<s>[INST] Number of digits of $xy$ is either $m+n$ or $m+n-1$ Given positive integers $x$ and $y$ with $m$ and $n$ digits respectively, must $xy$ have either $m+n$ or $m+n-1$ digits?
If $x$ has $m$ digits, then $10^{m-1} \le x < 10^m$. Likewise, if $y$ has $n$ digits, then $10^{n-1} \le y < 10^n$. Multiplying those two inequalities together (which is OK because all the numbers involved are positive), one gets $10^{m+n-2} \le xy < 10^{m+n}$.
So, from $10^{m+n-2} \le xy < 10^{m+n}$, one could conclude that $xy$ has either $m+n$ or $m+n-1$ digits, right?
 [/INST] Yes, you are right.
From $10^{m-1} \le x$ and $10^{n-1} \le y$ it follows that $10^{m+n-2} \le xy$, so $xy$ has at least $m+n-1$ digits. Also, from $x<10^m$ and $y<10^n$ it follows that $xy < 10^{m+n}$. Since $10^{m+n}$ is the least integer having $m+n+1$ digits, $xy$ has less than $m+n+1$ digits and thus $xy$ has at most $m+n$ digits.
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