No options
![Razz :P](./images/smilies/razz.png)
Regards.
The reason I chose a large number is so that once you solve this you should be able to do the smaller ones easily.leovonp wrote:You mean all 42 of them?
Doesn't seem like a feasible exam question
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.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)
beeparoo wrote:Oooh.. It's like watching a battle of the experts.
I'm going to go and seek out a giant foam finger...
Start something? Heavens - YES! I'd like to be a spectator of a battle of wits. Luckily, both sides are are equipped.gabriel wrote:![]()
. 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![]()