The_Blade Posted July 30, 2004 Share Posted July 30, 2004 (edited) Sorry, you've got the correct answer but a BAD solution, as if you domn't know anything about math. The answer is easy, but it's the way you solved it. A don't see any progression formula. When you say a 3-digit number, than it should be written as:Let x = 100's digit, y = 10s didigt, and z = 1s digitSo the number would be 100x+10y+z = the number This is how you solve a math problem... Ok?I think r3mu503 you presented the only answer. 246 + 596 = 842 solution no1. write all possible 3 digit #s which when you add to 596 will give you a 3 digit answer. the easiest solution, T&R. (trial and error). 123 596 719135 596 731147 596 743159 596 755234 596 830246 596 842258 596 854321 596 917345 596 941357 596 953369 596 965 solution no2. solve backwards. any 3 digit number when you add to 596 will always be greater than 596. now think of an integer larger than 5 but less than 9 (becoz the answer should be lesss than 999) which will give at least 3 factors. and could only be 8...4...2. or vice versa start with the last sensible digit. 1 is eliminated. 3 is disqualified becase it will give you 3..9...27 already. so 2 is the base number. 2..4..8. (process of elimination). now check your answer. 842-596=246, therefore 246 is the answer. solution no.3 using algebra Let x = 100's digit, y = 10s didigt, and z = 1s digitSo the number would be 100x+10y+z = N1(#in arithmitic progression) Let 100a+10b+c =N2 (# in geometric progression) where a=c^3; b=c^2 (now this is tricky, i assumed the 100digits is > 10 digit> 1 digit bcoz its the only sensible number > 596) tf: 100c^3+10c^2+c=N2 100x+10y+z+596= 100c^3+10c^2+c ; 1 equation & 3 unknowns = indeterminate but since 2 is the only sensible integer for c, by substitution: 100x+10y+z+ 596 = 100(8) + 10(4) + 2 = 842 100x+10y+z = 842-596 = 246 = 100(2)+10(4)+ 6 100x +10y + z = 100(2) + 10(4) + 6 therfore x=2 ; y=4 ; z=6 now boss xerxes and PageDown may i have your thoughts? Edited July 30, 2004 by edzabu Quote Link to comment
ukyo_batusai Posted July 30, 2004 Share Posted July 30, 2004 (edited) to dali lang.... find the product of.... ( X - a ) ( X - b ) ( X - c )......( X - z )? Edited July 30, 2004 by ukyo_batusai Quote Link to comment
The_Blade Posted August 2, 2004 Share Posted August 2, 2004 (edited) to dali lang.... find the product of.... ( X - a ) ( X - b ) ( X - c )......( X - z )?X^26 +.................+ abc...z = 0 elementary! the expression is also equal to: (x-a) (x- (x-c) .........(x-x) (x-y) (x-z) but (x-x) = 0 therefore: any number multiplied by zero is equal to zero. itaas ko lang meron pang tanong si oric91 na di nasasagot. Edited August 2, 2004 by edzabu Quote Link to comment
r3mu503 Posted August 2, 2004 Share Posted August 2, 2004 hey, is the above post from ukyo a solution to a different question??? I still haven't seen the pure equation solution to the original question? you know, starting from: 100x + 10(x+k) + (x+2k) + 596 = 100y + 10yc + yc*c and manipulating the above equation to prove the answer 246... hi edzabu, I think you'll need to use kc, kc^2, kc^3 for your geom. progression b/c that is strictly speaking the definition of a geom. progress. and the original poster did not say we could assume k=1 (hell, he actually didn't even say we could assume the order of the digits would be left to right in increasing value). so where is the pure equation solution to this anyway??? Quote Link to comment
D_Engineer Posted August 2, 2004 Share Posted August 2, 2004 oppss..padaan lang po... Im an engineer by profession but mahina sa math... Quote Link to comment
The_Blade Posted August 2, 2004 Share Posted August 2, 2004 hey, is the above post from ukyo a solution to a different question??? I still haven't seen the pure equation solution to the original question? you know, starting from: 100x + 10(x+k) + (x+2k) + 596 = 100y + 10yc + yc*c and manipulating the above equation to prove the answer 246... hi edzabu, I think you'll need to use kc, kc^2, kc^3 for your geom. progression b/c that is strictly speaking the definition of a geom. progress. and the original poster did not say we could assume k=1 (hell, he actually didn't even say we could assume the order of the digits would be left to right in increasing value). so where is the pure equation solution to this anyway??? so common page down. show us your solution. Quote Link to comment
Jourdan Posted August 2, 2004 Share Posted August 2, 2004 hey, is the above post from ukyo a solution to a different question??? I still haven't seen the pure equation solution to the original question? you know, starting from: 100x + 10(x+k) + (x+2k) + 596 = 100y + 10yc + yc*c and manipulating the above equation to prove the answer 246... hi edzabu, I think you'll need to use kc, kc^2, kc^3 for your geom. progression b/c that is strictly speaking the definition of a geom. progress. and the original poster did not say we could assume k=1 (hell, he actually didn't even say we could assume the order of the digits would be left to right in increasing value). so where is the pure equation solution to this anyway??? worthwhile to point out that in that eqn: x,k, y are all integersand c is a Real no... m looking for a solution na rin. I think you should write 596 as 100(5) + 10(9) + 6 so that the eqn becomes: 100(x+5-y) + 10 (x+k+9-yc) + (x+2k+6-yc*c) = 0 hopefully i'll have time tonite to solve this one. Quote Link to comment
The_Blade Posted August 3, 2004 Share Posted August 3, 2004 hopefully i'll have time tonite to solve this one. asan na? Quote Link to comment
katips Posted August 4, 2004 Share Posted August 4, 2004 oppss..padaan lang po... Im an engineer by profession but mahina sa math... haha.. pero sa chicks malakas hehehe... *howlers Quote Link to comment
The_Blade Posted August 5, 2004 Share Posted August 5, 2004 hehehe. This thread's days are numbered. Moderators need not be a math wizard to close this. Calling the topic starter to save your thread! Quote Link to comment
author Posted August 11, 2004 Share Posted August 11, 2004 This is what happens when the thread starter doesn't know much about the topic he/she started. Quote Link to comment
cruiser007 Posted August 18, 2004 Share Posted August 18, 2004 did the thread starter een attempt to solve the math problems posted? Quote Link to comment
boy男孩 Posted August 23, 2004 Share Posted August 23, 2004 FIND THE SUM: 1+1/4+1/9+1/16+...+1/n^2 Quote Link to comment
floppydrive Posted August 24, 2004 Share Posted August 24, 2004 FIND THE SUM: 1+1/4+1/9+1/16+...+1/n^2 As n approaches infinity in the sum of 1/1 + 1/4 + ... + 1/(n^2),the sum approaches 1.645, but will never equal it. (Assymptotic graph) :upside: Quote Link to comment
boy男孩 Posted August 25, 2004 Share Posted August 25, 2004 As n approaches infinity in the sum of 1/1 + 1/4 + ... + 1/(n^2),the sum approaches 1.645, but will never equal it. (Assymptotic graph) :upside: somehow youre ryt. Actually its Euler's Formula: the sum is equal to 1/6 Pi^2 where Pi = 3.1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095 5058223172 5359408128 4811174502 8410270193 8521105559 6446229489 5493038196 4428810975 6659334461 2847564823 3786783165 2712019091 4564856692 3460348610 4543266482 1339360726 0249141273 7245870066 0631558817 4881520920 9628292540 9171536436 7892590360 0113305305 4882046652 1384146951 9415116094 3305727036 5759591953 0921861173 8193261179 3105118548 0744623799 6274956735 1885752724 8912279381 8301194912 9833673362 4406566430 8602139494 6395224737 1907021798 6094370277 0539217176 2931767523 8467481846 7669405132 0005681271 4526356082 7785771342 7577896091 7363717872 1468440901 2249534301 4654958537 1050792279 6892589235 4201995611 2129021960 8640344181 5981362977 4771309960 5187072113 4999999837 2978049951 0597317328 1609631859 5024459455 3469083026 4252230825 3344685035 2619311881 7101000313 7838752886 5875332083 8142061717 7669147303 5982534904 2875546873 1159562863 8823537875 9375195778 1857780532 1712268066 1300192787 6611195909 2164201989N..... from: http://www.math.com/tables/constants/pi.htm#euler Now new question, What is N?( the 1001st digit of Pi? Quote Link to comment
floppydrive Posted August 25, 2004 Share Posted August 25, 2004 Now new question, What is N?( the 1001st digit of Pi? Can you give us 10 guesses? Quote Link to comment
Huddaf Posted August 25, 2004 Share Posted August 25, 2004 Duh! m not sure about this! cge, i'll just watch na lang... Quote Link to comment
boy男孩 Posted August 27, 2004 Share Posted August 27, 2004 Duh! m not sure about this! cge, i'll just watch na lang... hehe nagpaparami ka lang ng post mo eh. Quote Link to comment
rstlne999 Posted August 29, 2004 Share Posted August 29, 2004 di naman talaga dapat ganyan kahirap ang map. pinapahirap lang nila.di mo naman gagamitin sa buhay mo karaminhan ng mahihirap na tinuturo nila. Quote Link to comment
boy男孩 Posted September 2, 2004 Share Posted September 2, 2004 Show the mathematical solution showing a person's chance to winnning a 6/42, 6/45 & 6/49 lotto. How about 2, 3 or 4 persons winning the same draw? Quote Link to comment
floppydrive Posted September 2, 2004 Share Posted September 2, 2004 6/42 lotto: For n person(s), the chances of getting the 6 numbers are:n * 42*41*40*39*38*37 = n*42!/36! so for 1 person it's 42!/36!, for 2 people, it's 2*42!/36!, etc. 6/45 lotto: For n person(s), the chances of getting the 6 numbers are:n * 45*44*43*42*41*40 = n*45!/39! 6/49 lotto: For n person(s), the chances of getting the 6 numbers are:n * 49*48*47*46*45*44 = n*49!/43! For lotto 6/X, where X is the number of numbers to choose from, and n is the number of winners for the same draw, the general equation is: n * X!/(X-6)! Quote Link to comment
boy男孩 Posted September 2, 2004 Share Posted September 2, 2004 6/42 lotto: For n person(s), the chances of getting the 6 numbers are:n * 42*41*40*39*38*37 = n*42!/36! so for 1 person it's 42!/36!, for 2 people, it's 2*42!/36!, etc. 6/45 lotto: For n person(s), the chances of getting the 6 numbers are:n * 45*44*43*42*41*40 = n*45!/39! 6/49 lotto: For n person(s), the chances of getting the 6 numbers are:n * 49*48*47*46*45*44 = n*49!/43! For lotto 6/X, where X is the number of numbers to choose from, and n is the number of winners for the same draw, the general equation is: n * X!/(X-6)! hmmn. i believe something is wrong with your equations. for 1 thing, the chance or probabity of one person in winning a lottery decreases as the numbers of winners increases. wat u think? Quote Link to comment
floppydrive Posted September 2, 2004 Share Posted September 2, 2004 I thought the question was what is the probability of 2 people winning the lotto on the same draw. If my understanding is correct, the chances for 2 people winning at the same time becomes slimmer than for one person winning it. one person has a one in 42!/36! chance of winning the draw. two people winning at the same time is one chance in 2*42!/36! to occur. If we examine the probability of one person winning, the equation is still 42!/36!, regardless of the number of people entering the lotto. I would assume that a person's chance of winning is the same. He would receive a smaller sum, but his chances are still the same. What do you think, Sir Boy? Off topic, your picture looks very familiar... Are you from Jakarta? Quote Link to comment
floppydrive Posted September 2, 2004 Share Posted September 2, 2004 I thought the question was what is the probability of 2 people winning the lotto on the same draw. If my understanding is correct, the chances for 2 people winning at the same time becomes slimmer than for one person winning it. one person has a one in 42!/36! chance of winning the draw. two people winning at the same time is one chance in 2*42!/36! to occur. The above is from the perspective of the lotto organizer, checking what are the chance of having multiple winners. Let's now take it from the perspective of the chances of ONE person winning. For him or her to win, he or she has to choose the correct 6 numbers out of 42. It doesn't matter how many people enter, becuase what matters is for the entrant to get all 6 numbers correct. If we examine the probability of one person winning, the equation is still 42!/36!, regardless of the number of people entering the lotto. I would assume that a person's chance of winning is the same. He would receive a smaller sum, but his chances are still the same. What do you think, Sir Boy? Off topic, your picture looks very familiar... Are you from Jakarta? Quote Link to comment
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