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{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?!?!?

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{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.

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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 :)

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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

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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 :)

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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:

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Guest -=xerxes=-
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=-
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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 digit

So the number would be 100x+10y+z = the number

 

This is how you solve a math problem... Ok?

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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 digit

So 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

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