what will be the numbers of zeroes in the expension of a) 100!×25! b) 100!+25! please help


Question and Answers Forum

All Questions Topic List
Permutation and Combination Questions
Previous in All Question Next in All Question
Previous in Permutation and Combination Next in Permutation and Combination
Question Number 55192 by Learner last updated on 19/Feb/19

$w h a t w i l l b e t h e n u m b e r s o f z e r o e s i n t h e e x p e n s i o n o f$ $a) 100! \times 25!$ $b) 100! + 25!$ $p l e a s e h e l p$
Commented by MJS last updated on 19/Feb/19

$computed and counted$ $a) 45$ $b) 15$ ${there}^{'} s no other way to know a l l zeros$
	Answered by kaivan.ahmadi last updated on 19/Feb/19
	$n! has [(n/5)]+[(n/5^2 )]+[(n/5^3 )]+... zero in right hand so 100! has [((100)/5)]+[((100)/(25))]+[((100)/(125))]=20+4=24 and 25! has [((25)/5)]+[((25)/(25))]=6 100!×25! has 24×6=144 zero in right hand 100!+25! bas 6 (min{24,6})zero in right hand$
	$n! h a s [\frac{n}{5}] + [\frac{n}{5^{2}}] + [\frac{n}{5^{3}}] + . . . z e r o i n r i g h t h a n d$ $s o 100! h a s [\frac{100}{5}] + [\frac{100}{25}] + [\frac{100}{125}] = 20 + 4 = 24$ $a n d 25! h a s [\frac{25}{5}] + [\frac{25}{25}] = 6$ $100! \times 25! h a s 24 \times 6 = 144 z e r o i n r i g h t h a n d$ $100! + 25! b a s 6 (m i n {24, 6}) z e r o i n r i g h t h a n d$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum