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#1 st

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Posted 12 July 2004 - 12:02 PM

Mahirap ba talaga masolve yong problem mo sa Math? Post it here! Thanx!

#2 Brando_Matias

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Posted 17 July 2004 - 01:04 AM

Hmmmm...mukhang walang nahihirapan sa math ahhh????

#3 naked_angel

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Posted 17 July 2004 - 12:38 PM

{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?!?!?

#4 st

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Posted 17 July 2004 - 08:37 PM

{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.

#5 naked_angel

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Posted 19 July 2004 - 07:08 AM

Just visit google-blog.dirson.com, the answer s are there.

Ngek. Kala ko kse share your math problems for everyone to solve. :rolleyes:


:)

#6 rich beem

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Posted 19 July 2004 - 11:53 AM

mga math majors ba kayo??

#7 st

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Posted 19 July 2004 - 12:21 PM

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

#8 r3mu503

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Posted 19 July 2004 - 04:59 PM

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 :)

#9 st

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Posted 19 July 2004 - 05:23 PM

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

#10 st

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Posted 19 July 2004 - 05:25 PM

Ngek. Kala ko kse share your math problems for everyone to solve. :rolleyes:


:)

okey. if there has another way of solving it...then post it.

#11 r3mu503

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Posted 19 July 2004 - 05:45 PM

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?

#12 st

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Posted 19 July 2004 - 06:13 PM

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

#13 xerxes

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Posted 19 July 2004 - 07:35 PM

i am an engineer by profession

maybe i could help you with this thread ;)


#14 st

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Posted 19 July 2004 - 07:52 PM

i am an engineer by profession

maybe i could help you with this thread ;)

THANX ENGINEER.....YOU ARE WELCOME HERE

#15 rich beem

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Posted 19 July 2004 - 08:51 PM

ano mag tatayo na ba tayo ng mtc math club?

#16 st

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Posted 20 July 2004 - 02:17 PM

ano mag tatayo na ba tayo ng mtc math club?

Kayo po bahala

#17 PageDown

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Posted 20 July 2004 - 08:29 PM

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.

#18 r3mu503

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Posted 21 July 2004 - 12:12 PM

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^3
b*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=1
b*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^2
b*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 >= 6
9 >= 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 :)

#19 Chinese Chicken

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Posted 21 July 2004 - 06:12 PM

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^3
b*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=1
b*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^2
b*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 >= 6
9 >= 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. :lol:

#20 xerxes

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Posted 21 July 2004 - 08:35 PM

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^3
b*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=1
b*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^2
b*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    >= 6
9 >= 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 :lol:

Edited by -=xerxes=-, 24 August 2004 - 01:30 PM.





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