The last digit of $(1 !+2 !+\ldots \ldots+2009 !)^{500}$ is

JEE MATHEMATICS MCQ: PERMUTATIONS

1 Easy

1 of 1

Application +4 | -1

The last digit of

(1 !+2 !+\ldots \ldots+2009 !)^{500}

View answer and explanation

Correct option: A

Option A (1) is correct because the last digit is determined by the last digit of the sum of factorials, which simplifies to the last digit of 33 raised to the power of 500.

More Questions from PERMUTATIONS

Practice All Chapter MCQs →

How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions? The number of ways in which

r

letters can be posted in

n

letterboxes in a town is The number of four-digit numbers that can be made with the digits

1,2,3,4

and 5 in which atleast two digits are identical, is The number of ways in which a pack of 52 cards can be divided equally into four groups, is Five digited numbers with distinct digits are formed by using the digits, 5, 4, 3, 2, 1, 0. The number of those numbers which are multiples of 3, is A man invites a party to

\left( {m + n} \right)

friends to dinner and places m at one round table and

n

at another. The number of ways of arranging the guests is