Sorry all.Amber G. wrote: First, if a person was born in 2001, the sum will be 11 and not 111 ...
But if a person was born in 19xx .. big deal? x+(111-x) is always 111.
(Next year this will be 112 .. and so on..)
Suitably contrite.
Sorry all.Amber G. wrote: First, if a person was born in 2001, the sum will be 11 and not 111 ...
But if a person was born in 19xx .. big deal? x+(111-x) is always 111.
(Next year this will be 112 .. and so on..)
Hello,matrimc wrote: Given a number n and k, how many different k-dimensional boxes of volume n with integral sides can be got? Assume that the sides of the boxes are parallel to the coordinate axes.
Better term for 'side' probably is 'edge'. 'Integral' because we want to limit the length of the edge to be a member of the set of natural numbers.Amber G. wrote:Hello,matrimc wrote: Given a number n and k, how many different k-dimensional boxes of volume n with integral sides can be got? Assume that the sides of the boxes are parallel to the coordinate axes.
Can you define the problem clearly please. What is a k-dimensional box? What is meant by integral side? What is a side?
Or by box, you mean, say for 2 dimensional, a rectangle of side a, and b, where n=ab, (which just means how many ways you can factor n)..( and for 3 dimensional space, how many ways you can write n as product of 3 factors?)?
Amber JiAmber G. wrote:In my opinion, It does not matter much what you use, "side" or "edge", both are vague, unless you define what you mean by "volume", "box" etc.. in k-dimension..Also meaning of "different" (as in - is 1 x 3 box different than 3 x 1 box..In three dim lot of people will call a tall building "different" than a long building while criteria may be different for Lego blocks where one can easily rotate a block..).. FWIW, unless one knows the audience very well, at least in math, one should use a language which is more precise...
Can this problem be restated, such as :
How many ways a positive integer n, can be factored in k factors?
Or am I missing something? (Also, does order count here? - if not, one ought to restate that explicitly).
Thanks.
Code: Select all
#!/usr/bin/env python
# Note I'm using python 3. For python 2.xx users, you may need to remove the parens around the print statements.
from decimal import Decimal, getcontext
x = [0] * 2017
getcontext().prec = 1000
x[1] = Decimal('1')
x[2] = Decimal.sqrt(Decimal('1') / Decimal('3'))
for n in range(1, 2015):
x[n + 2] = (x[n+1] + x[n]) / (Decimal(1) - x[n] * x[n+1])
# Print last few numbers
for n in range(2001, 2016):
print(n, " => ", x[n])
# Print x[2011]
print(x[2011])
Nice computer assited pattern recognition. This is how computers can help - similar to in Physics natural phenomenon obeserved/measured leading to theories or hypothesis by theoretical physicist-conducting a repeatable experiment-experimental Physicist gets a nobel prize for validating/invalidating the hypothesis-theoretical physicist might get a nobel prize cycle.ArmenT wrote:^^^
tan(a + b) = (tan(a) + tan(b)) / (1 - tan(a) * tan(b))
This looks a heck of a lot like your formula, where x[n] = tan(a) and x[n+1] = tan(b) and x[n+2] = tan(a + b). Now the rest of it is pretty easy to figure out
There are n weights. Their values are (1,2,4,8,....2^(n-1)). You are also given a balance. You are to place each of the weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step you choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.
Determine the number of ways in which this can be done.
Ah, I see. I was counting 1 & 2 on the left pan as the same as 2 & 1 on the left pan, in which case my formula works out.Amber G. wrote:^^^
When n=2, answer is 3, (If you first put '1' (has to be on left), '2' has to be on left, but if you first put '2' on left, now 1 can be on placed on either side -- total 3 ways)
ArmenT: Thanks but I think the answer is still incorrect. Hint for n=1, answer is 2, n=2, answer is 3 (not 2) Remember, that order does matter. For example, in case of n=3 (weights 1,2,4) if you start with '4' (which has to be placed on left) ..there are still 8 ways you can put the rest (1, and 2) (1(L)2(L), 1(L),2(R),1(R)2(L), 1(R),2(R) and 4 more when you start putting 2 first then 1 )ArmenT wrote:Ah, I see. I was counting 1 & 2 on the left pan as the same as 2 & 1 on the left pan, in which case my formula works out.Amber G. wrote:^^^
When n=2, answer is 3, (If you first put '1' (has to be on left), '2' has to be on left, but if you first put '2' on left, now 1 can be on placed on either side -- total 3 ways)
Whoops, also screwed my logic a bit. I noticed the weights were from (1, 2, 4...2^(n-1) and I went off-by-one on the factorial bit. Should have been (n-1)! + 1
In case you're wondering about my reasoning, I worked like this:
1. Once n'th weight is on the left pan, remaining (n - 1) weights can be placed in any combination on either the right or left pan (it doesn't matter where they go, since their total can never exceed 2^(n - 1).
2. Ergo this means that the remaining (n-1) weights can be arranged in (n-1)! ways (assuming that order doesn't matter and there's at least one weight on the right pan)
3. We also have the case where all weights are on the left pan. Therefore we add 1 more to the answer.
4. Hence (n-1)! + 1.
Brilliant!gakakkad wrote:AMBER SIR, heres solution for first one. h&d blows to us apduls when high school kids do these more efficiently than us But its true . I was far better when I was in school than now. Maybe because I took up medicine. But i try to keep my maths alive.
there are only 4 good pairs possible. Let {a1,a2,a3,a4} be a set . Substitute a1=1,a2,=f-1,a3=v-1,a4=e-1 where we assume that these are edge sides and vertices of a regular polyhedron without holes (platonic solid) , let G= A1+A2+A3+A4. detailed solution as soon as i have access to scanner. feeling lazy to type , But briefly G = V+E+F -2. nOW EULARS formula is v-e+f=2 adding 2e to both sides of eulars formula .we get g=2e. Now a1+a2=f. , a1+a3=v ,for a solid pf=2e=qv where p = the number of edges of each face,q = the number of faces meeting at each vertex but g=2e ,ergo (a1+a2)|g ,(a1+a3)|g. a1+a4 = e which obviosly is divisible by 2e, a3+a4 can easily be proved to be not divisible,a2+a3=e. so we have four possible good pairs. {1,5,7,11} is one pair a1+a2 , a1+a3 etc are divisible but a3+a4 are not. other pair is {1,11,19,29}which is similar divisibility prop to previous one. QED
Here is one solution.. Stop reading further, if you are working on it.Amber G. wrote:Here is a problem from today's International Math Olympiad.
(There were 3 problems - 4.5 hours)
...
There are 4 numbers, all positive integers, and all distinct (no two are equal). Let there sum be S. Now you take a pair (two numbers) from these four and add them , if this sum divides S it is a "good pair". Choose the four numbers such that you can get maximum number of "good pairs" .
<snip>
.
The answer is none. a) The answer is close to 3.1035..*10^64svenkat wrote: 2)1! + 2! +... + 50!=?
a)3.1035*10^64 b)2.1021*10^65 c)3.1035*10^63 d)3.1035*10^62
Again, with multiple choice, one can easily rule out 27, or 0. 18 is easily reachable (select 9,9,9). So to reach 20 is trivial. ( Alok chooses 7, if it is Z one can easily add 18 and reaches 20. If it is X, then select 5(4 will work too), so net gain could be (9-5) or more than 5.. In either case you can reach 20.svenkat wrote:The following are CAT(MBA giri) questions.But I find them sufficiently interesting to post them here.
Re: Alok and Bhanu play the following min-max game. Given the expression N = 9 + X + Y - Z Where X, Y and Z are variables representing single digits (0 to 9), Alok would like to maximize N while Bhanu would like to minimize it. Towards this end, Alok chooses a single digit number and Bhanu substitutes this for a variable of her choice (X, Y or Z). Alok then chooses the next value and Bhanu, the variable to substitute the value. Finally Alok proposes the value for the remaining variable. Assuming both play to their optimal strategies, the value of N at the end of the game would be one of the following 27, 0.0 ,20 ,18
Same problem with N = 12 + X*(Y - Z)
Going through archives to find one unsolved problem I could solve .Amber G wrote:A ticket for the cinema costs Rs 1 (well it's an old problem when ticket did cost 1 Rs), There are n people in the line, each person has either Rs1 note(bill) of Rs 2 note, the cashier has no chance when he opened the window, how many ways these people have the notes so that cashier will not run out of change) .(For example for n=3, (1,1,1) is ok so is (1,2,1) but nothing else - 2,1,1 is not possible because for the first person cashier can't give change)
If order does not matter, it should be 2^(n-1), not (n-1)! + 1. But since order matters ...ArmenT wrote:Ah, I see. I was counting 1 & 2 on the left pan as the same as 2 & 1 on the left pan, in which case my formula works out.Amber G. wrote:^^^
When n=2, answer is 3, (If you first put '1' (has to be on left), '2' has to be on left, but if you first put '2' on left, now 1 can be on placed on either side -- total 3 ways)
Whoops, also screwed my logic a bit. I noticed the weights were from (1, 2, 4...2^(n-1) and I went off-by-one on the factorial bit. Should have been (n-1)! + 1
In case you're wondering about my reasoning, I worked like this:
1. Once n'th weight is on the left pan, remaining (n - 1) weights can be placed in any combination on either the right or left pan (it doesn't matter where they go, since their total can never exceed 2^(n - 1).
2. Ergo this means that the remaining (n-1) weights can be arranged in (n-1)! ways (assuming that order doesn't matter and there's at least one weight on the right pan)
3. We also have the case where all weights are on the left pan. Therefore we add 1 more to the answer.
4. Hence (n-1)! + 1.
If order matters ....skumar wrote:If order does not matter, it should be 2^(n-1), not (n-1)! + 1. But since order matters ...
Only if scans of 'rough sheets' are not archived. Just joking!skumar wrote: Do the kids get points for finding the solution without knowing how to get there?
Thanks, yes I was sloppy . (1,1,2) is ok.skumar wrote: <snip above post >
It also highlights one error in example given, for n=3, (1,1,1), (1,2,1) & (1,1,2) are ok.
skumar wrote:
learnt about double factorial, so f(n) = (2n-1)!! = (2n)! / (n!)(2^n)
Do the kids get points for finding the solution without knowing how to get there?