Easy math puzzles
Poll
 | 25 votes (49.01%) | ||
| | 16 votes (31.37%) | ||
| | 7 votes (13.72%) | ||
| | 4 votes (7.84%) | ||
| | 12 votes (23.52%) | ||
| | 3 votes (5.88%) | ||
| | 6 votes (11.76%) | ||
| | 5 votes (9.8%) | ||
| | 12 votes (23.52%) | ||
| | 10 votes (19.6%) |
51 members have voted
There are only seven partitions of six using 1-3 digits:Quote: WizardHere is something a bit easier.
How many ways are there to put six distinct balls into three identical boxes?
Is there a general formula for n distinct balls into three identical boxes?
link to original post
6
15
24
33
114
123
222
It's easy to calculate the number of combinations for any of them. For instance, there are combin(6,1) * combin(5,2) * combin(3,3) = 60 combinations of 123. The combinations of all seven partitions sum to the answer of 122 combinations for boxes containing 0-6 balls.
If all boxes must contain at least one ball, then we subtract the 32 combinations of 6, 15, 24 and 33 to get the answer of 90 combinations
Incidentally, regarding the “random board cuts” problem: You guys analyzed it for days, eventually concluding there was no direct solution for n cuts, then I post the direct solution using nothing but basic combinatorics/calculus, and apparently no one cares?
Quote: Ace2The combinations of all seven partitions sum to the answer of 122 combinations for boxes containing 0-6 balls.
I agree!
Sorry for the tardy response.
Quote:
Incidentally, regarding the “random board cuts” problem: You guys analyzed it for days, eventually concluding there was no direct solution for n cuts, then I post the direct solution using nothing but basic combinatorics/calculus, and apparently no one cares?
link to original post
Sorry but I was out of town for almost a month and didn't see it. I'll have a look.
Quote: Ace2
Let's say you have 100 baseballs lined up on the ground and you randomly divide them into 5 groups (a group can contain 0-100 balls). There are combin(101,4) ways to insert 4 dividers among 101 positions to create the 5 groups. If you want to ensure all 5 groups contain at least 3 balls, then you remove 5*3 balls, then randomly insert 4 dividers into the remaining 85 balls ((combin(86,4) ways), then add 3 balls back to each group. Therefore, the probability that each of the 5 random groups contains at least 3 balls is 86*85*84*83/(101*100*99*98). As the total number of balls (n) goes to infinity, the probability that each group contains at least 0.03n balls is obviously 0.85^4. Therefore, as n goes to infinity, the probability that all 5 groups contain least xn balls is (1 - 5x)^4
So if we make 4 random cuts into a 1 foot long board, we know that the probability of all 5 pieces being at least x long is (1 - 5x)^4. Now we apply the very useful property that the expected value of an event to happen is the cumulative probability of it never happening. The "event" is finding a piece shorter than x. For this problem, this means: add the probabilities that all pieces are longer than x for all x from 0 to 0.20 (obviously all 5 pieces can't be longer than 0.20). We get this sum by integrating (1 - 5x)^4 dx from 0 to 0.20, which gives us the answer of 1/25 feet =expected length of shortest piece
Putting it all together, the expected length of the shortest piece when making (s - 1) cuts into a 1 foot board to get (s) pieces is the integral from 0 to 1/s of (1 - sx)^(s-1) dx, which evaluates to 1/s^2. So, for example, if we are cutting the board into s=8 pieces, the expected length of the shortest piece is 1/8^2
Note: the probability of all segments having a minimum size can be explained without the discrete baseball example as follows. But IMO, the discrete example is more intuitive. Starting with a board 1 foot long, you want to make 4 random cuts into 5 pieces with a minimum length of 0.03. You accomplish this by drawing a line at length 5 * 0.03 =0.15, then make the 4 random cuts irrespective of the line. If they all fall on the 0.85 side then subdivide the 0.15 side into 5 pieces of 0.03 and fasten one of them to each of the 5 pieces cut on the 0.85 side. So there's a (1-0.15)^4 chance they all fall on the 0.85 side and no pieces are <0.03
link to original post
This is really good. Thank you. I think I'll make an Ask the Wizard question out of it. Sorry it took so long to reply. I was out of town.
You take a 100 foot board, make one random cut and keep the shorter piece. Then you take another 100 foot board, make two random cuts and keep the shortest piece. Then you take another 100 foot board, make three random cuts and keep the shortest piece. This pattern continues forever.
What's the expected length of all the shortest pieces combined?
Quote: Ace2Now for a truly easy math puzzle.
You take a 100 foot board, make one random cut and keep the shorter piece. Then you take another 100 foot board, make two random cuts and keep the shortest piece. Then you take another 100 foot board, make three random cuts and keep the shortest piece. This pattern continues forever.
What's the expected length of all the shortest pieces combined?
link to original post
The expected length of cutting a 100-foot board into N pieces is 100 / N^2, so the sum is 100 / 2^2 + 100 / 3^2 + 100 / 4^2 + ...
= 100 (1 / 2^2 + 1 / 3^2 + 1 / 4^2 + ...)
= 100 ((1 + 1 / 2^2 + 1 / 3^2 + 1 / 4^2 + ...) - 1)
= 100 (PI^2 / 6 - 1)
= 50 PI^2 / 3 - 100, or about 64.4934.
Height = 2 + sqrt(24) / 3
Consider a tetrahedron with vertices (1,1,0), (1,0,1), (0,1,1), and (0,0,0).
The coordinates of the center of a face of the tetrahedron are the average of coordinates of its vertices. The average of (1,1,0), (1,0,1), and (0,1,1) is (2/3, 2/3, 2/3). The distance between this center and (0,0,0) is sqrt(4/9 + 4/9 + 4/9), which is sqrt(12) / 3.
Place a sphere at each vertex. Touching spheres would have radii of sqrt (2) / 2. The height of the group of spheres would be sqrt(12) / 3 + sqrt(2).
Scale the ball pyramid height such that the radii are 1 instead of sqrt (2) / 2. The object’s height would then be sqrt (24) / 3 + 2.
$40,000 in cash today or $40,000 financed over five years at 12% APR compounded CONTINUOUSLY. So there would be 5*12 = 60 equal payments (first payment due one month from today).
If you decide to finance, how much would the monthly payment be?
If the APR is 12%, then the amount the principle is multiplied by each time for monthly payments is 1.12^(1/12).
Let I = 1.12^(1/12), P = 50,000, and N = 60.
Applying the interest/loan formula P I^N = M (I^N - 1) / (I - 1), rewritten as M = P I^N (I - 1) / (I^N - 1), we get the monthly payment = $1096.79.
Of course, if it's me, I pay the $40,000 up front. I haven't made payments on a car since 1988.
Also, doesn't "APR" always mean that the interest is compounded continuously? A 12% annual rate, compounded monthly, has an APR of about 12.6825%.
Quote: Ace2If you decide to finance, how much would the monthly payment be?
link to original post
If the interest is 12% per year, then it's 1% per month, so it's .001% per 1/1000th month, etc.
As a comparison if 12% were applied each year and you never paid anything back, then on $40k after 5 years I get it you owe $70,493.67 (1.12^5*40k). Whereas the 1/1000th method gives it at 72,884.53.
APR means annual percentage rate. The compounding interval is another variable. A normal car/house loan is paid back monthly and also compounded monthly. So for a 12% APR, 12% / 12 months per year = 1% of interest is added to the outstanding balance every month. This would be an APY (annual percentage yield) of 1.01^12 - 1 =~ 12.68%. Compounded daily would be an APY of (1 + .12/365)^365 - 1 =~ 12.75%. For this problem, interest is compounded CONTINUOUSLY.Quote: ThatDonGuy
If the APR is 12%, then the amount the principle is multiplied by each time for monthly payments is 1.12^(1/12).
Let I = 1.12^(1/12), P = 50,000, and N = 60.
Applying the interest/loan formula P I^N = M (I^N - 1) / (I - 1), rewritten as M = P I^N (I - 1) / (I^N - 1), we get the monthly payment = $1096.79.
Of course, if it's me, I pay the $40,000 up front. I haven't made payments on a car since 1988.
Also, doesn't "APR" always mean that the interest is compounded continuously? A 12% annual rate, compounded monthly, has an APR of about 12.6825%.
link to original post
Your answer is for an APR of (1.12^(1/12) - 1) * 12 =~ 11.39% compounded MONTHLY for an APY of 12%, applied to a $50,000 loan instead of $40,000. Disagree
A few dollars off. DisagreeQuote: WizardQuote: Ace2If you decide to finance, how much would the monthly payment be?
link to original post$905.01
link to original post
You have the right idea and are within a penny of the answer. I verified my answer by compounding one million times per month in excelQuote: charliepatrickUsing a spreadsheet I get about $890.99. This was using the assumption where you're initially charged 40c on $40,000 for the first 1/1000th of a month, and similar logic thereafter.
If the interest is 12% per year, then it's 1% per month, so it's .001% per 1/1000th month, etc.
As a comparison if 12% were applied each year and you never paid anything back, then on $40k after 5 years I get it you owe $70,493.67 (1.12^5*40k). Whereas the 1/1000th method gives it at 72,884.53.
link to original post
For full credit, you must give an exact expression of the answer. It cannot be answered exactly to two decimal places.
History and Hint: The mathematical constant e was discovered by Bernoulli in 1683 while studying compound interest
Agree. 🍺Quote: ThatDonGuyQuote: Ace2Now for a truly easy math puzzle.
You take a 100 foot board, make one random cut and keep the shorter piece. Then you take another 100 foot board, make two random cuts and keep the shortest piece. Then you take another 100 foot board, make three random cuts and keep the shortest piece. This pattern continues forever.
What's the expected length of all the shortest pieces combined?
link to original post
The expected length of cutting a 100-foot board into N pieces is 100 / N^2, so the sum is 100 / 2^2 + 100 / 3^2 + 100 / 4^2 + ...
= 100 (1 / 2^2 + 1 / 3^2 + 1 / 4^2 + ...)
= 100 ((1 + 1 / 2^2 + 1 / 3^2 + 1 / 4^2 + ...) - 1)
= 100 (PI^2 / 6 - 1)
= 50 PI^2 / 3 - 100, or about 64.4934.
link to original post
I’m still amazed that the solution to this problem uses pi.
