Raymond B. answered 05/27/23
Math, microeconomics or criminal justice
L=3+s
3 more Large (100) boxes than small (10) boxes
2 less left remaining , r clips than L
r=L-2
total clips = 100L+10s +r= 100L+10(L-3) +L-2= 111L-32
where L=number of 100 clip Large boxes
r,s and L are all whole numbers, positive integers, L>3, r>0, s>0
let s=1, r=2, L=4
111(4)-32= 444-32
= 412 = minimum total paper clips