+0

GCD/LCM

0
59
3

If the least common multiple of two 6-digit integers has 10 digits, then their greatest common divisor has at least how many digits?

Nov 27, 2021

#1
0

four

Nov 27, 2021
#2
0

LCM(123450, 987650)=2438507850

GCD(123450, 987650)==50

Nov 27, 2021
#3
+115920
+1

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.

Nov 28, 2021