Question

Find the highest power of $3$ which is contained in $|\underset{–}{100}$.

Answer 1

Of the first $100$ integers, as many are divisible by $3$ as the number of times that $3$ is contained in $100$, that is, $33$; and the integers are $3,6,9,...99$.
Of these, some contain the factor $3$ again, namely $9,18,27,....99$, and their number is the quotient of $100$ divided by $9$.

Some again of these last integers contain the factor $3$ a third time, namely $27,54,81$, the number of them being the quotient of $100$ by $27$. One number only, $81$, contains the factor $3$ four times.
Hence the highest power required $=33+11+3+1=48$.
This example is a particular case of the theorem investigated in the next article.