Question

Find the sum of the divisors of the number $21600$

• A. $78122$
• B. $78120$
• C. $74462$
• D. $64221$

Answer 1

The correct option is B $78120$

$21600:{\left(2\right)}^5\times {\left(3\right)}^3\times {\left(5\right)}^2$

Number of factors

$=6\times 4\times 3=72$

The factors of $21600$ are

$1,2,3,4,5,6,7,8,9,10,12,15,16,18,20,$

$24,25,27,30,32,36,40,45,48,50,54,$

$60,72,75,85,90,96,100,108,120,135,$

$144,150,160,180,200,216,225,240,270,$

$288,300,360,400,432,450,480,540,600,$

$675,720,800,864,900,1080,1200,1350,1440,$

$1800,2160,2400,2700,3600,4320,5400,7200,$

$10800,21600$

Sum $=78120$