Question

The last digit of ${{\left( 1!+2!+...............+2005! \right)}^{500}}$ is :
(A). 9
(B). 2
(C). 7
(D). 1

Answer 1

Hint: Here we are looking for units digits of a large power number. So, we just need units’ digits of numbers inside. So, if we add all the units’ digits we get the units digit of sum. By that we can find units digit of greater power of the units digit. Whenever we multiply a number many times its units’ digit only depends on the unit’s digits of numbers multiplying. Remaining digits doesn’t affect the product we get. So by this we get the value of the unit's digit of the power, which is the required result in the question. For factorials after a point, all next factorials have units placed as 0. Find that point and then solve.

Complete step-by-step solution -
Factorial:- In mathematics, factorial is an operation, denoted by “$\left( ! \right)$ “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: $n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1$
For example: $5!=5\times 4\times 3\times 2\times 1=120.$
Note that we assume $0!=1$ . It is standard value. It has a wide range of applications in combinations.
Given expression in the question can be written as follows:
$I={{\left( 1!+2!+3!+...................+2005! \right)}^{500}}$ last digit of I.
If you observe carefully in $5!$ , we get a product of $5\times 2$ which is 10.
This term repeats here after in every factorial greater than 5.
So, after 5 every term has 0 as units digit in the number.
So, it doesn’t affect one result. We need to consider up to 5.
By this we can write our required expression as follows:
Last digit of ${{\left( 1!+2!+3!+4! \right)}^{500}}$ .
So, we can tell values of these factorials, by definition, we get –
$1!=1,2!=2,3!=6,4!=24$ .
So, By adding these all we get the last digit 3. So, we write the condition in the simple form as below:
Last digit of ${{3}^{500}}$ .
We know cyclicity of 3 is 4. So, digits of units placed for power repeats the pattern for every 4 powers. So, we take $\dfrac{500}{4}$ and the remainder as 0. So, we set units digits the same as ${{3}^{4}}=81$ .
So, the last digit of ${{\left( 1!+2!+...............+2005! \right)}^{500}}$ is 1.
Therefore, option (d) is the correct answer.

Note: The idea of taking units digits of the factorial after 5 is always 0 for all numbers is very important. So follow that step carefully. The idea of projecting the last digit is very important. The concept of cyclicity is very important. Each number has its own cyclicity value. This property is very crucial. So, use it carefully. Remember that 0 to power at units digits gives 0. So, it doesn’t affect me.