Problem 10 · Cards in boxes with no multiples
A chain $2 \mid 4 \mid 8 \mid 16$ forces it; four boxes suffice.
Integer answer, at most 4 digitsRamon has 20 cards numbered 2 to 21 and wants to place them in boxes so that, if one card's number is a multiple of another card's number, those two cards go in different boxes. What is the minimum number of boxes he needs?
Reasoned solution
Key idea: the cards $2, 4, 8, 16$ are successive multiples of one another: every pair among these four must be separated → at least $4$ boxes are needed.
And four boxes suffice. The key: if $a < b$ are in the same box and $b$ is a multiple of $a$, then $b \ge 2a$. We group into “zones” where this cannot happen:
Within each box, the largest number is less than twice the smallest ($5 < 2\cdot 3$, $10 < 2 \cdot 6$, $21 < 2 \cdot 11$), so no card can be a multiple of another in the same box.