Problem
COM-B1-M11-P024 Sum Divisible by \(20\)
#24
★★★★★ Level 5 of 5
Prove that among any \(20\) integers, one can choose several consecutive numbers in the given order whose sum is divisible by \(20\).
Consider partial sums modulo \(20\).
Let the numbers be \(a_1,\ldots,a_{20}\). Consider partial sums \(S_k=a_1+\cdots+a_k\) for \(k=1,\ldots,20\). If some \(S_k\) is divisible by \(20\), then the segment \(a_1,\ldots,a_k\) works. If none is divisible by \(20\), then the \(20\) partial sums have only \(19\) nonzero residues modulo \(20\). Hence two of them, \(S_i\) and \(S_j\) with \(i
Strong mixed problem: pigeonhole plus partial sums.