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# The sum of all factors of 100000 which are divisible by 10 i

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Senior Manager
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The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 12:39
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How to solve this PS?

The sum of all factors of 100000 which are divisible by 10 is
200000
305000
275000
231250
242110

[Reveal] Spoiler:
OA is E
Manager
Joined: 10 Aug 2009
Posts: 130
Followers: 3

Kudos [?]: 62 [1] , given: 10

Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 13:55
1
KUDOS
the sum of all factors of 100000=$$\frac{2^6-1}{2-1}\times \frac{5^6-1}{5-1}=246078$$

and the sum of div. by 10 will be

$$246078-(2^1+2^2+2^3+2^4+2^5)-(5^1+5^2+5^3+5^4+5^5)-1=242110$$

I get E
Senior Manager
Joined: 23 May 2008
Posts: 428
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 14:04
LenaA wrote:
the sum of all factors of 100000=$$\frac{2^6-1}{2-1}\times \frac{5^6-1}{5-1}=246078$$

and the sum of div. by 10 will be

$$246078-(2^1+2^2+2^3+2^4+2^5)-(5^1+5^2+5^3+5^4+5^5)-1=242110$$

I get E

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

I am getting 2^5 * 5^5
Manager
Joined: 10 Aug 2009
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Kudos [?]: 62 [1] , given: 10

Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 14:10
1
KUDOS
gurpreet07 wrote:
LenaA wrote:
the sum of all factors of 100000=$$\frac{2^6-1}{2-1}\times \frac{5^6-1}{5-1}=246078$$

and the sum of div. by 10 will be

$$246078-(2^1+2^2+2^3+2^4+2^5)-(5^1+5^2+5^3+5^4+5^5)-1=242110$$

I get E

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
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 14:15
thanks i got it now.....
Senior Manager
Joined: 22 Sep 2005
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 14:32
Can you explain this formula?
It is the first time that I see it!!

Futhermore, why do you substract the powers of 2 and 5?
Manager
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 18:04
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, 18:09, edited 1 time in total.
Manager
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Kudos [?]: 62 [2] , given: 10

Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 18:06
2
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 :

$$\frac{2^{3+1}-1}{2-1}\times \frac{3^{1+1}-1}{3-1}=15\times 4=60$$

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 $$63\times 3906$$...

Hope this helps
Manager
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Location: UK
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 18:54
This is the factor table am referring too. Construct the table using the factors

Total factors = 36 (since 6*6)

2^0 2^1 2^2 2^3 2^4 2^5

5^0 1 2 4 8 16 32

5^1 5 10 20 40 80 160

5^2 25 50 100 200 400 800

5^3 125 250 * * * *

5^4 625 1250 * * * *

5^5 3125 6250 * * * *

* indicates ends with 00 ( unit digit and tens digit are zeroes)

Add the terms divisible by 10 ( again add only the last 2 digits ) and you will see the sum ending with 10 , so the final answer is 242110 (e)

Give me a kudos if found useful

note : try solving using factor table and you can find the answer fastly ( although it looks like tedious)

ps : the table doesnt seem to look like a table after posting it .

Row I - 2^0 until 2^5
Col I - 5^0 until 5^5

cell value = Ri * Cj
_________________

FEB 15 2010 !!

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

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

SVP
Joined: 29 Aug 2007
Posts: 2492
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Kudos [?]: 681 [0], given: 19

Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 21:59
LenaA wrote:
the sum of all factors of 100000=$$\frac{2^6-1}{2-1}\times \frac{5^6-1}{5-1}=246078$$

and the sum of div. by 10 will be

$$246078-(2^1+2^2+2^3+2^4+2^5)-(5^1+5^2+5^3+5^4+5^5)-1=242110$$

I get E

Thats wonderful method if the method works for every case.

_________________

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GT

Senior Manager
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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05 Sep 2009, 23:18
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 :

$$\frac{2^{3+1}-1}{2-1}\times \frac{3^{1+1}-1}{3-1}=15\times 4=60$$

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 $$63\times 3906$$...

Hope this helps

Awesome explanation Lena ...

Kudos to you
Senior Manager
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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06 Sep 2009, 17:32
I agree, KUDOS to you because of your explanation....
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Re: The sum of all factors of 100000 which are divisible by 10 i [#permalink]

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07 Sep 2009, 02:05
Yup...that is a GENERIC formula for SUM of factors...it is true for every case even for factors with a unity power.

For eg. lets take 10 = 5*2
sum of factors = ( 5^2-1)(2^2-1)/(5-1)(2-1) = 24*3/4 = 18
Re: The sum of all factors of 100000 which are divisible by 10 i   [#permalink] 07 Sep 2009, 02:05
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