Subsequences Summing to Sevens

Problems and Contests
#1

In this problem how will you say that (b-a)%7 = b%7 - a%7
b,a are 2 values in prefix sum

#2

It’s basic modular arithmetic: is there a counterexample you came up with or anything?

#4

(b-a)%7 = (b%7 - a%7)%7 right ?

#5

I believe so. It might not be true in C++/Java because their % operator is a remainder operator rather than a true modulus operator, but I’m not sure.

#6

Let there be a subarray (i, j) whose sum is divisible by k
sum(i, j) = sum(0, j) - sum(0, i-1)
Sum for any subarray can be written as q*k + rem where q
is a quotient and rem is remainder
Thus,
sum(i, j) = (q1 * k + rem1) - (q2 * k + rem2)
sum(i, j) = (q1 - q2)k + rem1-rem2

We see, for sum(i, j) i.e. for sum of any subarray to be
divisible by k, the RHS should also be divisible by k.
(q1 - q2)k is obviously divisible by k, for (rem1-rem2) to
follow the same, rem1 = rem2 where
rem1 = Sum of subarray (0, j) % k
rem2 = Sum of subarray (0, i-1) % k