Find the number of trailing zeros in (23!+24!+26!).

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

Though I understood my mistake, I want your help to solve such question.

23! = a0000
24!= b0000
25!= c000000

Now, while summing up all these, I understood my mistake that a+b can lead to another zero.

So to get the number of zero of an expression, get the expression in multiplication/division form only. No addition or subtraction.

Yes absolutely...to get number of zeros, better solve the factorial and bring it to one expression...and in this example we can see 24! and 25! both have 23! in its expanded form ...so anywhere we come through similar expression first simplify and get to mostly simpler version that will help minimize errors

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