64. The number 100 needs to be divided twice so that the larger part from the first division is twice the smaller part from the second division, and the larger part from the second division is three times the smaller part from the first division.
64. Diophantus' Solution. Let the smaller part of the second division be $x$. Then the larger part of the first division will be $2 x$. The smaller part of the first division is $100-2 x$, and the larger part of the second division is $300-6 x$. Both parts of the second division should add up to 100, hence $300-5 x=$ $=100$, from which $x=40$.