A student is solving combinatorics problems. Each day he solves at least one problem. He solves no more than 500 problems a year. Prove that there is an interval of days in which he solves 229 problems.
This is my approach.Lets consider partial sums. A(x) is a sum of problems after x days. Now what is the remainder of A(x) and 229? A(x) changes as days go by, there are only 229 possible remainders and 365 days - so there must be two days x and y, where A(x) and A(y) have the same remainder. A(x) and A(y) represent sums of problems of two overlapping periods. A(x) is the larger period (you can assume this). A(x)-A(y) is divisible by 229 and is positive. A(x)-A(y) also represents a sum of problems of a certain period (since the two periods are overlapping, just take the larger period and disregard the part of the larger period that is the smaller period.
However, I can only show that A(x)-A(y) is divisible by 229, I didnt show that A(x)-A(y) = 229. Is there any way how I can salvage my approach?
The only other value for $A(x)-A(y)$ is $458$, because we know it is less than $500$. Repeat your argument for the first $310$ days of the year, as he must have done less than $458$ problems in that span.