st Posted July 12, 2004 Share Posted July 12, 2004 Mahirap ba talaga masolve yong problem mo sa Math? Post it here! Thanx! Quote Link to comment
Brando_Matias Posted July 16, 2004 Share Posted July 16, 2004 Hmmmm...mukhang walang nahihirapan sa math ahhh???? Quote Link to comment
naked_angel Posted July 17, 2004 Share Posted July 17, 2004 {First 10 digit prime in consecutive digits of e }.com if you get the right number, go to that site and you'll be presented another math problem and a username. the answer to this second problem is the password. go to Linux.Org, login and you'll be redirected to google labs where they will give you the email address for you to send your resume. math wizards???? ANYONE?!?!? Quote Link to comment
st Posted July 17, 2004 Author Share Posted July 17, 2004 {First 10 digit prime in consecutive digits of e }.com if you get the right number, go to that site and you'll be presented another math problem and a username. the answer to this second problem is the password. go to Linux.Org, login and you'll be redirected to google labs where they will give you the email address for you to send your resume. math wizards???? ANYONE?!?!? Just visit google-blog.dirson.com, the answer s are there. Quote Link to comment
naked_angel Posted July 18, 2004 Share Posted July 18, 2004 Just visit google-blog.dirson.com, the answer s are there. Ngek. Kala ko kse share your math problems for everyone to solve. Quote Link to comment
rich beem Posted July 19, 2004 Share Posted July 19, 2004 mga math majors ba kayo?? Quote Link to comment
st Posted July 19, 2004 Author Share Posted July 19, 2004 mga math majors ba kayo?? no . I am not math majors. i love numbers so much that's it......kahit anong field...maka math, chem, physics, accounting,calculus...etc Quote Link to comment
r3mu503 Posted July 19, 2004 Share Posted July 19, 2004 no . I am not math majors. i love numbers so much that's it......kahit anong field...maka math, chem, physics, accounting,calculus...etc wow, now that's intriguing, you post on quite a wide range of topics and now are a self-confessed math geek, I'm impressed what was you're course in college anyway- maybe philo? refuge for the restless intellectual Quote Link to comment
st Posted July 19, 2004 Author Share Posted July 19, 2004 wow, now that's intriguing, you post on quite a wide range of topics and now are a self-confessed math geek, I'm impressed what was you're course in college anyway- maybe philo? refuge for the restless intellectual self study lang...mahilig lang talaga me sa numbers.that's it Quote Link to comment
st Posted July 19, 2004 Author Share Posted July 19, 2004 Ngek. Kala ko kse share your math problems for everyone to solve. okey. if there has another way of solving it...then post it. Quote Link to comment
r3mu503 Posted July 19, 2004 Share Posted July 19, 2004 self study lang...mahilig lang talaga me sa numbers.that's it self-taught in math, all the way up to calculus? yikes, you're scary just kidding, I am mighty impressed what's the most out of this world, obscure math you studied? Quote Link to comment
st Posted July 19, 2004 Author Share Posted July 19, 2004 self-taught in math, all the way up to calculus? yikes, you're scary just kidding, I am mighty impressed what's the most out of this world, obscure math you studied? basta tipo ko lang numbers Quote Link to comment
Guest -=xerxes=- Posted July 19, 2004 Share Posted July 19, 2004 i am an engineer by profession maybe i could help you with this thread Quote Link to comment
st Posted July 19, 2004 Author Share Posted July 19, 2004 i am an engineer by profession maybe i could help you with this thread THANX ENGINEER.....YOU ARE WELCOME HERE Quote Link to comment
rich beem Posted July 19, 2004 Share Posted July 19, 2004 ano mag tatayo na ba tayo ng mtc math club? Quote Link to comment
st Posted July 20, 2004 Author Share Posted July 20, 2004 ano mag tatayo na ba tayo ng mtc math club? Kayo po bahala Quote Link to comment
PageDown Posted July 20, 2004 Share Posted July 20, 2004 A 3-digit number has its digit in arithmetic progression.If 596 is added to this 3-digit number, the sum will have its digit in geometric progression.What is the number?Please post your solution. Quote Link to comment
r3mu503 Posted July 21, 2004 Share Posted July 21, 2004 246 I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies ): The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression): b b*r b*r^2 b*r^3b*r^3 b*r^2 b*r b (where each digit <=9) With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number). So now we have either of these two conditions (since the leftmost digit is a 1): b=1b*r^3=1 In the first case, if b=1 then r=1 or 2 (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false. If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well. So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number. Ok, so now we know the sum is a 3-digit number, hence in the following form: b b*r b*r^2b*r^2 b*r b We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus: 9 >= b >= 69 >= b*r^2 >= 6 Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r: b*r^2 b (r=1) b (r=2) b (r=3)------- -------- --------- --------- 6 6 X X 7 7 X X 8 8 2 X 9 9 X 1 ('X' marks non-valid combinations) Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596: 666 (b=6,r=1)777 (b=7,r=1)888 (b=8,r=1)999 (b=9,r=1)842 (b=2,r=2)931 (b=1,r=3) Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress. Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer Quote Link to comment
Chinese Chicken Posted July 21, 2004 Share Posted July 21, 2004 246 I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies ): The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression): b b*r b*r^2 b*r^3b*r^3 b*r^2 b*r b (where each digit <=9) With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number). So now we have either of these two conditions (since the leftmost digit is a 1): b=1b*r^3=1 In the first case, if b=1 then r=1 or 2 (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false. If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well. So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number. Ok, so now we know the sum is a 3-digit number, hence in the following form: b b*r b*r^2b*r^2 b*r b We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus: 9 >= b >= 69 >= b*r^2 >= 6 Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r: b*r^2 b (r=1) b (r=2) b (r=3)------- -------- --------- --------- 6 6 X X 7 7 X X 8 8 2 X 9 9 X 1 ('X' marks non-valid combinations) Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596: 666 (b=6,r=1)777 (b=7,r=1)888 (b=8,r=1)999 (b=9,r=1)842 (b=2,r=2)931 (b=1,r=3) Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress. Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer first paragraph pa lang nawala na ako. Quote Link to comment
Guest -=xerxes=- Posted July 21, 2004 Share Posted July 21, 2004 (edited) 246 I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies ): The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression): b b*r b*r^2 b*r^3b*r^3 b*r^2 b*r b (where each digit <=9) With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number). So now we have either of these two conditions (since the leftmost digit is a 1): b=1b*r^3=1 In the first case, if b=1 then r=1 or 2 (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false. If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well. So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number. Ok, so now we know the sum is a 3-digit number, hence in the following form: b b*r b*r^2b*r^2 b*r b We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus: 9 >= b >= 69 >= b*r^2 >= 6 Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r: b*r^2 b (r=1) b (r=2) b (r=3)------- -------- --------- --------- 6 6 X X 7 7 X X 8 8 2 X 9 9 X 1 ('X' marks non-valid combinations) Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596: 666 (b=6,r=1)777 (b=7,r=1)888 (b=8,r=1)999 (b=9,r=1)842 (b=2,r=2)931 (b=1,r=3) Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress. Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer it can be solved in a much easier manner Edited August 24, 2004 by -=xerxes=- Quote Link to comment
oric91 Posted July 21, 2004 Share Posted July 21, 2004 Write 1/2 as a decimal number without using the digit 5 nor use any signs of operator - plus, minus, and such in your solution. For example,1/2= 0.6-0.1, is not considered a right solution. Quote Link to comment
r3mu503 Posted July 22, 2004 Share Posted July 22, 2004 (edited) it can be solved in a much easier manner well dang man! don't just say it, post it here I did say at the top of my post it was a clunky sol'n Edited August 24, 2004 by -=xerxes=- Quote Link to comment
goku Posted July 26, 2004 Share Posted July 26, 2004 please post your answers na para ma-post ko bagong puzzle about balls Quote Link to comment
PageDown Posted July 26, 2004 Share Posted July 26, 2004 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? Quote Link to comment
r3mu503 Posted July 27, 2004 Share Posted July 27, 2004 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? hi pagedown, actually that's what I started with: 100x + 10(x+k) + (x+2k) + 596 = 100y + 10yc + yc*c arith. progress. on the left (base of x, constant increment of k), geom. progress. on the right (base of y, constant multiplier of c) but I wasn't getting anywhere using the above. I figured I lacked some know-how in number theory or sequences so I tried another approach. I didn't bother including the above anymore because it got me nowhere and don't tell me I don't know math unless you know me, I have a BS in physics so I know enough math at least for that degree Quote Link to comment
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