Finding an integer that will go into three other integers

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nami

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Hello. The problem I was given was to find the greatest integer that will divide three numbers: 13,511, 13,903, and 14,589 with each having the same remainder.

I started out by saying a = 13,511, b = 13,903, and c = 14,589. From there, I constructed three equations: a = xd +r, b = yd+r, and c = zd+r. D is the number that will divide into each number, and r is the remainder.

I cannot figure out where to go from here. If any one can provide help, I would GREATLY appreciate it.
 
No. This is my first time. Is there someone else that has the same problem as me? If so, are they having any luck?
 
Re: Finding an integer that will go into three other integer

nami said:
Hello. The problem I was given was to find the greatest integer that will divide three numbers: 13,511, 13,903, and 14,589 with each having the same remainder.

I started out by saying a = 13,511, b = 13,903, and c = 14,589. From there, I constructed three equations: a = xd +r, b = yd+r, and c = zd+r. D is the number that will divide into each number, and r is the remainder.

I cannot figure out where to go from here. If any one can provide help, I would GREATLY appreciate it

Well, your first set of equations are not quite correct.

If the greatest divisor is X and the remainders r, you have

1--13511/D = A + r
2--13903/D = B + r (Sorry for the typo)
3--14589/D = C + r (Sorry for the typo)
4--Solve all three for r and equate to each iother in 3 equations resulting in
5--X = 398/(B - A)= 1078/( C -- A) = 686/(C - B)

See what you can do with these.
Click to expand...
 
As I said in the other posting of this problem: What a dumb problem!
13903 is a prime number, so the question is over: The GCF is 1.
I bet some one with a degree in mathematics education wrote this problem.
 
I'm sorry, but I am not grasping how you obtained the equations X = 398/(B - A)= 1078/( C -- A) = 686/(C - B). For the first equation, when I solve for r, I obtain -A + 13511/D.

The same goes for the others.I think there is something I am not seeing. Thank you so much for your help.
 
nami, the answer is 1.
Because one of the given numbers is prime, the answer is 1.
You need to confront your teacher.
Ask why you were given such a meanless question?
Tell that person that this pointless question.
 
Our teacher told us that the answer isn't one. She said that each number can be divided by the same number and give the same remainder.
 
Re: Finding an integer that will go into three other integer

nami said:
Hello. The problem I was given was to find the greatest integer that will divide three numbers: 13,511, 13,903, and 14,589 with each having the same remainder.

I started out by saying a = 13,511, b = 13,903, and c = 14,589. From there, I constructed three equations: a = xd +r, b = yd+r, and c = zd+r. D is the number that will divide into each number, and r is the remainder.

I cannot figure out where to go from here. If any one can provide help, I would GREATLY appreciate it

Sorry for the earlier typos.

Well, your first set of equations are not quite correct.

If the greatest divisor is X and the remainders r, you have

1--13511/D = A + r
2--13903/D = B + r (Sorry for the typo)
3--14589/D = C + r (Sorry for the typo)
4--Solve all three for r and equate to each iother in 3 equations resulting in
5--X = 398/(B - A)= 1078/( C -- A) = 686/(C - B)

See what you can do with these.

Does anything prevent a prime number from being divided by a number and leaving a remainder?
Click to expand...
 
Thank you so much! I came up with the divisor being 98 and the remainder being 85. I hope the other individual got the same answers I did as well.
 
Re: Finding an integer that will go into three other integer

nami said:
Hello. The problem I was given was to find the greatest integer that will divide three numbers: 13,511, 13,903, and 14,589 with each having the same remainder.

I started out by saying a = 13,511, b = 13,903, and c = 14,589. From there, I constructed three equations: a = xd +r, b = yd+r, and c = zd+r. D is the number that will divide into each number, and r is the remainder.

I cannot figure out where to go from here. If any one can provide help, I would GREATLY appreciate it.

I don't know whether you 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--Slving 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 were right. Nice going.
Click to expand...
 
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