Remaining Remainder

The four numbers $1234 - r, 2340 - r, 3446-r, 5500-r$ are all divisible by $n$ .

Then, the difference between any 2 of these 4 numbers above are all also divisible by $n$ .

We have $(2340-r) - (1234- r) = 1106 \mod n = 0 , \quad \quad (5500-r) - (3446-r) = 2054 \mod n = 0$

In other words, both 1106 and 2054 are multiples of $n$ .

Since $1106 = 2\times7\times79,$ all the factors of 1106 are $1,2,7,14,79,158,553, 1106$ .

Similarly, $2054 = 2\times13\times79,$ all the factors of 2054 are $1,2,13,26,79,158,1027,2054$ .

We have narrowed down the possible values of $n$ to $1,2,79,158$ .

However, if $n = 1$ , then $r = 0$ is forced, but this we want a positive remainder $r$ . So $n=1$ is rejected. We reject $n=2$ for the same reason.

We are left with $n = 79,158$ .

Performing long division shows that when $1234,2340,3446,5500$ are all divided by 79, the remainders are all equal to 49.

Analogously, performing long division shows that when $1234,2340,3446,5500$ are all divided by 158, the remainders are all equal to 128.

Hence, all the possible values of $(r,n)$ are $(49,79), (128,158)$ . The answer is $49+79+128+158 = \boxed{414} .$

Too much work? Just want to write a script? Here you go:

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# The divisor N cannot exceed 5500, otherwise, the remainders will not all be distinct.

answer = 0
for n in range(1, 5500):
    if 1234 % n == 2340 % n == 3446 % n == 5500 % n:
        r = 1234 % n
        if r > 0 :
            answer += n + r

print(answer) # Output: 414