PDA

View Full Version : Find the greatest integer that will divide....



kia
05-24-2006, 11:53 AM
Hi. I have a problem where I have to find the greatest integer that will divide into 13,511, 13,903, and 14,589.

I first thought to find the greatest common factor in each number, and divide it by each to see if it returns the same remainder. Didn't work. So, I have this other idea. I could put each number over x and have them equal to r. Then, I figure I could use system of equations to solve for x by canceling r.

Is this a good route to take :?: Thanks for any help :D

pka
05-24-2006, 12:02 PM
You may be trying too hard!
13,903 is a prime number.
Thus the GCF is 1.

kia
05-24-2006, 03:08 PM
Yeah. Would solving it by systems of equations work?

pka
05-24-2006, 03:23 PM
Yeah. Would solving it by systems of equations work?
I am not sure what system one would use.

The whole point is factorization of natural numbers.
Once factored, it is easy to see common factors.
If one number is prime then the GCF is 1.

kia
05-24-2006, 07:50 PM
I'm sorry, I didn't mean system of equations. I meant if I put (13,511/x) = (13,903/y) = (14589 /z). I just tried this way, but I end up with x = x. Any hints would greatly be appreciated.

pka
05-24-2006, 08:00 PM
I'm sorry, I didn't mean system of equations. I meant if I put (13,511/x) = (13,903/y) = (14589 /z). I just tried this way, but I end up with x = x. Any hints would greatly be appreciated.
There is absolutely nothing for you to be sorry about.
The truth is that you should have been expected to use prime-factorization.
It appears that you may not have been exposed to this idea.
You need to peruse that concept. It is basic to GCF.

TchrWill
05-28-2006, 02:54 PM
[quote="kia"]Hi. I have a problem where I have to find the greatest integer that will divide into 13,511, 13,903, and 14,589.

I first thought to find the greatest common factor in each number, and divide it by each to see if it returns the same remainder. Didn't work. So, I have this other idea. I could put each number over x and have them equal to r. Then, I figure I could use system of equations to solve for x by canceling r.

Is this a good route to take :?: Thanks for any help

There was another post with the same problem where the writer gave what answers he finally arrived at. That series of posts seems to have disappeared but I hope the fllowing will be of some help to you.

I don't know whether you,or the other person derived the answer algebraically or not but here is my most recent path to the answer.

From the given information:

1--13511/D = A + r/D or 13511 = DA + r
2--13904/D = B + r/D or 13903 = DB + r
3--14589/D = C + r/D or 14589 = DC + r
4--Solving for r and equating the resulte in 3 steps yield
....13511 - DA = 13903 - DB or D = 392/(A - B)
....13903 - DB = 14589 - DC or D = 1078/(C - A)
....12903 - DB = 14589 - DC or D = 686/(C - B)
5--Therefore, 392/(A - B) = 1078/(C - A) = 686/(C - B)
6--The common divisor is contained within the three numerators.
7--The prime factorization of 392 is 392 = 2^3(7^2)
8--The total number of factors of 392 is therefore f(392) = (3 + 1)(2 + 1) = 12
9--These 12 factors are 1, 2, 4, 7, 8, 14, 28, 56, 49, 98, 196 and 392.
10--The prime factorization of 686 is f(686) = 2^1(3^3)
11-The total number of factors of 686 is therefroe (1 + 1)(3 + 1) = 8
12--These 8 factors are 1, 2, 7, 14, 47, 98, ,343 and 686
13--The prime factorization or 1078 is f(1078) = 2^1(7^2)11^1
14--The total number of factors of 1078 is f(1078) = (1 + 1)(2 + 1)(1 + 1) = 12
15--These factors are 1, 2, 7, 11, 14, 22, 49, 77, 98, 154, 539 and 1078.
16--Note that the highest common factor of all three numbers is 98.
17--13511/98 = 137 + an 85 remainder
18--13903/98 = 141 + an 85 remainder
19--14589/98 = 148 + an 85 remainder.

Looks you, or the other person were right.