If P = (n)(n - 1)(n - 2) . . . (1) and n > 2, what is the largest valu

It's a simple formula.
Divide the no you are given with 5 and it's '5 times'.

Case 1
23 which is under 25 coz 5*5
23/5 gives you 4.6. Strike out the decimal part. Therefore, 23! have 4 trailing zeroes

Case 2
39 which is more than 25 but less than 125
39/5 is 7.8. Do the needful.
Now we have to divide the no again by 25 because it is above 25 and 25 have 2 5s.
39/25=1.7
Add the answers i.e 7+1=8. therefore 39 have 8 trailing zeroes.

Case 3
187 which is more than 125 but less than 625
187/5=37.blah
187/25=7.blah
187/125=1.blah
Add=> 37+7+1= 45 trailing zeroes.

Scientific logic and reason
23 have 4 no which are multiple of 5 and 2 combined. 10, 20, 5*2 or any even no within 23 and 15*any even no. hence 4 zeroes

This is basically a trailing zero question asking which number for n! will the trailing zeros be 6?

\(P = (n)(n - 1)(n - 2) . . . (1)\) = n!

if n=29, n!= 29! ; Formula trailing zero is straight forward.

\(n/ 5 + n/5^2+ n/5^3...... n/5^k\), such that\( 5^k < n\)

\(29/5 + 29/ 25\) =\( 5+1 = 6\). Rest all the option gives a value >6 which is not required.

Basically, P = n!
Out of these options, we need to find the largest number, which:
(i) has 6 zeroes, i.e. (2^6)*(5^6)
(ii) has 9 as the first non-zero digit; we'll think about this later

One point, however, is clear that 2s are in abundance. Number of zeroes factorial would carry would be determined by the number of 5s it has.

(E) 39! has eight 5s. Nope
(D) 35! has eight 5s. Nope
(C) 34! has seven 5s. Nope
(B) 30! has seven 5s. Nope
(A) 29! has six 5s. This must be the answer.

Bunuel Can you confirm if OA is correct?

29! = 8841761993739701954543616000000
30! = 265252859812191058636308480000000

So I think 29 is the correct answer

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