Find the sum and number of divisors of 544 excluding 1 and 544.
Options
A) 587  B) 588  C) 589  D) 590  E) 591
if N = p^a * q^b *r^c *...   total no. of factors of N = (a+1)*(b+1)*(c+1)*... &amp;   sum of all of the factors of N =(1+p+p^2+...+p^a)*(1+q+..+q^b)*(1+r+r^2+...+r^c)   544 = 2^5 * 17   number of divisors of 544 excluding 1 and 544 = (5+1)*(1+1) - 2= 
<gadget id="calculator">(5+1)*(1+1) - 2</gadget>
<output>10</output>
10   sum of divisors of 544 excluding 1 and 544= (1+2+2^2+2^3+2^4+2^5)*(1+17)-1-544   = 589   ANSWER:589
<result>C</result>
Find the sum and number of divisors of 544 excluding 1 and 544.