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What is the largest integer n such that 3n is a factor of 1 x 3 x 5 x... x 97 x 99?

 Feb 16, 2020
 #1
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+1

We can use Legendre's formula: https://en.wikipedia.org/wiki/Legendre%27s_formula

 

By Legendre's formula, the number of factors of 3 is

\(\left\lfloor \frac{99}{3} \right\rfloor + \left\lfloor \frac{99}{3^2} \right\rfloor + \left\lfloor \frac{99}{3^3} \right\rfloor + \left\lfloor \frac{99}{3^4} \right\rfloor = 33 + 11 + 3 + 1 = 48\)

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 Feb 16, 2020
 #2
avatar+29257 
+5

I get the following:

 

Edit:  That last line should say 3n not 3n.

 Feb 16, 2020
edited by Alan  Feb 16, 2020
 #3
avatar+29257 
+3

As a check, here is WolframAlpha's factorisation:

 

 Feb 16, 2020
 #4
avatar+108776 
0

Thanks Alan, I had not seen this of course or I would not have answered the new thread.

 

Also looked at here.

https://web2.0calc.com/questions/help_59968

 Feb 16, 2020

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