# Math Problems!........share It Here!

### #1

Posted 12 July 2004 - 12:02 PM

### #2

Posted 17 July 2004 - 01:04 AM

### #3

Posted 17 July 2004 - 12:38 PM

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

Posted 17 July 2004 - 08:37 PM

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

### #5

Posted 19 July 2004 - 07:08 AM

Ngek. Kala ko kse share your math problems for everyone to solve.Just visit google-blog.dirson.com, the answer s are there.

### #6

Posted 19 July 2004 - 11:53 AM

### #7

Posted 19 July 2004 - 12:21 PM

no . I am not math majors. i love numbers so much that's it......kahit anong field...maka math, chem, physics, accounting,calculus...etcmga math majors ba kayo??

### #8

Posted 19 July 2004 - 04:59 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 intellectualno . I am not math majors. i love numbers so much that's it......kahit anong field...maka math, chem, physics, accounting,calculus...etc

### #9

Posted 19 July 2004 - 05:23 PM

self study lang...mahilig lang talaga me sa numbers.that's itwow, 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

### #10

Posted 19 July 2004 - 05:25 PM

okey. if there has another way of solving it...then post it.Ngek. Kala ko kse share your math problems for everyone to solve.

### #11

Posted 19 July 2004 - 05:45 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?self study lang...mahilig lang talaga me sa numbers.that's it

### #12

Posted 19 July 2004 - 06:13 PM

basta tipo ko lang numbersself-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?

### #13

Posted 19 July 2004 - 07:35 PM

*i am an engineer by profession*

maybe i could help you with this thread

maybe i could help you with this thread

### #14

Posted 19 July 2004 - 07:52 PM

THANX ENGINEER.....YOU ARE WELCOME HERE

i am an engineer by profession

maybe i could help you with this thread

### #15

Posted 19 July 2004 - 08:51 PM

### #16

Posted 20 July 2004 - 02:17 PM

Kayo po bahalaano mag tatayo na ba tayo ng mtc math club?

### #17

Posted 20 July 2004 - 08:29 PM

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

Posted 21 July 2004 - 12:12 PM

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

Posted 21 July 2004 - 06:12 PM

first paragraph pa lang nawala na ako.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

### #20

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*

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

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