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Peegee
03-30-2005, 03:07 PM
xx xxx xxx / xxx = xx8xx

Solve for the numbers (no, x is not the same digit throughout). Here, I'll give some hints:

http://img.photobucket.com/albums/v250/Pureghetto/omghi2u.jpg

Megami
03-30-2005, 03:12 PM
Sorry I am not understanding :confused:

Peegee
03-30-2005, 03:13 PM
I talking good english for you. Please to solving the probleming you giving a clue for win.

Zeldy
03-30-2005, 03:13 PM
...?
does x =... x?

Peegee
03-30-2005, 03:17 PM
No, like I said, x is not a digit.

?? ??? ??? / ??? = ??8??

better?

Polaris
03-30-2005, 03:17 PM
is that relativce with math or smth?

Megami
03-30-2005, 03:17 PM
I can't understand what x is

Polaris
03-30-2005, 03:18 PM
x are numbers that u have to figure out!

Caspian
03-30-2005, 03:23 PM
Hahahaha, the fact that you posted this here of all places is pretty amusing.

Zeldy
03-30-2005, 03:23 PM
Like..where do u begin..?

Megami
03-30-2005, 03:27 PM
Like..where do u begin..?

yeah! :(

Peegee
03-30-2005, 03:28 PM
Here, I'll give you a start:

xxx * x = xxx

xxxx - xxx = xx

A four digit number subtracting a three digit number equals a two digit number? The first digit in the numerator must be a one.

Zeldy
03-30-2005, 03:30 PM
The first digit in the numerator must be a one.
Oh ok.. if you say so..?

Megami
03-30-2005, 03:32 PM
I think I give up :cry:

Peegee
03-30-2005, 03:33 PM
...what do they teach you people at school?

Clout and co, please save me :(

Megami
03-30-2005, 03:36 PM
...what do they teach you people at school?

Clout and co, please save me :(

Some things I can't remember right now

Caspian
03-30-2005, 03:45 PM
Well, so far I've got

xx xxx xxx / xxx = x080x

based on the dropping of two digits in the long division part. I can also make some educated guesses on the first few digits of the divisor.

Zeldy
03-30-2005, 03:47 PM
Some things I can't remember right now
Same.. :rolleyes2

chionos
03-30-2005, 05:18 PM
You do realize that there is more than one answer, right? like
10 408 000/500 = 20816
14 327 901/369 = 38819
et cetera
et cetera
The first 5-digit number multipilied by a 3-digit number that produces an 8-digit number is 10011. 10 000 989/999 = 10011
every increase in 10 by the 5-digit number allows the 3-digit number to drop by 1 at first.
10 000 958/998 = 10 021
10 000 907/997 = 10 031
...
10 000 181/991 = 10 091
then it jumps by 11, then back to an increase of 10
10 000 980/990 = 10 102
10 000 768/989 = 10 112
@985 the pattern changes again as:
9 999 720/985 = 10 152
the lowest possible product that fulfills your equation is
10 000 800/926 = 10 800

by the time we get to 20 000 the 3-digit number has fallen to 500
10 000 000/500 = 20 000
so any number > 20,000 multiplied by any number > 500 produces an 8-digit number. now all that is required is to pop an 8 into that 3rd digit.

the only 3-digit number insidentally that cannot create an 8-digit product when multiplied by a 5-digit number is 100.
100 * 99 999 = 9 999 900
but 101 * 99 999 = 10 099 899
a better question would be, How many 5-digit numbers with 8 in the hundredths place can create an 8-digit product when mulplied by a 3-digit number.
which is 9801, I believe.
I think maybe what you were going for if you wanted just one answer is something like ax bzd ydc / azb = by8ex where each letter stands for a specific number. And that one, I don't feel like solving. =P

Megami
03-30-2005, 05:19 PM
I feel so... :confused: :cry:

Caspian
03-30-2005, 05:22 PM
You do realize that there is more than one answer, right? like
10 408 000/500 = 20816
14 327 901/369 = 38819
et cetera
et cetera
The first 5-digit number multipilied by a 3-digit number that produces an 8-digit number is 10011. 10 000 989/999 = 10011
every increase in 10 by the 5-digit number allows the 3-digit number to drop by 1 at first.
10 000 958/998 = 10 021
10 000 907/997 = 10 031
...
10 000 181/991 = 10 091
then it jumps by 11, then back to an increase of 10
10 000 980/990 = 10 102
10 000 768/989 = 10 112
@985 the pattern changes again as:
9 999 720/985 = 10 152
the lowest possible product that fulfills your equation is
10 000 800/926 = 10 800

by the time we get to 20 000 the 3-digit number has fallen to 500
10 000 000/500 = 20 000
so any number > 20,000 multiplied by any number > 500 produces an 8-digit number. now all that is required is to pop an 8 into that 3rd digit.

the only 3-digit number insidentally that cannot create an 8-digit product when multiplied by a 5-digit number is 100.
100 * 99 999 = 9 999 900
but 101 * 99 999 = 10 099 899
a better question would be, How many 5-digit numbers with 8 in the hundredths place can create an 8-digit product when mulplied by a 3-digit number.
which is 9801, I believe.
I think maybe what you were going for if you wanted just one answer is something like ax bzd ydc / azb = by8ex where each letter stands for a specific number. And that one, I don't feel like solving. =P

My thoughts exactly.

Chemical
03-30-2005, 05:50 PM
xx xxx xxx / xxx = xx8xx

Solve for the numbers (no, x is not the same digit throughout). Here, I'll give some hints:

http://img.photobucket.com/albums/v250/Pureghetto/omghi2u.jpg

Perhaps to clarify the problem, you could substitute values for letters.
That would create a better understanding of what is similar and disimilar.

IE)

af dba ehj / jij = hc8ha

So that a = all the same number and so on.

mateoff8
03-30-2005, 05:56 PM
"lmao"

udsuna
03-30-2005, 06:10 PM
No, what are they teaching YOU in school... you cannot solve for any numbers without establishing the difference of the numbers. Basic arithmatic, which is what this problem's supposed to be, doesn't work that way. You have to create a different value for each number, or provide at least two actual digit values, to make this formula work.

Peegee
03-30-2005, 06:10 PM
Arbitary solutions will not fit into my non-arbitary diagram.

edit: it is solvable. Do it.

Samuraid
03-30-2005, 06:55 PM
Working on the solution now. I hopefully have something.

Misfit
03-30-2005, 06:58 PM
Arbitary solutions will not fit into my non-arbitary diagram.
Dork.








jkjkjk! ^__^

Samuraid
03-30-2005, 07:13 PM
Take your pick:

12 256 256 / 256 = 47876
15 320 320 / 320 = 47876
18 384 384 / 384 = 47876
13 625 625 / 625 = 21801
18 625 625 / 625 = 29801
14 640 640 / 640 = 22876

Any of them satisfy the original equation. (EDIT: and PG's statement that the first digit in the numerator had to be a 1)

Peegee
03-30-2005, 08:00 PM
Does it satisfy the digit placement of the image?

SeeDRankLou
03-30-2005, 09:01 PM
10,020,316/124=80,809

Here's how I got this:

First, I see that you drop two numbers instead of one after the first and second subtraction, indicating that the numbers before and after the 8 in the quotient are 0. In both the first and second subtraction, a 4-digit number minus a 3-digit number results in a 2-digit number. The only way that's possible is if the first two numbers of the of the 4-digit number are 10 and the first number of the 3-digit number is 9. Furthermore, since 8 is being multiplied by a 3-digit number to get a 3-digit number, the first number of the denominator must be 1. And, since 8*the denominator results in a 900 number, the second digit of the denomintor must be a 1 or 2. If it is a 1, then the third digit must be at least a 3 to get a 900 number from the product, and if it is a 2 the third digit can be at most a 4. The first two subtractions must result in a 2-digit number, meaning that the product must be at least 920, so that at least 1010-920=10. Thusly, 113 and 114 are two low for the denominator. So the denominator is a number from 115 to 124. Alright, since the last multiplication results in a number larger than the first two, the last digit of the quotient must be a 9. And in the highest possibility, 7*124=868, not a 900 number. So the first digit of the quotient must also be 8. So now I know the quotient is 80,809, and the possible denominators range from 115 to 124. Now simply multiply the quotient by the possible denominators, and the only one that gives an 8-digit product is 124. 80,809*124=10,020,316. If you do the long division you will see that all the numbers line up correctly.

Psychotic
03-30-2005, 09:04 PM
42.

Levian
03-30-2005, 09:32 PM
I hate this thread's existance.

Caspian
03-30-2005, 09:47 PM
SeeDRankLou confirmed what I wagered earlier, so I'll agree with him.

Samuraid
03-30-2005, 11:26 PM
SeeDRankLou's method seems to be good. I assumed that all instances of xxx in the equation had to be the same number. (likewise with xx, x, and xxxx) If that were the case, any of my solutions would work not counting the extra long division "hints".

Dixie
03-30-2005, 11:57 PM
I know the answer, but I'm not going to ruin everything.:)

Edit: And I won't kill the thread.

Peegee
03-31-2005, 11:29 AM
yay to seedranklou!

Meat Puppet
03-31-2005, 11:38 AM
7

Rase
03-31-2005, 05:52 PM
42.


Darn it, beat me to it!