| Author | Message | ||
Frodo Junior Artist Username: Frodo Post Number: 423 Registered: 11-2009 Posted From: 219.64.65.109 Rating: |
Calvin : good one. the key thing to realise is that for a given set of 6 digits, there is only one way of arranging them in a descending order or non-descending order. same holds true obviously for ascending or non-ascending. anyway for non descending, no repetition of digits allowed, so it is just choosing 6 digits out of 10, so = 210 (you dont have to worry about digits numbers starting with zero, because descending order means zero will always end at the units place. for descending, the way to pick 6 digits out of 10 possible from 0-9 to form a 6digited number with any number of repitions, is given by what vjavasi was suggesting : let x0 be no. of zeros and x1 be no of 1s and so on then the answer for above is given by the no. of non-negative integer solutions to x0+x1+x2.....+x9 = 6. the answer for that is 15C6. vjavasi @ 10:31: the answer to that is (n+m-1)C m . if you are interested that is an interesting problem to solve too. Calvin: in the problem it states the guy is stuck in the 1st ditch. so F(2) = 1, F(3) = 2 and we need to find F(100) which equals fib(100). if you choose to call the 1st ditch 0th, then since there are 100 ditches only, you have to find F(99). anyways it is just a matter of semantics, i guess Dal tadka, dum-biryani and Deccan chargers - always no.1  | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2289 Registered: 11-2009 Posted From: 117.195.206.153 Rating: N/A |
excellent bro....you got it | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2288 Registered: 11-2009 Posted From: 117.195.206.153 Rating: N/A |
sorry it's 10C1+5*10C2+7*10C3+8*10C4+5*10C5+10C6 -1 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2287 Registered: 11-2009 Posted From: 117.195.206.153 Rating: N/A |
is it 10*10C1+5*10C2+7*10C3+8*10C4+5*10C5+10C6 -1 i am trying to evaluate possible six digit number combinations using 1,2,3,4,5,6 digits out of ten...one is deducted to remove 6 digit number formed by zero(oooooo)... if there is a general formula to get number of possible integer solutions for an equation of form x1+x2+.....+xn = m then we can generalize the above solution I have to catch flight to US, i will work on this later | ||
Calvin Junior Artist Username: Calvin Post Number: 39 Registered: 03-2008 Posted From: 66.129.224.36 Rating: |
Looks like f(k) = 5c(6-k) First lets see strictly descending - selection of 6 digits from 10, and arrange them in any order. There is only arrangement possible that causes it to have strictly descending order for these 6 digits. From this order, if we swap any 2 places to create a new number, its no longer strictly descending order.. so the answer for that is 10c6 Now extending that to descending- where a(i) >= a(i+1) this can also be written as a(i) = a(i+1) = a(i+2) ... = a(i+k-1) > a(i+k) that is if you pick unique digits in the number, they will always be in strict descending order. So in descending - the number of unique digits in the number can be from 6 to 1. When the unique digits is 6, its case 1. 10c6 Now create 6 digit numbers with 5 unique digits. select 5 digits - 10c5. Now we need to fill the position of 6th digit which is going to be a duplicate of one of the 5 digits selected.. for example - 654322 or 654432 or 654332.. the duplicate can occur at any place from 1 to 6. Now there is one place to fill with a digit selecting from 5= 5c1 therefore the number of 6-digit numbers with 5 unique digits = 5*10c5 now for 4 digits = pick 4 unique digits , now we have 2 places to fill. This is where I got wrong earlier.. This again recursively - (not sure if it can be done much simpler way) in the 2 places, there can be 1 digit or 2 digits... this can be done in 4c1 + 4c2 = 5c2 For 3 unique digits - 3 duplicate places to fill - 3c1 + 3c2* 2c1 + 3c3 = 5c3 For 2 unique digits - 4 duplicate places to fill - 2c1 *1c1*1c1*1c1 + 2c2 * (2c1+2c2) = 5c4 for 1 unique digit - 5 duplicate places to fill - 1c1 * 1c1*1c1*1c1*1c1 = 5c5 F(2) = 2, considering he can reach 2nd ditch in 2 1-hops or 1- 2hops if he starts from 0th ditch (no ditch) | ||
New_user Side Hero Username: New_user Post Number: 9918 Registered: 02-2008 Posted From: 174.136.55.4 Rating: N/A |
At any given ditch, other than 100th, only 2 choices untayi. So answer 99x2 + 1 choice at 100th ditch = 199 anukuntunna. Khan Dada - Chiru. Manikyam - Pans. | ||
Frodo Junior Artist Username: Frodo Post Number: 420 Registered: 11-2009 Posted From: 115.117.197.7 Rating: N/A |
calvin @ 9:32 : that expression does not evaluate to the right answer. Could you explain what you mean by selection of digits with duplicates from 1 to 6? also @ 8:49 F(2) = no of ways to get to ditch 2 from ditch 1 = 1way only 1hop. so F(2) = 1 and not 2, right? Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Calvin Junior Artist Username: Calvin Post Number: 38 Registered: 03-2008 Posted From: 66.129.224.36 Rating: N/A |
Nah.. got it wrong again - Found to be wrong when I was driving home from office.. Descending = Summation of f(k)* 10ck -1 for k=6 to 1. f(k) = k^(6-k) I think I got it right this time.... its selection of digits with duplicates from 1 to 6. | ||
Calvin Junior Artist Username: Calvin Post Number: 37 Registered: 03-2008 Posted From: 66.129.224.36 Rating: N/A |
correction - Descending = 10 c6+ 5*10c5+ 4*10c4 + 3*10c3+ 2 *10c2+ 1*10c1 - 1 (we need to subtract the number with all zeros) | ||
Calvin Junior Artist Username: Calvin Post Number: 36 Registered: 03-2008 Posted From: 66.129.224.36 Rating: N/A |
strictly descending - 10c6 = 10c4 = 210 descending = 10c6 + 5*10c5 + 4*10c4 + 3*10c3 + 2 * 10c2 + 1*10c1 The other problem - should not it be 101 th term of fibonacci series as opposed to 100 th term? F(100) = F(99) + F(98) and F(2) = 2 In fibonacci = Fib(2) = 1 F(100) = Fib(101) | ||
Frodo Junior Artist Username: Frodo Post Number: 416 Registered: 11-2009 Posted From: 59.161.177.142 Rating: N/A |
no dude. unless my matlab programming is wrong, that expression is way off the right answer. your logic is obviously correct, but there is a line of thought, that will simplify the solution a lot. want to try another way of solving this? i dont know if this is much of a hint, but the answer can be expressed as nCr. Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2285 Registered: 11-2009 Posted From: 202.133.58.84 Rating: N/A |
is it 9*10^5 - sigma of n (2(n^5+n^4+n^3+n^2+n) +(n^4*(9-n)+n^3*(9-n)^2+n^2*(9-n)^3+n(9-n)^4))+9^5 where n ranges from 1 to 9 , 9^5 is not included in the sigma, the logic behind this is for each decimal place i tried to find the count of numbers having bigger digits to the right or smaller digits to the left or bigger to the right and smaller to the left, subtracted the sum of these counts from number of six digit numbers...the last term 9^5 is to compensate for numbers that have higher digits to the right but starts with zero | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2284 Registered: 11-2009 Posted From: 202.133.58.83 Rating: N/A |
i am trying to subtract
i want to subtract all possible combinations of digits that are greater to the right and lesser to the left for each decimal place for all possible digits from number of six digit numbers...the expression below just subtracts combinations of greater digits to the right from the number of six digit numbers...let me try for some more time to get the right expression | ||
Thirtyplus Junior Artist Username: Thirtyplus Post Number: 417 Registered: 03-2009 Posted From: 174.103.243.218 Rating: N/A |
thread motham choosthe ardham ayindi.....order of jumping kooda teesukunnarani 2500 is when the order is not considered anyways TDP - TODU DONGALA PARTY | ||
Thirtyplus Junior Artist Username: Thirtyplus Post Number: 416 Registered: 03-2009 Posted From: 174.103.243.218 Rating: N/A |
101 - 2x where x = 1 to 50 answer - 2500 TDP - TODU DONGALA PARTY | ||
Frodo Junior Artist Username: Frodo Post Number: 415 Registered: 11-2009 Posted From: 59.161.177.142 Rating: N/A |
You are way off @ 9:29 but spot-on @ 9:35. what is your logic behind 9:29? oka concept grasp chesthe this problem has a straight forward solution, though it does need a not-so-beginner level combinatorial concept to get the eventual answer. The prof i mentioned used to give his prospective students this problem to see if they get that concept. the prof had the total works - caltech ph.d, total loner, talking to himself as he walks in the corridors, punching the walls in excitement, bachelor @ 45. i dont think he talked to any non- communications guy other than me. many a time we used to be the only 2 guys @ 3 in the morning and that was how we became friends.... anyway, back to the problem, tell me the logic for that 2nd part. Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2283 Registered: 11-2009 Posted From: 202.133.58.83 Rating: N/A |
my answer is 210 for this ....what's the right answer | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2282 Registered: 11-2009 Posted From: 202.133.58.83 Rating: N/A |
9*10*10*10*10*10 -(8^5+7^5+6^5+5^5+4^5+3^5+2^5+1^5) -(9^4+8^4+........+1) -(9^3+.........+1^3) - (9^2+.........+1^2) -(9+8.....+1) | ||
Frodo Junior Artist Username: Frodo Post Number: 414 Registered: 11-2009 Posted From: 59.161.177.142 Rating: N/A |
strictly descending means a(i) > a(i+1) dscending means a(i) >= a(i+1) Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2281 Registered: 11-2009 Posted From: 202.133.58.78 Rating: N/A |
sorry my answers are for numbers descending from lakhs to units, if strictly descending means difference between any two consecutive digits should be constant then my answer is 5, the numbers are 543210 654321 765432 876543 987654 if the difference between the digits need not be constant but digits can't repeat then answer is 210 which also includes the above five numbers. | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2280 Registered: 11-2009 Posted From: 202.133.58.78 Rating: N/A |
strictly descending means difference between any two consecutive digits should be same right? | ||
Frodo Junior Artist Username: Frodo Post Number: 413 Registered: 11-2009 Posted From: 115.117.204.188 Rating: N/A |
it is strictly descending or descending from lakhs positions to units. (left to right that is...) even when you do right to left, the answers are not correct. Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2279 Registered: 11-2009 Posted From: 202.133.58.75 Rating: N/A |
it's only 5 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2278 Registered: 11-2009 Posted From: 202.133.58.75 Rating: N/A |
if it is from right..it's 6
from the right 5C5+6C5+7C5+8C5+9C5 = 1+6+21+56+126 = 210 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2277 Registered: 11-2009 Posted From: 202.133.58.75 Rating: N/A |
from the left or right? | ||
Frodo Junior Artist Username: Frodo Post Number: 412 Registered: 11-2009 Posted From: 115.117.204.188 Rating: N/A |
ok google tells me F(100) = 354 224 848 179 261 915 075. and incidentally 354 224 848 179 261 915 075 / (2^50) = 314 614.866 so I was about 6orders of magnitude away in my calculation. thanks for the puzzle. If guys are interested here is a combinatorics problem that earned me 1000$ from a prof of mine. How many 6digited numbers have their digits A) in strictly descending order B) in descending order Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Frodo Junior Artist Username: Frodo Post Number: 411 Registered: 11-2009 Posted From: 115.117.204.188 Rating: N/A |
" it is actually the 100 term of a fibanacci series..i,e the sum of the first 99 terms " 100th term is sum of only 99th and 98th term right, not all the 1st 99 terms. yes? Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Frodo Junior Artist Username: Frodo Post Number: 410 Registered: 11-2009 Posted From: 115.117.204.188 Rating: N/A |
wait, the answer is F(100). right? i goofed up again. it is not sum of 1st 100 fibonacci terms, but just F(100). Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5620 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
it is actually the 100 term of a fibanacci series..i,e the sum of the first 99 terms and it is of the order 2^50 but not equal to it | ||
Frodo Junior Artist Username: Frodo Post Number: 409 Registered: 11-2009 Posted From: 115.117.204.188 Rating: N/A |
@ Der @ 8:11 is it the sum of the 1st 100 terms in the fibonacci sequence? I have no idea how to calculate that without a computer program..... is the answer right? if yes, how do you sum up the 1st 100 terms of the fibonacci? Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2276 Registered: 11-2009 Posted From: 202.133.58.74 Rating: N/A |
i think number of possible combinations is greater than 2^50 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2275 Registered: 11-2009 Posted From: 202.133.58.74 Rating: N/A |
emi change avvali bro? | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5614 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
Yes. this is about right but for some integer change. But u are on the right track...keep persevering with that line of thought and see if u see some elegance in the way that terrible falling sum will surface into sum beautiful series?? | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5613 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
seems right....there is a beautiful way to solve this with no combinatorics at all | ||
Frodo Junior Artist Username: Frodo Post Number: 408 Registered: 11-2009 Posted From: 115.117.251.42 Rating: N/A |
slumdawg: did we interact before. pardon me for being dense, too much stress on my grey-matter last 1month or so.... vjavasi : both your expressions are the same as the one i put up with k..... i think your is a more elegant and concisely put. Stig: another way to look at this is, getting 100 = no of ways to get to 99 + number of ways to get to 98. You can then keep going recursively as follows : number of ways to get to 99 = number of ways to get to 98 + number of ways to get to 97... similarly number of ways to get to 98 = number of ways to get to 97 + number of ways to get to 96... you can continue this series till you get number of ways to get to 2 which is equal to 2. I am not good with math jargon at undergrad level, but that might be some kind of recursive series..... this was my thought intuitively, but as usual my intuition led me to a monster  but then this seems to qualify the only addition tag better than the combinatorial solution i mentioned in the morn.... i have to work on what that summation simplifies to.... provided we have confirmation that is the way to go.... later... Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Slumdawg Junior Artist Username: Slumdawg Post Number: 560 Registered: 04-2010 Posted From: 204.74.221.58 Rating: N/A |
Frodo kurrod how r u Sombabu | ||
Frodo Junior Artist Username: Frodo Post Number: 407 Registered: 11-2009 Posted From: 115.117.251.42 Rating: N/A |
@ stig : forget that, i made an @ss of U and ME (i @SSUMEd something wrong). here is what i thought.... no. of ways = no of ways person can make 50 2hops and 0 1hops + no of ways person can make 49 2hops and 2 1hops + no of ways person can make 48 2hops and 4 1hops + no of ways person can make 47 2hops and 6 1hops + .................. + ........... + no of ways person can make 0 2hops and 100 1hops In short it is the summation of ways in which a person can make 50-k 2hops and 2k 1hops, with k ranging from zero to 50. to find the kth term, let us look at how many ways person can make 50-k 2hops and 2k 1hops. basic combinatorics ( guess i am violating the counting problem and/or only addition operator funda here), anyhow the number of ways that can be done = (50-k + 2k) C 2k which simplifies to (50+k) C 2k. therefore the answer should be summation of [ (50+k) C (2k) ] for k = 0 to 50. Hope i did not goof it this time around. There is another way of doing this, i will tell you that when i take my break from my next class. Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2274 Registered: 11-2009 Posted From: 202.133.58.71 Rating: N/A |
sorry it should be sum over 100-n(C)n where n ranges from 0 to 50 | ||
Vjavasi Side Hero Username: Vjavasi Post Number: 2273 Registered: 11-2009 Posted From: 202.133.58.71 Rating: N/A |
let's say no of two hops is n no of one hops is m relation between n and m is m = 100 - 2n n could range from 0 to 50 so total number of combinations is 51 if you have to consider the order of jumping also then you have to look for different combinations for a given m,n it will be sum of m+n(C)m where m changes from 0 to 50 | ||
Frodo Junior Artist Username: Frodo Post Number: 406 Registered: 11-2009 Posted From: 115.117.251.42 Rating: N/A |
by the way, this part of the question is very ambiguous : "As an example: U making 50 2 ditch hops is one way same with 100 "1 ditch hops" Do you mean 50 2ditch hops is one way of jumping and another way is 100 1 ditch hops? the way you worded it makes it sound like 50 2ditch hops is equivalent to 100 1 ditch hops. I am guessing you meant the former interpretation of mine, but can you elaborate? Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Stig Side Hero Username: Stig Post Number: 2169 Registered: 01-2010 Posted From: 67.80.101.77 Rating: N/A |
Ela ?? ------- Only seven people have looked The Stig straight in the eyes. They are all dead now !! | ||
Frodo Junior Artist Username: Frodo Post Number: 405 Registered: 11-2009 Posted From: 115.117.251.42 Rating: N/A |
if i understand the question correctly and if i can trust my 8th grade math, the answer is 2^50 ~ 1e15. right? compliments to frodo, brickbats to 8th grade math prof! j/k, i had a totally nutty prof (that is what i will call him, he was just way too impatient to qualify as a middle-school teacher! all the students except me and 2 other friends of mine used to hate him with a vengeance! and we used to call him nattu) and by the time i finished 12th, i could crack math problems faster and more elegantly than him, but am yet to beat him in a game of chess. for good or bad, he made me the competitive bar-stud i am....... Dal tadka, dum-biryani and Deccan chargers - always no.1 | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5612 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
Didn't get u? Proline......well when I say there are 100 ditches then is it not implicit that they are unique?? | ||
Maverick Hero Username: Maverick Post Number: 15898 Registered: 01-2008 Posted From: 24.1.171.91 Rating: N/A |
did u purposefully misguide ppl here? if u r in ditch1 and make a 2 ditch hop, u r in ditch 4..so u need only 25 2 ditch hops+ 1 more to passs 100 ditches??something like that..am i wrong? | ||
Abcdefghij Side Hero Username: Abcdefghij Post Number: 7285 Registered: 02-2007 Posted From: 76.226.22.167 Rating: N/A |
are you saying starting 1hop, 49 2hops is different than 49 2 hops and 2 1 hops etc? | ||
Hail_the_labour Side Hero Username: Hail_the_labour Post Number: 2148 Registered: 06-2008 Posted From: 75.185.82.44 Rating: N/A |
der: do you mean hops are not horizontal straight. if they are spread in rectangular,square,geometric fashion they 3rd dimension of cross-hopping will add more possible ways to my number series so my previous answer+extra possible ways is my revised answer | ||
Proline Side Hero Username: Proline Post Number: 4509 Registered: 06-2008 Posted From: 173.3.73.246 Rating: N/A |
if permutations/order does matter, then the problem will be stated as there are 100 distinct ditches or there are 100 ditches numbered from 1 to 100 etc and thats when the order does matter. ... | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5611 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
The only hint I can afford to give with out bleeding the life of the problem is that it is classic computer science problem | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5610 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
I never mentioned the word combinations or permutations, I posed this as a counting problem and it unless specified can't be classified as a combinatoric problem....its another issue that u can cast it as one | ||
Tnmnm Junior Artist Username: Tnmnm Post Number: 6 Registered: 04-2010 Posted From: 199.71.214.205 Rating: N/A |
correct annai. problem antha simple ga ledu naaku. manakekkaledu...correct answer kosam wait sestha. | ||
Hail_the_labour Side Hero Username: Hail_the_labour Post Number: 2147 Registered: 06-2008 Posted From: 75.185.82.44 Rating: N/A |
I understood the condition as any comnination of hops is allowed (1 or 2). best hop : 50 (2 ditches) worst hop: 100(1 ditch) any combination of hops will fall in between 50 to 100. ex. he can do 45 2 hops =90 + 10 1hops = 55 hops . like wise | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5609 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
Hmm let me clarify...I said what are possible ways in which u can reach the 100th ditch...implicit to it is the assumption that intrinsic order is important | ||
Proline Side Hero Username: Proline Post Number: 4508 Registered: 06-2008 Posted From: 173.3.73.246 Rating: N/A |
order does not matter as per the quiz. so it does not matter how they are distributed ... | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5608 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
agreed but there 1 other dimension...how are these 1 and 2 hops distributed per individual way?? | ||
Proline Side Hero Username: Proline Post Number: 4507 Registered: 06-2008 Posted From: 173.3.73.246 Rating: N/A |
note: order does not matter (as it is not given in the quiz, so i can safely ignore the permutations) ... | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5607 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
well thats not a difficult series it is just (50*51/2)+50 = 25*53 not the right answer...gimme the line of attack | ||
Proline Side Hero Username: Proline Post Number: 4506 Registered: 06-2008 Posted From: 173.3.73.246 Rating: N/A |
simple. you have to use only one addition for combining the ways: so let us start with 2hops. min two hops can be 0 and max can be 49 (as per the hint, 0 two hops combination and 50 two hops combination are the same so ignoring 50 two hop) 0 two hops + 100 1 hops same as 50 2 hops + 100 0 hops 1 two hop + 99 1hops and so on 49 two hops + 2 1 hops so total combinations = 50 ... | ||
Hail_the_labour Side Hero Username: Hail_the_labour Post Number: 2146 Registered: 06-2008 Posted From: 75.185.82.44 Rating: N/A |
sum of series : starting 50,51,52...100 answers is : some thousand combinations | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5604 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
elaborate please...I need to see ur line of reason?? | ||
Abcdefghij Side Hero Username: Abcdefghij Post Number: 7284 Registered: 02-2007 Posted From: 76.226.22.167 Rating: N/A |
50 | ||
Proline Side Hero Username: Proline Post Number: 4504 Registered: 06-2008 Posted From: 173.3.73.246 Rating: N/A |
inlcuding the hint, it will be 50 ways ... | ||
Proline Side Hero Username: Proline Post Number: 4503 Registered: 06-2008 Posted From: 173.3.73.246 Rating: N/A |
49 ways you can solve this (assuming the hint given is not considered in 49) is it correct  ... | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5603 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
The problem with this approach are many fold: 1.) the right hand side of the above equation is not 100 (why??) 2.) y= 2x is a false relationship (??) | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5602 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
Navvu endhukandi....remember einstein's quote, " if u think you have problems with math, rest assured as I have even bigger ones with it" answer to me is trivial, I am more keyed to how original u are in ur thoughts 20 is wrong | ||
Tnmnm Junior Artist Username: Tnmnm Post Number: 5 Registered: 04-2010 Posted From: 76.73.2.10 Rating: N/A |
naa calculation septhaa... 2x+1y = 100 eskunte... y=2x substitution sesthe...20 ways vathaayi...  | ||
Tnmnm Junior Artist Username: Tnmnm Post Number: 4 Registered: 04-2010 Posted From: 76.73.2.10 Rating: N/A |
20 total possible ways aa annai? nenu maths la eek. navvukoka | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5601 Registered: 01-2009 Posted From: 68.46.21.1 Rating: N/A |
umm let me know how u arrived at this number...the answer is not the important thing but the process (49 tho is wrong) | ||
Stig Side Hero Username: Stig Post Number: 2163 Registered: 01-2010 Posted From: 67.80.101.77 Rating: N/A |
49 ?? ------- Only seven people have looked The Stig straight in the eyes. They are all dead now !! | ||
Slumdawg Junior Artist Username: Slumdawg Post Number: 526 Registered: 04-2010 Posted From: 204.74.221.12 Rating: N/A |
resume mottam seppak tamud...sigguto sachi savam aipota Sombabu | ||
Rakesh Junior Artist Username: Rakesh Post Number: 18 Registered: 04-2010 Posted From: 98.210.96.94 Rating: N/A |
4years intermediate and 3 times 10th class....sallagaa nokkesaav gaa aatini | ||
Slumdawg Junior Artist Username: Slumdawg Post Number: 525 Registered: 04-2010 Posted From: 204.74.221.12 Rating: N/A |
lekkallo eek.. donasion katti BA sesa....6 ears lo complete sesesa Sombabu | ||
Der_schuler Side Hero Username: Der_schuler Post Number: 5600 Registered: 01-2009 Posted From: 68.46.21.1 Rating: |
Super day recruiting ki monna variety scheme ettar.....people who came to interview are supposed to pose a question to the interviewer opposed to the normal way...the intent was to see whether people are active with their math/logic and also on how the prospects will make others think along with them when they are in the driving seats.. Oka hebrew university math undergrad kurrod ee question adigad nannu.... "Suppose there are 100 ditches and you are stuck in ditch 1 and need to go past the 100th ditch. Your physical capacity allows you to hop over 1 ditch or 2 ditches in one go i.e you can either make 50 "2 ditch hops" or 100 "1 ditch hops" or a combination of both. If you are allowed to solve the problem using only one arithmetic operation of addition, how do you calculate all the possible ways in which u can go past the 100 ditches?? As an example: U making 50 2 ditch hops is one way same with 100 "1 ditch hops" The time to solve this is 5-7 mins roughly |
