Find the sum and number of divisors of 544 excluding 1 and 544.
Find the sum and number of divisors of 544 excluding 1 and 544.
Answer/Solution
589
Steps/Work
if N = p^a * q^b *r^c *...
total no. of factors of N = (a+1)*(b+1)*(c+1)*... &
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 = 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:A
total no. of factors of N = (a+1)*(b+1)*(c+1)*... &
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 = 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:A