Question

k个连续正整数之和为2012则k的最大值是?

Answer 1

设最小数为k,则连续k个正整数为k,k+1,k+2...,2k-1

k(1+2+...+k-1)=2012

k^2(k-1)/2=2012

k^2(k-1)=4024

k最大为16.