(91!-90!+89!)/89!=

break up the fraction;

(91!)/(89!) - (90!)/(89!) - (89!)/(89!)
which reduces to (91)(90) - (90) +1 = 90^2 + 1 (D)

take out the common factor from numerator and denominator which in this case would be 89!

Your numerator should look like

91*90* 89! - 90*89! + 89!

89![ 91*90 - 90 +1]/89!

Which leaves you with

91*90 - 89

\(\frac{91! - 90! + 89!}{89!} = \frac{91!}{89!} - \frac{90!}{89!} + \frac{89!}{89!}\)

\(\frac{91!}{89!}\) > Units place = 0

\(\frac{90!}{89!}\) > Units place = 0

\(\frac{89!}{89!} = 1\)

Answer should have 1 in the units place; only option D stands out

Answer = D

Factor out 89!: 91!-90!+89!/89! = 89!(90*91-90+1)/89! = 90*91-90+1.

Now, factor out 90: 90*91-90+1=90(91-1)+1=90^2+1.


Where does the 89! on the top come from?? Is it not true that if we want to get rid of the 89! at the bottom we multiply the top by 89! ? But in this answer we multiply the top by 89! but the 89! remains also at the bottom??

Dear JeroenReunis,
I'm happy to respond.

First of all, my friend, in your first line of text, you have a mathematical mistake that reflects a misunderstanding of mathematical grouping symbols. See this blog for more information:
https://magoosh.com/gmat/2013/gmat-quan ... g-symbols/

Now, I believe you are misunderstanding the nature of factorials. We did NOT multiply the numerator by 89! in order to cancel it--- you are perfectly correct that this move would have been quite illegal.

Instead, we factored out numbers from the factorial. This blog, the blog from which this question is taken, explains all this in detail:
https://magoosh.com/gmat/2012/gmat-factorials/

Think about what, for example, (91!) means. This is the product of all the positive integers from 91 down to 1. That product would be a very large number, much larger than 10^100 (a googol). We could represent this as

91! = 91*90*89*88*87* ..... *5*4*3*2*1

We have 91 factors all multiplied together. Well, we can group multiplication into any arrangement we like (technically, this is known as the associative property of multiplication). For example,

91! = 91*90*(89*88*87* ..... *5*4*3*2*1)

Well, that set of terms grouped in the parentheses equal 89! Thus,

91! = 91*90*(89!)

Similarly,

90! = 90*(89!)

Thus,

91! + 90! + 89! = 91*90*(89!) + 90*(89!) +(89!) = (91*90 + 90 + 1)*(89!)

That's the precise origin of the (89!) factor in the numerator. We factor it out from the rest of the numerator, and cancel it legitimately with the denominator.

Does all this make sense?
Mike

Hello Mike,

Thank you so much Mike! I really have broken my brains on this one for at least an hour or so..
Now I see my mistake it all makes way more sense.

Thanks again for your clear answer!!

nice approach!
but would you mind to explain how 91!/89! and 90!/89!, units place = 0?

I am going to try to explain what I think Paresh is trying to say.

For \(\frac{91!}{89!}\) - the numerator has a factor of 89! which can be cancelled out with the denominator leaving out 91*90 and hence the units place is 0.
For \(\frac{90!}{89!}\) - the numerator has a factor of 89! which can be cancelled out with the denominator leaving out 90 and hence the units place is 0.

Does anyone know where I can find a section of similar problems?

Check our questions' bank: https://gmatclub.com/forum/search.php?view=search_tags

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