
Can any of you redo the problem with a general distance d ?)
 
b) He would have to be infinitely fast. This is true no matter what speed he was travelling.
Anyone travelling x mph over a road nx long, would take n hours one direction. To
average 2x, over 2nx (the total road length), 2nx/n+?=2x, ? must equal zero, so it took
him zero hours, or he went infinitely fast.
(Bach's note: So in other words, it's already too late.)

WINNERS - Problem 20 . . . (back to top) . . . leader board
Andy Murdock - 10 pts - Nice explanation! (see above)
C R M . . . . . . . . - 3 pts - (afraid you guys fell for . . .
Trevor Bird . . . .- 2 pts - . . . the speed-averaging trap.)
 
 THANKS to all of you who have entered, or even just clicked and looked.
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YOU CAN ALWAYS FIND ME AT dansmath.com - Dan the Man Bach - 3*23*29 A.D.
 
Problem Archives Index
 
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