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Lena how did u get the prime factors of 100000 as 2^6 * 5^6

I am getting 2^5 * 5^5

You are getting the right prime powers. There is a formula for the total sum of all factors: present the number n in prime factorization: \(n=a^i\times b^j\)

the the sum of all factors of \(n = \frac{a^{i+1}-1)}{a-1}\times \frac{b^{j+1}-1}{b-1}\) i just gave an example with n with only 2 unique primes but it can be generalized to any number of primes

Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]
05 Sep 2009, 17:04

LenaA ..I have got a question for you..we can solve this using the formula..but how did you simplify the equation .

\frac{2^6-1}{2-1}\times \frac{5^6-1}{5-1}=246078

Seems solving this equation itself is 600 score question and will consume more time..I tried solving this using the factor table method and it took around 2 mins to solve this . _________________

FEB 15 2010 !!

well I would not disturb you after the D-day ..so please !!!

It Will Snow In Green , One Day !!!!

Last edited by Snowingreen on 05 Sep 2009, 17:09, edited 1 time in total.

Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]
05 Sep 2009, 17:06

2

This post received KUDOS

1) Formula The formula gives you the sum of all factors of a number. Take example, \(n=24=2^3\times 3\) 1,2,3,4,6,8,12,24. They sum is 60. Now try to use the formula :

2) why deduct powers of 2 and 5 and integer1? we are asked not about the total sum of factors but only a subset of factors, i.e. multiples of 10. since powers of 5 and powers of 2 and 1 are not multiples of 10 you have to deduct them from the total sum of factors. Basically I calculated the total sum of factors and then deducted not relevant factors to get the sum of factors that are multiples of 10...I could not see the other way to do that.

3) As for simplifying the expression, you just have to calculate it... you just have to remember some powers...if you remeber that 2^5=32 and 5^4=625...it makes it easier to calculate, even though it was still time consuming , you get \(63\times 3906\)...

Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]
05 Sep 2009, 22:18

LenaA wrote:

1) Formula The formula gives you the sum of all factors of a number. Take example, \(n=24=2^3\times 3\) 1,2,3,4,6,8,12,24. They sum is 60. Now try to use the formula :

2) why deduct powers of 2 and 5 and integer1? we are asked not about the total sum of factors but only a subset of factors, i.e. multiples of 10. since powers of 5 and powers of 2 and 1 are not multiples of 10 you have to deduct them from the total sum of factors. Basically I calculated the total sum of factors and then deducted not relevant factors to get the sum of factors that are multiples of 10...I could not see the other way to do that.

3) As for simplifying the expression, you just have to calculate it... you just have to remember some powers...if you remeber that 2^5=32 and 5^4=625...it makes it easier to calculate, even though it was still time consuming , you get \(63\times 3906\)...

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