Different divisors gets the same remainder!

The difference of the two numbers must be divisible by that three digit number, since the remainders have eliminated.

$34369-31513=2856$

Three digit factors of 2856 include 168, 357, 714... all of them yields the same remainder when dividing 34369 which is $\boxed{97}$

(Method without using a calculator.)

If $31513 \equiv 34369$ mod $n$ , then $n$ must be a divisor of their difference, which is $2856$ . It is easy to see that this goes about 11 times in the first given number and 12 times into the second given number; indeed, $31513 - 11\cdot 2856 = 97; \\ 34369 - 12\cdot 2856 = 97.$ Therefore the remainder after division by $n$ will be $97\ \text{mod}\ n$ ; because $n > 97$ , the remainder will be exactly $\boxed{97}$ .

Others have written about the nice property of modulo (subtract the numbers and the remainder must be a divisor of the difference). But for the curious, those 3 digit numbers are 102,119,136,168,204,238,357,408,476,714 and 952.

34369 - 31513 = 2856

Factorising we get 3 digit no. as .... 714 , 357 , 119. All are multiple or sub-multipleof each other. Remainder for 31513 / 714 = 97

Background logic :

Let x be the three digit number. Let

mx + r = 31513 .............. (1)

And nx + r = 34369 ........ (2)

Where m , n , r >=0 & m , n , r € Z+

Subtracting (2) from (1) : x * (n-m) = K (say) Here K = 2856 = 2 x 2 x 2 x 3 x 119

There may be many such x-s with three digits that make K as a factor of it.. they all are multiples or submultiples of that particular x.

And all such numbers will hv same remainder r, with both given numbers.

I didn't use any fancy mathematics here. We know that r + xd = 31513 and r + yd = 34369. Subtracting the first equation from the second gives (y-x) d = 2856. (This is where the people using modulo arithmetic get 2856 from.)

2856 factors into 2^3 * 7 * 51, and we know d divides that. The possible divisors are therefore 357 (7 * 51), 408 (8 * 51) and 714 (2 * 7 * 51). Any of those .divisors yields 97 as the remainder. Not as elegant a solution as some, but also not dependent on any mathematical knowledge beyond algebra.

The difference of the numbers 34369 and 31513 must be divisible by the three digit number. The difference is 2856. So we get 2856=(the three digit number )*(an integer). The prime factorization of 2856 is 3X8X7X17.. As we know that 31513 and 34369 are not divisible by 3 and 8. So the three digit number must be 7X17 which is 119. The remainder is therefore 31513 mod 119=97.

34369 mod x= 31513 Subtract 31513 from the left side to yield 2856 mod x=0. All of the three digit factors of 2856 will result in the same remainder, which is 97.

31513=r(mod a)
34369=r(mod a) Subtracting, 2855=0(mod a) Therefore, 2855*11=0(mod a) 31416=0(mod a) Therefore, 34369=97(mod a)

There are 11 numbers that fulfill this condition, they are 102, 119, 136, 168, 204, 238, 357, 408, 476, 714, 952. Interestingly remainder in all cases is 97.

It is possible to write a simple program in order to solve this problem in python:

for i in range(100,1000):
    rem1=31513%i
    rem2=34369%i
    if rem1 == rem2:
        print(rem1)
        break

I wrote a code in C# to solve this problem :

for(int i=100 ; i<=999; i++){

 if(31513%i == 34369%i ){

       the_answer=31513%i ; 
       breack;

} }

I have found the number using a graph:

• define $x_{n}$ = mod(34369,n) and $y_{n}$ = mod(31513,n)

• define $z_{n}$ = $x_{n}$ - $y_{n}$

Then run a cycle for $n$ =100 until $n$ =999 to generate a matrix of $z_{n}$ . Plot all $z_{n}$ on y-axis and $n$ on x-axis. Every $n$ that gives $z_{n}$ = 0 is a 3-digit number divided by which the reminders of 34369 and 31515 are equal. Pick one of the numbers (zoom in), for example 102, to get mod(34369,102)=97 and mod(31513,102)= $\boxed{97}$ .

Here is the plot with y-axis limits set to (-1,1):

Question reformatted as a modular equavaluence is $31513 \equiv 34369\ (\textrm{mod}\ n)$ . Substract left side from the eqaution for: $2856 \equiv 0\ (\textrm{mod}\ n)$ , which means that $n$ can be $2856$ as $x \equiv 0\ (\textrm{mod}\ x)$ . Since we are looking for a 3-digit number, we can use $714$ because $2856 = 714 \cdot 4$ and if two numbers are congruent modulo $x\cdot y$ , they will be congruent modulo $x$ . The remainder of the division $31513 \div 714$ is $97$ .