5. Initially, the number 10 is written on the board. In each subsequent moves, you can either (i) erase the number 1 and replace it with a 10 , or (ii) erase the number 10 and replace it with a 1 and a 25 or (iii) erase a 25 and replace it with two 10 . After sometime, you notice that there are exactly one hundred copies of 1 on the board. What is the least possible sum of all the numbers on the board at that moment?
5. Suppose the number of times that operations (i), (ii) and (iii) have been performed are $x, y$ and $z$, respectively. Then the number of 1,10 and 25 are $y-x, 1+x-y+2 z$ and $y-z$, respectively, with $-x+y=100$. Thus the sum is
$$
S=y-x+10(1+x-y+2 z)+25(y-z)=-890+5(5 y-z) .
$$

Since we want the minimum values of $S, y$ has to be as small as possible and $z$ as large as possible. Since
$$
y-x=100,1+x-y+2 z \geq 0, \quad y-z \geq 0
$$
we get, from the first equation, $y \geq 100$, from the second inequality, $2 z \geq 99$ or $z \geq 50$ and $y \geq z$ from the third. Thus the minimum is achieved when $y=100, x=0$ and $z=100$. Thus minimum $S=1100$.