M46-23

Question

The function  f(n)  is defined for all positive integers  n , where  n > 1 , as the product of  n  consecutive positive multiples of  n . How many zeros does  f(100)  end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

Bunuel · Answer

Official Solution:

The function  f(n)  is defined for all positive integers  n , where  n > 1 , as the product of  n  consecutive positive multiples of  n . How many zeros does  f(100)  end with?

A. 24
B. 26
C. 143
D. 224
E. 2400

Let's break down the question:
 
According to the problem,  f(100)  is the product of 100 consecutive positive multiples of 100. So:
   
 f(100) = 100 * 200 * 300 * ... * 10,000 = 
 
 = (100 * 1) * (100 * 2) * (100 * 3) * ... * (100 * 100) = 
 
 = 100^{100} * 100! 
   
Next:
 
 100^{100} = (10^2)^{100} = 10^{200} , which ends with 200 zeros.
 
 100!  ends with  \frac{100}{5} + \frac{100}{25} = 20 + 4 = 24  zeros.
 
Thus, the total number of zeros at the end of  f(100) = 10^{200} *  (something with 24 zeros at the end) is:
 
 200 + 24 = 224  zeros.

Answer: D

Vivek1707 · Answer

This is a great question that’s helpful for learning.

AyuShar · Answer

I like the solution - it’s helpful.

DimK · Answer

I like the solution - it’s helpful.

thisyearboom · Answer

Why are we finding largest power of 5 in 100! ? Shouldn't we be finding largest power of 10 in 100! which would be 11 ?

Bunuel · Answer

Trailing zeros refer to a sequence of 0's in a decimal representation of a number, after which no other digits are present.

For instance, 125,000 has 3 trailing zeros.
 
The number of trailing zeros in the factorial of a non-negative integer,  n! , can be determined using this formula:

\frac{n}{5}+\frac{n}{5^2}+\frac{n}{5^3}+...+\frac{n}{5^k} , where  k  must be selected such that   5^k \leq n 
 
Let's consider an example:

How many zeros are at the end of 32!?

\frac{32}{5}+\frac{32}{5^2}=6+1=7 . Observe that the last denominator ( 5^2 ) must be less than 32. Also, note that we only consider the quotient of the division, that is  \frac{32}{5}=6 .

So, there are 7 zeros at the end of 32!.

Another example: how many trailing zeros does 125! have?

\frac{125}{5}+\frac{125}{5^2}+\frac{125}{5^3}=25+5+1=31 ,

The formula essentially counts the number of factors of 5 in n!, but since there are at least as many factors of 2, this is equivalent to counting the number of factors of 10, each of which contributes one more trailing zero.

thisyearboom · Answer

Thanks for your quick reply. I got it now!

M46-23

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