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…

#1

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.

#2