The Counting Competition

A teacher writes a whole number and its square on the board and asks three pupils to count the total number of digits in the two numbers together. Anil says $999$, Brijesh says $1000$, Chandra says $1001$. The teacher says that one of them cannot possibly be right, whatever the number was. Which one, and why?

Solution

Brijesh, with $1000$, must be wrong.

Suppose the number has $d$ digits. Then its square has either $2d$ or $2d - 1$ digits: squaring roughly doubles the length, dropping one digit only when the leading figures are small enough not to carry into a new place. So the combined count is either $$d + 2d = 3d \qquad \text{or} \qquad d + (2d - 1) = 3d - 1.$$ Every possible answer is therefore a multiple of three, or one less than a multiple of three. Now $999 = 3 \times 333$ is a multiple of three, and $1001 = 3 \times 334 - 1$ is one less than a multiple of three, so both could occur. But $1000$ is neither: it is one more than the multiple $999$, a count of the forbidden kind. So Brijesh’s total can never arise, and he gets no prize.