Number Theory - Same remainder no division by multiple numbers

Question

How many natural numbers divide 2500, 4250 and 6700 leaving the same remainder in each case?

My approach to this question was

2500 = N.f1 + r 4250 = N.f2 + r 6700 = N.f3 + r

Hence N has to be a multiple of (6700 - 4250) , (6700-2500), and (4250-2500)

Hence, N should be a multiple of 2940 (LCM of the above mentioned numbers)

I'm stuck here.

Any help will be appreciated

Thank You

Answer 1

Think about the path you are on: $10$ satisfies this property and is not a multiple of $2940$.

We are looking for values of $n$ for which $2500\equiv 4250$ and $4250\equiv 6700$ mod $n$.

The first congruency implies $n|1750$, and the second implies $n|2450$. Thus $n|2450-1750$ or $n|700$.

There are only $18$ divisors of $700$, so you should be able to check them all.

Answer 2

I'd made a mistake

The approach leads to the fact that N has to be a factor of 1750 , 2450 and 4200

That means 350 is the largest value that N can take(HCF of the above mentioned numbers)

Since 350 has 12 factors, 12 is the answer