Find the sum and number of divisors of 544 excluding 1 and 544.

• 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
  
• 10 years agoHelpfull: Yes(59) No(8)
• 544 can be written as 2^5*17
  no. of divisors of 544 will be 6*2 i.e. 12
  and the sum will be equal to
  (2^0+2^1+2^2+2^3+2^4+2^5)(17^0+17^1)
  i.e. 1134
  but as the question says 1 and 544 are to be excluded so 1134-(1+544)=589(ans.)
• 10 years agoHelpfull: Yes(14) No(1)
• 544 CAN BE WRITTEN AS 2^5*17^1 SO ADD 1 TO POWERS AND MULTIPLY THEN (5+1)(1+1)=12 EXCLUDE 1 AND 544 THEN 12-2=10.
• 10 years agoHelpfull: Yes(8) No(4)
• if N = x^a * y^b *z^c *...
  total no. of factors of N = (a+1)*(b+1)*(c+1)*... &
  sum of all of the factors of N =(1+x+x^2+...+x^a)*(1+y+..+y^b)*(1+z+z^2+...+z^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
• 10 years agoHelpfull: Yes(1) No(0)
• HOW??? number of divisors of 544 excluding 1 and 544 = (5+1)*(1+1) - 2 = 10
• 9 years agoHelpfull: Yes(0) No(0)