(30)*(31)*(32)**(99) is divisible by 5n. What is the maximum possible

Question

(30)*(31)*(32)*…*(99)  is divisible by  5^n . What is the maximum possible value of  n ?

A. 11
B. 13
C. 14
D. 16
E. 18

Maria240895chile · Accepted Answer

all multiples of 5 ends with 0 or five so in 99!/29! we would have

30,35,40,45,50,55,60,65,70,75,80,85, 90, 95 all have one 5 as factor except 50 and 75 which have 2

so 14+2=16

gmatophobia · Answer

(30)*(31)*(32)*…*(99)

\frac{99!}{29!}

The highest power in of 5 in 99! = 22

Quotient (99/5) = 19
Quotient (19/5) = 3
Highest Power of 5 in 99! = 19+3 = 22
 
The highest power in of 5 in 29! = 06

Quotient (29/5) = 5
Quotient (5/5) = 1
Highest Power of 5 in 29! = 5+1 = 06
 
Highest Power of power in  \frac{99!}{29!}  = 22 - 6 = 16

Hence, the highest power of 5 in  (30)*(31)*(32)*…*(99)  is 16.

Option D

aanisha15belur · Answer

Why did we not take 99/5 + 99/25 = 19 + 1 = 20? The other examples don't take the quotient in the numerator.. they take the same number (factorial) in numerator.

Bunuel · Answer

That method works only when you're finding trailing zeros of a factorial, like 99!. In this question, we’re not dealing with 99!, so you can’t apply 99/5 + 99/25 here. Instead, we need to count how many multiples of 5 are in the product directly, as shown in the discussion above.

(30)(31)(32)**(99) is divisible by 5n. What is the maximum possible

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