Suppose you multiply the first 50 positive whole numbers. How many zeros will there be in the string of zeros at the end of the product?

Question

Suppose you multiply the first 50 positive whole numbers. How many zeros will there be in the string of zeros at the end of the product?
a.
10 zeros
b.
12 zeros
c.
5 zeros
d.
25 zeros

Answer 1

 

 

 

 

 

The product 1*2*3*4*5*…*48*49*50 is:
30,414,093,201,713,378,043,612,608,166,064,768,844,377,641,568,960,512,000,000,000,000.
That is 12 zeros at the end.

It helps to think about factorials:
n! = 1*2*3*4*5*…*n
For factorials of 1<=n<=4, there are no zeros at the end.
For factorials of 5<=n<=9, there is 1 zero at the end.
For factorials of 10<=n<=14, there are 2 zeros at the end.
For factorials of 15<=n<=19, there are 3 zeros at the end.
For factorials of 20<=n<=24 there are 4 zeros at the end.
For factorials of 25<=n<=29 there are 6 zeros at the end. **
For factorials of 30<=n<=34 there are 7 zeros at the end.
For factorials of 35<=n<=39 there are 8 zeros at the end.
For factorials of 40<=n<=44 there are 9 zeros at the end.
For factorials of 45<=n<=49 there are 10 zeros at the end.
For factorials of 50<=n<=54 there are 12 zeros at the end. **

Each multiple of 5 produces a zero when multiplied by an even number. So, 5, 10, 15, 20, 25, 30, 35, 40, 45, and 50 each produce a zero when each is multiplied by one of the many even values. That could account for 10 zeros.

** But, wait! Products of 25 and 50 produce 2 zeros each. For example, 4*25=100 and 6*50=300. That adds two more zeros to the total.

 

There are 12 zeros at the end.

Answer 2

Eu, let's think about it. Any muliple of 5 multiplied by an even number is going to give is a number that ends in 0. Because the second number we are multiplying by is 2, then our product will be even, as well as each product along the way. There are 10 multiples of 5, so there will be 10 zeroes.