Legendre's Theorem

Number Theory Level 2

Find the number of trailing zero's in 2014 ! 2014!

Note:

2014 ! = 2014 × 2013 × 2012 × 2011 × . . . × 2 × 1 2014! = 2014 \times 2013 \times 2012 \times 2011 \times...\times 2\times 1

or, n ! = n × ( n − 1 ) × ( n − 2 ) . . . × 3 × 2 × 1 n! = n\times (n-1)\times (n-2)...\times 3\times 2\times 1

501 495 485 490

Roland Copino by Roland Copino , 50, Philippines

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1 solution

Roland Copino
Sep 20, 2014

In Legendre's Theorem .. You divide always the number by factors of 5 to get the number of trailing zero's, therefore :

2014/5 = 402

2014/25= 80

2014/625= 3

402 + 80 + 3 = 485

0 Helpful 0 Interesting 1 Brilliant 0 Confused

Dude, why haven't you added 16 to 485, you missed 2014/125.

jaiveer shekhawat - 6 years, 8 months ago

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Thanks, I have updated the answer to 501.

Calvin Lin Staff - 6 years, 8 months ago

Oh thanks sir calvin for the update.

Roland Copino - 6 years, 8 months ago

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