Latest TCS Aptitude Question SOLUTION: Find the sum and number of divisors of 544 excluding 1 and 544.

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

Asked In TCS Subrata Patra (10 years ago)

Solved RAKESH Read Solution (5)

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Read Solution (5) :

RAKESH  (10 years ago)

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

Is this solution Helpfull? Yes (59) | No (8)  

himanshu shekhar  (9 years ago)

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.)

Is this solution Helpfull? Yes (14) | No (1)  

prem kumar  (9 years ago)

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.

Is this solution Helpfull? Yes (8) | No (4)  

Tarun Sinha  (9 years ago)

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

Is this solution Helpfull? Yes (1) | No   

Surajit  (9 years ago)

HOW??? number of divisors of 544 excluding 1 and 544 = (5+1)*(1+1) - 2 = 10

Is this solution Helpfull? Yes | No   

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