dips1122 wrote:
pk123 wrote:
nick1816 wrote:
Find the number of trailing zeros in (23!+24!+26!).
A. 2
B. 4
C. 5
D. 8
E. 10
(23!+24!+26!)
=23!*(1+24+26*25*24)
=23!*(25+26*25*24)
=23!*25*(1+26*24)
=23!*25*(1+(25+1)*(25-1))
=23!*25*(1+25*25-1)
=23!*25*625
=23!*25^3
no of zeros in 23!= [23/5]+[23/25]=4
no of zeros in 25^3 or 5^6 = 6
so total no of zeros =4+6=10
Hi Could you plz explain how "no of zeroes in 5^ 6 = 6??" Isn't 5^6 = 15625?
Also what method did u use for zeroes in 23!?
To calculate the number of zeros we need to see how we could get 10s which is possible only when we have combination of 2s and 5s as prime factors
So for 23!= 23*22*21*20*19*18*17*........3*2*1
Think as when you multiply each item how many zeros you will get depends on how many 5's and 2's are there in the multiplication....and how would you find how many 2's and 5's are can be found from [ number/5] + [ number /5*5] +[number /5*5*5]+ so on
Why do we need to do this...think of 5!= 5*4*3*2*1=120...how many zeros are there ...1 right but 2s are plenty ....total 3 2's.
so any consecutive multiplication 2's outnumber 5's
coming to 23!, number of zeros can be found by number of 5 it has so [23/5] +[23/5*5]+[23/5*5*5] =4 +0 +0 =4
Number of 5 in 5^6 means how many times 5 appear in 5^6 , which is 6
Hope this answers your question