Recurrence solution for Tower of Hanoi help?!!?
817569
Is that the game? I had solve it a few time with no problem. So what you need to know?
1157976
0000, 0001, 0002, 0003, 0004, 0005, 0006, 0007, 0008, 0009, 0010, 0011... etc, etc  untill u get to 9998, 9999\n\ntry it out, might take a long time but one combination will work :))
48530
Looking at...\n\n>>>>>>>>> 0\n>>>>>>>>> 1\n>>>>>>>>> 2\n>>>>>> etc.\n>>>> 999999\n\n...in the ones or 10^0 column you repeat\n\n>>>> 0+1+2+3+4+5+6+7+8+9 = 45\n\n100,000 times.  That gives you 4,500,000.\n\nIn the tens or 10^1 column you repeat\n\n>>>> 10*0 + 10*1 + 10*2 + ... + 10*9 = 10*45\n\n10,000 times.  (i.e., group ten 0's in a row, then ten 1's in a row, ..., ten 9's in a row, and repeat that 10,000 times.)  That also gives you 4,500,000.\n\nLikewise with the 10^2, 10^3, 10^4, and 10^5 columns.\n\nSo far that totals 6 times 4,500,000 = 27,000,000.\n\nAll that remains is to add in the 1 from 1,000,000 to get the final total of 27,000,001.\n\nCheck: compare to previous reply. :-)\n\n(How do you get this thing to indent a line?)
970878
OOOOOOOOOh i LUV LUVLUVLUV challenge 24!!!! ok her are some ideas:\n7,2,2,1=(7-1)(2+2)\n6,8,2,1=(8/4)(6*1)              ***All taken from Challenge 24 cards\n3,3,6,2=(3+3)(6-2)                     (single digit)\n9,2,5,1=(9-1)(5-2)\n~* ah what the heck i dont have time to finish this. I'llget back to you later, soon. and for future reference, DO buy challenge 24 cardsand so you can use them often. the only annoying thing is, it doesnt come w/anser key u have to figure it out yourself:(
129538
The next number is 30651.\nIt has a recursion, a_n=a_(n-1)^2+2*a_(n-2).
504740
After several attempts, it seems that it is rather impossible.  If you start in the bottom corner, you can get all but 3 squares rather easily.  If you're careful, you can have 1 remaining if you start diagonal to the corner.\n\nWhat are you trying to do this for?  Is it a website?  I'd love to try it again.
177946
just try this out:\n\nfor(couter=n; counter>=0; counter--)\n{\n   printf("&#92;n %d", counter);\n   sleep(1000); //or might be Sleep(1000) \n   //check out delay() also in place of sleep()\n   //sleep(1) suggests a delay of 1 millisecond\n}\nprintf("&#92;n That is..!");\n\n\n----------\nI think tht shud be simple n sweet. :-)\n\nI think the answerer below really needs to refer back to his expert programming.. as sleep takes milliseconds only.
61458
Your answer is correct, and may be fine for your professor, but for readers who are not already familiar with the problem, both the question and the solution are confusingly worded.  I suggest that in the future, you copy the original question here exactly.\n\n\nI would precede your solution with the following:\n\n\nYou can find the floor in a maximum of 14 attempts.\n\nDrop a bulb from the floors in this order:\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100\n\nIf the bulb breaks, stop and test starting from just above the last floor that it didn't break on, and go up one floor at a time.  If the bulb breaks on the first attempt, resume the test from the first floor.\n\nExamples:\n14 (breaks), 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13\n\n14, 27, 39 (breaks), 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38\n\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99 (breaks), 96, 97, 98\n\n14, 27, 39, 50, 60, 69, 77, 84, 90, 95, 99, 100 (breaks)
408271
take the cube root of 729 u get 9
464622
Use any recursive algorithm.\n\nFor instance:\n\nf(x) = f(x) + 1
849879
here you go...\n(it'll work with up to 255 different disks, if you need more, you'll have to replace chars with ints)\n\nm1(char *a,b) {memove(b+1,b,strlen(b)+1); *b=*a; memmove(a+1,a, strlen(a);}\nm(char *a,b,c) {if (*a)  {m(a+1,b,c);m1(a,c);m(b,c,a);}
631878
If you are only concerned about the ways that you can place the 15 women between the 10 men, then you have 9 positions to place the 15 women.  (You don't have to worry about the arrangements for the men since there has to be one in between there is only one way to arrange.) These are the ways to do it (regardless of order).\n\na) 1 1 1 1 1 1 1 1 7\nb) 1 1 1 1 1 1 1 2 6\nc) 1 1 1 1 1 1 1 3 5\nd) 1 1 1 1 1 1 1 4 4\ne) 1 1 1 1 1 1 2 2 5\nf) 1 1 1 1 1 1 2 3 4\ng) 1 1 1 1 1 1 3 3 3\nh) 1 1 1 1 1 2 2 2 4\ni) 1 1 1 1 1 2 2 3 3\nj) 1 1 1 1 2 2 2 2 3\nk) 1 1 1 2 2 2 2 2 2\n\nNow to rearrange these, you would have to use the multinomial coefficient.  You have to use this because you have repeats: For example in a), you can switch two 1's around and get the same result, but moving the 7 will give you a different result.\n\nFor a), there are 9!/8!1! = 9 ways to arrange.\nFor b) and c), there are 9!/7!1!1! = 72 ways to arrange.\nFor d), 9!/7!2! = 36 ways.\nFor e), 9!/6!2!1! = 252 ways.\nFor f), 9!/6!1!1!1! = 504 ways.\nFor g), 9!/6!3! = 84 ways.\nFor h), 9!/5!3!1! = 504 ways.\nFor i), 9!/5!2!2! = 756 ways.\nFor j), 9!/4!4!1! = 630 ways.\nFor k), 9!/3!6! = 84 ways.\n\nSo in total, there are 9 + 72 + 72 + 36 + 252 + 504 + 504 + 84 + 756 + 630 + 84 = 3003 ways to place the 15 women between the 10 men.\n\nNow if after that you want to know how many ways there are to arrange if you do make a difference between men and between women, then multiply by the number of ways to arrange the men and women.\n\n3003*10!*15! = 14,250,102,106,034,995,200,000 ways.\n\nNotice this is less than the number of ways to arrange all 25 people regardless of gender.  That would have been 25! = 15,511,210,043,330,985,984,000,000.
381290
I have an iterative method that is good for at least 80% of these types of problems.\n\nFirst, write down the series.\n\nNext, above the line you just wrote, write the difference between each pair of elements. In this case, it will be 1, 2, 4, 8, .... In this case, the answer is obvious - they are powers of two. The next one in the differences line should be 16, so the next on the original line should be 15+16=31. This will allow you to get any member in the sequence.\n\nIf the first order of differences doesn't give rise to a discernable pattern, try the differences of differences and so on.
960925
The answer you gave cannot possibly be right. Let n=1. Then 2^n * (n+2)C2 = 2^1 * 3C2 = 6, but [k=1, n]∑k²*nCk = 1² * 1C1 = 1, and 6≠1.\n\nI'll post a proof of the correct answer just as soon as I figure out what it is.\n\nEdit: after much rumination, and many used pieces of scratch paper, I have found a solution. \n\n[k=1, n]∑(k²nCk) = n(n+1)2^(n-2)\n\nIn proving this, I will assume that you are already familiar with the fact that (n+1)Ck = nCk + nC(k-1) and that [k=0, n]∑nCk = 2^n. If not, a good proof of the first is provided in the wikipedia article: http://en.wikipedia.org/wiki/Pascal%27s_rule . The second follows from the fact that this is the sum of the numbers on the nth row of my triangle, and each row has a sum twice that of the row before it (think about how each row is constructed).\n\nOur strategy in this proof will be to find a recurrence relation for each sum, and then solve this relation to obtain a general formula. First we will need to prove a lemma:\n\n[k=0, n]∑knCk = n2^(n-1)\n\nProof:\n\nFirst note that [k=0, 0]∑k0Ck = 0*0C0 = 0. Now we find a recurrence relation as follows:\n\n[k=0, n+1]∑k(n+1)Ck\nExtract the (n+1)th term:\nn+1 + [k=0, n]∑k(n+1)Ck\nExpand the binomial coefficient:\nn+1 + [k=0, n]∑k(nCk + nC(k-1))\nDistribute:\nn+1 + [k=0, n]∑knCk + [k=0, n]∑knC(k-1)\nExtract the 0th term from the right-hand sum (this term is 0):\nn+1 + [k=0, n]∑knCk + [k=1, n]∑knC(k-1)\nChange the indices of summation:\nn+1 + [k=0, n]∑knCk + [k=0, n-1]∑(k+1)nCk\nSince the term we extracted at the beginning is the nth term of the sum on the right, we can merge it into the sum:\n[k=0, n]∑knCk + [k=0, n]∑(k+1)nCk\nDistribute the k+1 term:\n[k=0, n]∑knCk + [k=0, n]∑knCk + [k=0, n]∑nCk\nSimplify:\n2[k=0, n]∑knCk + [k=0, n]∑nCk\nAnd finally:\n2[k=0, n]∑knCk + 2^n\n\nSo we are looking for an f that satisfies the recurrence relation:\n\nf(0)=0\nf(n+1) = 2f(n) + 2^n\n\nApplying this to itself:\n\nf(n+1) = 2(2f(n-1) + 2^(n-1)) + 2^n = 4f(n-1) + 2*2^n\nf(n+1) = 4(2f(n-2) + 2^(n-2)) + 2*2^n = 8f(n-2) + 3*2^n\nExtrapolating this pattern further:\nf(n+1) = 2^(n+1)f(0) + (n+1)2^n\nBut f(0)=0, so this is simply:\nf(n+1) = (n+1)2^n or f(n) = n2^(n-1)\n\nTo ensure our extrapolation is correct, we shall prove it rigorously by induction. First we note that f(0)=0, as it should. To show that n2^(n-1) satisfies f(n+1) = 2f(n) + 2^n in general, we assume it works for some n. Then:\n\nf(n+1) = 2f(n) + 2^n\nf(n+1) = 2n2^(n-1) + 2^n\nf(n+1) = n2^n + 2^n\nf(n+1) = (n+1)2^n\n\nWhich is what we get by plugging n+1 into the formula. So it works for all n.\n\nNow having proved this lemma, we proceed to [k=0, n]∑k²nCk (yes, your original formula had [k=1, n]∑k²nCk, however since the k=0 term is zero, including it does not change the sum). We'll use the function g to avoid confusing it with the previous function. Of course g(0)=0. Now we need our recurrence relation:\n\n[k=0, n+1]∑k²(n+1)Ck\nExtract the (n+1)th term:\n(n+1)² + [k=0, n]∑k²(n+1)Ck\nExpand the binomial coefficient and distribute:\n(n+1)² + [k=0, n]∑k²(nCk + nC(k-1))\n(n+1)² + [k=0, n]∑k²nCk + [k=1, n]∑k²nC(k-1)\nChange the indices of summation:\n(n+1)² + [k=0, n]∑k²nCk + [k=0, n-1]∑(k+1)²nCk\n(n+1)² is the nth term of the sum on the right, so we can merge them:\n[k=0, n]∑k²nCk + [k=0, n]∑(k+1)²nCk\nExpand the binomial:\n[k=0, n]∑k²nCk + [k=0, n]∑(k²+2k+1)nCk\nAnd distribute:\n[k=0, n]∑k²nCk + [k=0, n]∑k²nCk + 2[k=0, n]∑knCk + [k=0, n]∑nCk\nEmploying the formulas for [k=0, n]∑nCk and [k=0, n]∑knCk (the latter of which was proved as our lemma):\n2[k=0, n]∑k²nCk + 2n2^(n-1) + 2^n\nAnd simplifying further:\n2[k=0, n]∑k²nCk + (n+1)2^n\n\nSo g(n+1)=2g(n) + (n+1)2^n. Applying this to itself:\n\ng(n+1) = 2(2g(n-1) + n2^(n-1)) + (n+1)2^n = 4g(n-1) + ((n+1)+n)2^n\ng(n+1)=4(2g(n-2) + (n-1)2^(n-2)) + ((n+1)+n)2^n = 8g(n-2) + ((n+1) + n + (n-1))2^n\nExtrapolating this pattern further:
452780
use some tape and make 4 x's in a box shape with another x in the middle. the hop on one foot hitting all x's as quick as you can moving from bottom x's to top x's then back down and that is one set. then switch feet. i usually time myself and do 5 set continually trying to always beat my time.
977457
1 to 64k.. so there exists about 40 cubes (1^3, 2^3...  39^3, 40^3)..  and sum of 4 cubes. that's like. "40 choose 4" (combinatorics) so that comes out to be about 91390 combinations of 4 cubes and then you just find the cube roots of each. each should have its own cube root. and 91 thousand isn't too bad for an average computer to handle. i suggest you program a computer to do it for you. it wouldn't be hard, just basic programming language.\n\nif you cant do that, then i suggest you try to narrow down what you're looking for...
674697
For five consecutive, there are 34 possibilities because you could start with 1, 2, 3,..., or 34\nYou can't start with 35 because then you would have\n35, 36, 37, 38 which is only four numbers.\nI'm not sure about the other two.\nNote: The answer to the other two are not 35 and 36.