Question

The term digit of 1! + 2! + 3! + … + 49! Is

Solution

Correct option is

1

We have 1! + 2! + 3! + 4! = 33. Also 5! = 120, 6! = 720, 7! = 5040, 8! = 40320 and 9! = 326880. Thus, the tens digit of

1! + 2! + … + 9! is 1.  

Also, note that n! is divisible by 100 for all n ≥ 10, so that tens digit of 10! + 11! + … + 49! is zero. Therefore, tens digit of 1! + 2! + … + 49! is 1.  

SIMILAR QUESTIONS

Q1

In the certain test there are n questions. In this test 2k students gave wrong answers to at least (n – k) questions, where k = 0, 1, 2, …, n. If the total number of wrong answer is 4095, then value of n is 

Q2

If n > 1 and n divides (n – 1)! + 1, then 

Q3

In a group of 8 girls, two girls are sisters. The number of ways in which the girls can sit so that two sisters are not sitting together is

Q4

The number of words that can be formed by using the letters of the word MATHEMATICS that start as well as end with T is

Q6

The number of subsets of the set A = {a1a2, … an} which contain even number of elements is

Q7

The number of ways in which we can post 5 letters in 10 letter boxes is

Q8

The number of five digit telephone numbers having at least one of their digits repeated is

Q9

Let P be a prime number such that P  23. Let n = p! + 1. The number of primes in the list n + 1, n + 2, n + 3, … n + p – 1 is

Q10

Four dice are rolled. The number of possible outcomes in which least one die shows 2 is