#### Question

# When calculated, fifty-one factorial (51!) is quite a large number. If you calculate 51!, answer the following?

When calculated, fifty-one factorial (51!) is quite a large number. If you calculate 51!, answer the following two questions:

1. With what digit does 51! end (what is the ones digit – the rightmost digit)?

2. Beginning at the rightmost digit (ones place), how many consecutive places are represented by the digit found in the ones place?

#### Answers

1. 0. All factorials above 4! end with 0 in the one's place.

2. there are 52 zeros in a row.

51! = 1,551,118,753,287,380,000,000,000,000,00…

51! has 10 as a factor, so the right most digit is 0.

To find how many zeroes there are, we need to know the greatest k such that 10^k is a factor of 51!.

10 = 2*5, so we need to consider the prime factors 2 and 5. Clearly, 51! has more factors of 2 than 5, so we need only compute how many factors of 5 there are.

5 ==> 1

10 ==> 1

15 ==> 1

20 ==> 1

25 ==> 2

30 ==> 1

35 ==> 1

40 ==> 1

45 ==> 1

50 ==> 2

Add them up, there are 12. Thus, 10^12 divides 51!, so there are 12 zeroes at the right.