If the least common multiple of two 6-digit integers has 10 digits, then their greatest common divisor has at least how many digits?
+118608 Let the numbers be a and b
\(10^5 \le a,b \le 10^6-1\\ so\\ 10^{10} \le ab \le (10^6-1)^2\\ 10^{10} \le ab \le 9.99998*10^{11}\\ \frac{10^{10}}{10} \le\frac{ ab}{10}\qquad \qquad \frac{ ab}{100} \le \frac{9.99998*10^{11}}{100}\\ \text{I only care about the smallest one so ...}\\ 10^{9} \le\frac{ ab}{ \textcolor{red}{10}} \qquad \qquad \frac{ ab}{100} \le 9.99998*10^{9}\)
\(\frac{ab}{10 }\ge \text{10 digits}\qquad \frac{ab}{100}< 11\;\;digits\)
The smallest common divisor must be between 10 and 100 inclusive.
The smallest common divisor is bigger or equal to 10 and 10 has 2 digits.
