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I have no idea how to solve this question I have been trying to solve it for some time but I am unable to...

I tried to do this question by Brute force..but apparently there are 72 divisors of this number(well..if i could get the average then I could have multiplied it by $72$ to get the sum..but I guess that's more difficult) so writing all the divisors then adding them is surely not a good choice!!

My friend did solve it..first he wrote $38808=2^3×3^2×7^2×11$..then by using some kind of formula he wrote.. $$sum=(2^0+2^1+2^2+2^3)(3^0+3^1+3^2)(7^0+7^1+7^2)(11^0+11^1)$$ which on solving gives $$15×13×57×12=133380$$ which apparently is the right answer!!!

But he only knows the formula..(which I now know too!!) And nothing else....also the numbers above $(2^0+2^1+2^2+2^3)(3^0+3^1+3^2)(7^0+7^1+7^2)(11^0+11^1)$ are in multiplication if there was a sum then I might have figured out how to get it ...but here no clue!!!

Any suggestions/hints are appreciated... It would be really good to have an alternate solution..but if someone explains...and how to get/prove the formula then ..that is also fine..

EDIT:most of the responses here are about those brackets trying to explain how when we will expand it we will get the sum of all divisors..Thnx..I have got that..but the question is still not finished as how can we say that only this particular order of the factors that I wrote above is going to work and all others will not!! More specifically.. If someone is writing about formula please prove it..(preferably by combinations!!)

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I think Khan explains the concept behind the solutiom to such problems well. Have a look at his video here.

Essentially, he explains why

$$\sum_{a = 0}^3\sum_{b=0}^2\sum_{c=0}^2\sum_{d=0}^1 2^a3^b7^c11^d = \left(\sum_{i=0}^3 2^i\right)\left(\sum_{i=0}^2 3^i\right) \left(\sum_{i=0}^2 7^i\right) \left(\sum_{i=0}^1 11^i\right)$$

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  • $\begingroup$ Don't you think some brackets() are missing on the LHS?? $\endgroup$ – Freelancer Nov 15 '15 at 13:40
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    $\begingroup$ @Freelancer No, not really, since it's just a single term. But I'll add it in anyway. $\endgroup$ – Yiyuan Lee Nov 15 '15 at 13:46
  • $\begingroup$ @Freelancer Fixed the link, sorry about that. $\endgroup$ – Yiyuan Lee Nov 15 '15 at 13:48
  • $\begingroup$ Even if the answer was a bit short...but..Well..the video is really good!! ..by using this as the basis.. The problem becomes very easy!!!Thanks...and how did you get to know about this video..?? As it just discusses about a particular problem..and nothing else...what did you search for on you tube or Google?? $\endgroup$ – Freelancer Nov 15 '15 at 15:10
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When you distribute the multiplication over the addition, note there is a term for each choice of exponent for each prime divisor. Another way of saying that is that there is one term for each divisor of $38808$. So you are just adding up all the divisors of $38808$.

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  • $\begingroup$ well that is easy to say ..for someone who knows the formula that if you expand the term then you are going to get the sum of all the divisors I think that is obvious!! ..but can you please prove it?? because Anyone can see the numbers.... $\endgroup$ – Freelancer Nov 15 '15 at 9:17
  • $\begingroup$ $$(2^0+2^1+2^2+2^3)(3^0+3^1+3^2)(7^0+7^1+7^2)(11^0+11^1)$$ are arranged in a peculiar order.... and if we try to misplace the numbers like say..I write it as...$$ (2^0+3^0+7^0+11^0)(2^1+3^1+7^1+11^1)(2^2+3^2+7^2)(2^3)$$....but as expected this will not give the sum!!!..but prima-facie...anyone would have said ..yeah!!this should work!! all the possible factors are in the bracket and if we expand it...we will get the sum!! ...but that not true!!.. there is a particular way as to how the numbers are arranged in the bracket ..either you prove it..or solve the question in another way!!! ... $\endgroup$ – Freelancer Nov 15 '15 at 9:18
  • $\begingroup$ now just to be specific since you are going with the formula...you have to prove..(to answer the question)for any number $$N=p_1^a×p_2^b×p_3^c...$$ ...the sum of its divisors is $$=(p_1^0+p_1^1+p_1^2...p_1^a)×(p_2^0+p_2^1...p_2^b)..other/similar/terms.......$$ $\endgroup$ – Freelancer Nov 15 '15 at 9:20
  • $\begingroup$ @Freelancer If you distribute out your expression you will see that every possible factor is represented as an addend as you form each factor from its prime factorization when you do the distribution. $\endgroup$ – 1110101001 Nov 15 '15 at 9:35
  • $\begingroup$ @1110101001 please look at the question.. I have edited it.. $\endgroup$ – Freelancer Nov 15 '15 at 11:28
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Represent the number as powers of prime numbers, $n=p_1^{k_1} \cdots p_r^{k_r}$. We can give the sum of divisors by a common formula which is given by

$$ ( p_1^0+\cdots+p_1^{k_1}) \cdots (p_r^0+\cdots+p_r^{k_r}) $$ note $p_1,\ldots,p_r $ are primes raised to $k_1, \ldots,k_r$ respectively. Note you should also know formula for GP as terms are in GP, i.e. $a(r^n-1)/(r-1)$, where $r$ is the common ratio.

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  • $\begingroup$ Sorry i dont know all the LaTeX . $\endgroup$ – Archis Welankar Nov 15 '15 at 9:47
  • $\begingroup$ I have fixed the two major formulas. Please take a look, for future reference. $\endgroup$ – Andreas Caranti Nov 15 '15 at 9:53
  • $\begingroup$ @Andreas caranti Thanks for editing. $\endgroup$ – Archis Welankar Nov 15 '15 at 10:09
  • $\begingroup$ I think you are just talking about simplifying the formula $$( p_1^0..+p_1^{k_1}) \dots (p_r^0+..+p_r^{k_r})$$ using the formula of G.P....OK I have done it!!..I have written it as .. $\endgroup$ – Freelancer Nov 15 '15 at 11:06
  • $\begingroup$ $$(\frac{p_1^{k_1+1}-1}{p_1-1})×(\frac{p_2^{k_2+1}-1}{p_2-1})...$$..OK..but what now??..how to prove this is the sum of all the divisors..can you prove it?? ..please look at question I edited it. $\endgroup$ – Freelancer Nov 15 '15 at 11:26

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