Equal wills

If x is the number, then both 3201-2013 and 2013-1320 must be multiples of x. Therefore, we are looking for the GCD of 1188 and 693. Our answer is 99 .

We let assume that M-r represents the 3 numbers ( 2013, 1320, 3201) where M is divisible by N and r is the remainder.

We subtract the larger number to the number next to it to eliminate the remainder and get a larger value of divisible by N

(M2-r) - (M1-r) = M2-M1 , (M3-r) - (M2-r) = M3-M2

so

3201 - 2013 = 1188 , 2013 - 1320 = 693

Then we get both their prime factors

so 1188 = (11)(3)(3)(3)(2)(2) , 693 = (11)(3)(3)(7)

From these we can say that they have both (11)(3)(3) which is equals to 99

then N = 99

We can state the problem as:

2013\congr (mod N) ... (1)

3201\congr (mod N) ... (2)

1320\congr (mod N) ... (3)

Subtracting the three equations we get:

(2)-(1) ... 1188\cong0 (mod N)

(2)-(3) ... 1881\cong0 (mod N)

(1)-(3) ... 693\cong0 (mod N)

Now if we take the [GCD] of the three numbers we get 99 and that is the largest number that will give you a reminder of r when 2013, 1320 and 3201 are divided by it.

Let's generalize. We can find the largest integer value that leaves the same remainder between two numbers doing their difference.

So the problem goes about finding the gcd of the differences of the three numbers

$gcd(3201-2013,3201-1320,2013-1320)=$

$= gcd(1188,1881,693) = \boxed{99}$

We know that "N" must divide 2013-r, 1320-r, and 3201-r

From this, we can deduce that it can also divide 2013-1320, 3201-1320, and 3201-2013 because it fits an even amount of times (without extraneous remainder) 2013-r and 3201-r

The prime factorizations of 2013-1320 (693), 3201-1320 (1881), and 3201-2013 (1188) are, respectively, ) 3^2 x 7 x 11, 11 x 19 x 3^2, and 2^2 x 3^2 x 11.

The term 11x3^2, 99, appears in all three of these numbers.

Therefore, our solution is 99.

2013 - 1320 = 693

3201 - 2013 = 1188

3201 - 1320 = 1881

N = Greatest common divisor of 693, 1188, and 1881 = 99.

2013 ≡ 33 mod 99

1320 ≡ 33 mod 99

3201 ≡ 33 mod 99

2013 = 3 x 11 x 61 1320 = 2 x 2 x 2 x 3 x 5 x 11 = 3 x 11 x 40 3201 = 3 x 11 x 97 they have same product = 3 x 11 = 33 when the reminder of each product divided by 3, the equation will get the same reminder = 1 So, N = 33 x 3 = 99

We know $1320 = Nk+r$ $2013 = Np+r$ $3201= Nv + r , v, p, k$ are +integers whoose gcd is one.

our required answer is the $gcd (3201-2013, 3201-1320, 2013 - 1320 )$ as it eliminates r.

$2013-1320=693$

$3201-2013=1188$

To let the three numbers leave the same remainder,the largest possible value of $N=\gcd(693,1188)=99$

Ans= $99$

For a number N has the property to make that the divisions of 2013, 1320 and 3201 by N leave the smae remainder, then N must divide their differences; namely:

1. N must divide 2013 - 1320 = 693 = 3² x 11 x 7.
2. N must divide 3201 - 2013 = 1188 = 3³ x 11 x 2².
3. N must divide 3201 - 1320 = 1881 = 3² x 11 x 19.

Therefore the greatest number N for which the remainders are going to be the same is 3² x 11 = 99.

$2013 - 1320 = 693$ $3201 - 2013 = 1188$ $3201 - 1320 = 1881$

$N$ = Greatest common divisor of $693, 1188$ , and $1881 = \boxed{99}$ .