The Marble Conundrum

Suppose Rajiv uses $n>0$ bags. Let the number of marbles in each bag be $m_1, m_2, \ldots m_n$ . Reorder the numbers so that they are in strictly increasing order (since the number of marbles is distinct from each other), to obtain $1 \leq m_1 < m_2 < m_3 < \ldots m_n$ . Since each of these numbers are integers, we can conclude (inductively) that $m_i \geq i$ . Hence, we have

$1000 = m_1 + m_2 + \ldots + m_n \geq 1 + 2 + \ldots + n = \frac {n(n+1)}{2}.$

Since $44 \times 45 = 1980$ and $45 \times 46 = 2070$ , this gives that $n \leq 44$ (since $n$ is an integer).

To see that $44$ bags is possible, we have $m_1 = 1, m_2 =2 , \ldots m_{43} = 43$ , and then place the remaining $54$ marbles in the last bag. Hence, the maximum number of bags that Rajiv will use is $44$ .