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Number property

by gabriel » Fri Jun 27, 2008 12:32 pm
q.) Calculate the sum of all the possible factors of the number 14400.

No options :P . I just made up this question, I had seen a similar question elsewhere and thought I would share it with everyone here. There is something very interesting to learn over here.

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by leovonp » Fri Jun 27, 2008 2:06 pm
You mean all 42 of them?
Doesn't seem like a feasible exam question
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by Ian Stewart » Fri Jun 27, 2008 2:43 pm
It has 63 factors. The sum of the divisors should be 51,181. I'm only able to do that quickly using the multiplicative property of the sigma function, which no one would ever be expected to know on the GMAT. I can get there from scratch as follows:

14,400 = 2^6 * 3^2 * 5^2

It has nine odd divisors:
1, 3, 5, 3^2, 5^2, 3*5, 3^2*5, 3*5^2, 3^2*5^2

Let's add these up and call it S:

S = 1+3+5+3^2+5^2+3*5+3^2*5 +3*5^2 + 3^2*5^2
S = 403

Now, we get the rest of the divisors by multiplying each of the odd divisors by 2^1, 2^2, 2^3, 2^4, 2^5 and 2^6. That gives us all 63 divisors. When we multiply the odd divisors by 2^1 and add together the nine new divisors we get, the sum will just be 2^1 * S. When we do the same for 2^2, the sum will be 2^2 * S. And so on. So the sum of all the divisors must be:

S + 2^1*S + 2^2*S + ... + 2^6*S
= (1 + 2 + 4 + 8 + 16 + 32 + 64)*S
= 127*S
= 127*403
= 51,181

That's like a 900-level GMAT question though!
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by gabriel » Sat Jun 28, 2008 8:00 am
leovonp wrote:You mean all 42 of them?
Doesn't seem like a feasible exam question
The reason I chose a large number is so that once you solve this you should be able to do the smaller ones easily.

There is a pretty straight forward formula for calculating this.

Consider a number N which is divided into its prime factors as N=a^x*b*y*c^z . The sum of all the divisors of such a number is given by the formula

(a^0+a^1+a^2+......+a^x)(b^0+b^1+.....+b^y)(c^0+c^1+c^2+....+c^z)

So for 14400= 2^6*3^2*5^2 we have

(1+2+4+8+16+32+64)*(1+3+9)*(1+5+25) = 127*13*31 = 51181. This example can be extended to any number of primes with any index.

Let me know in case of any difficulty

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by Ian Stewart » Sat Jun 28, 2008 8:40 am
gabriel wrote: Consider a number N which is divided into its prime factors as N=a^x*b*y*c^z . The sum of all the divisors of such a number is given by the formula

(a^0+a^1+a^2+......+a^x)(b^0+b^1+.....+b^y)(c^0+c^1+c^2+....+c^z)
There's a simpler formula which saves you from needing to do those sums. sigma is the name given, in Number Theory, to the function that sums the divisors.

If N = a^x*b*y*c^z, where a, b and c are different primes:
then sigma(N) = [(a^(x+1) - 1)/(a-1)]*[(b^(y+1) - 1)/(b-1)]*[(c^(z+1) - 1)/(c-1)]

So for N = 2^6*3^2*5^2

sigma(N) = [2^7 - 1]*[(3^3 - 1)/2]*[(5^3 - 1)/4] = 127*13*31

Not that you'll ever need to use that on the GMAT.
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by gabriel » Sat Jun 28, 2008 11:29 am
@Ian it is the same formula. Except that you use the "sum of a GP" formula to make it a tad simpler.

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by beeparoo » Sat Jun 28, 2008 1:27 pm
Oooh.. It's like watching a battle of the experts.

I'm going to go and seek out a giant foam finger...
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by gabriel » Sat Jun 28, 2008 1:47 pm
beeparoo wrote:Oooh.. It's like watching a battle of the experts.

I'm going to go and seek out a giant foam finger...
:lol: :lol: . Are you trying to start something over here :twisted: . But seriously I have utmost respect for Ian and sincerely believe that he is very good at what he does. Moreover I am not an expert in GMAT, just a geek who likes math.

Regards

PS: - btw ... whom were you rooting for :wink: :P
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by beeparoo » Sat Jun 28, 2008 4:47 pm
gabriel wrote: :lol: :lol: . Are you trying to start something over here :twisted: . But seriously I have utmost respect for Ian and sincerely believe that he is very good at what he does. Moreover I am not an expert in GMAT, just a geek who likes math.

Regards

PS: - btw ... whom were you rooting for :wink: :P
Start something? Heavens - YES! I'd like to be a spectator of a battle of wits. Luckily, both sides are are equipped.

I take no sides here since I like to root for the underdogs and neither of you present to be such.

Cheers,
Sandra
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