IIT JEE , AIEEE Forum - Mathematics , Algebra

casio_ss

q. find the sum of all divisors of 9900.

soln. given at back is like dis:-

9900 = 3^2 . 2^2.5^2.11

therfore sum = (3^0+3^1+3^2)(2^0+..+2^2)(..and similarily other two)

now everything had gone above my head...plzzz help..i mean why they hv done like dis...what is the basic idea...there must be sum idea ....and why this expression will evaluate answer??plzz help expertse.g.nadeemoidu,hsbhatta sir?

casio_ss

please help me...i m in dire need...plzzzzz..

anchitsaini

i had learnt of this proof somewhere(most probably over the internet)--

lets start with proof for sum of divisors of p^k where p is a prime number

sum of its divisors should be (p^k+1)/ (p-1)

proof--
divisors of p^k = 1,p,p², ......p^k
hence sum of divisors is a GP =(p^k+1-1) / (p-1)

now if a number can be represented as

n = p₁^r1 . ...... p_k^rk

hence sum of its divisors would be --

( p₁^r1+1 - 1)/ (p₁-1) ... ......( p_k^rk+1 - 1) / (p_k-1)

anchitsaini

according to the above proof--

as
9900 = 3².2².5².11

so sum of divisors should be--

(3^3 - 1)/(3-1)*(2^3 - 1)/(2-1)*(5^3 - 1)/(5-1)* (11^2 - 1)/(11-1)

=33852

casio_ss

i don't understand why ppl post their views...i just asked u to explain the answer which i have already posted..but u guys are posting ur answers...............and only 1 reply....what happened to all others ???koni,hsbhatt,nadeem and many champs here???

hsbhatt

Madam, your champs were fast asleep while you were in distress. Please accept my apologies.

See, any number N can be uniquely factorised into its prime factors as

N = x₁^p x₂^q x₃^r ....

Any factor of N can now be constituted as x₁^p₁ x₂^q₁ x₃^r₁ .... where p₁

p, q₁

q, r₁

r

Another way is to say, that every factor of N is a term in the following expression:

(1+x₁+x₁²+...+x₁^p) (1+x₂+x₂²+...+x₂^q) (1+x₃+x₃²+...+x₃^r)....

Since every factor of N gets listed out, this expression is also the sum of all the factors of N, which is what is shown in your text book.

This can be simplified easily using GP summation formula as

$rac {(x_1^{p+1} -1)} {(x_1-1)} rac {(x_2^{q+1} -1)} {(x_2-1)} rac {(x_3^{r+1} -1)} {(x_3-1)} ...$

anchitsaini

what i posted above is not my own view
by the way
what is the answer to the qn casio_ss??

computer001

i think it is 33,852

nadeemoidu

The question is to explain the expression ( 3⁰ + 3¹ + 3²) ( 2⁰ + 2¹ + 2² ) ( 5⁰ + 5¹ + 5² ) ( 11⁰ +11¹) i.e. how do we get it , not how to simplify it.

Now,
Let us consider a smaller example .

12 = 2².3

Look at the factors of 12 i.e. 1 , 2 , 3 , 4 , 6, 12 or equivalently 1 , 2 , 3 , 2² , 2.3 , 2².3
Note that if we expand the expression ( 1+ 2 + 2²) ( 1 + 3 ) , all these terms are included.

The logic is that we can choose the no. of powers of each prime factor . So if all these powers are included , then we get all the factors. that is what we r doing.

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