The last four digits of natural number $7100$ are:

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2732

1301

2500

0001

answer is D.

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Detailed Solution

We are asked to find the last four digits of

7100

Now, let us convert the given number in the form

(x + 1) n

(x − 1) n

where

x

is a multiple of 10.
Here, we can see that the number 7 cannot be written in the required form.
We know that the formula of exponents that is

(a) c × b = (a c) b

By using the above formula to given number we get:

⇒ 7100 = 7 2 × 50 ⇒ 7100 = 7250 = 4950

Here, we can see that the number 49 can be written in the required form that is

⇒ 7100 = (50 − 1) 50

We know that the binomial expansion is given as

(x − 1) n = n C 0 x n − n C 1 x n − 1 + n C 2 x n − 2 − n C 3 x n − 3 + … … … + (− 1) n n C n

By using this expansion in above equation we get

⇒ 7100 = 50 C 0 (50) 50 − 50 C 1 (50) 49 + … … − 50 C 47 (50) 3 + 50 C 48 (50) 2 − 50 C 49 (50) + 50 C 50

Here we can see that the power of 50 is reduced from 50 to 0
Now, we can see that the last four digits of the terms up to

504

will be zeros.
So, we can represent the terms that are up to

504 a s 10000 × k

where

k

is some integer.
Now, let us rewrite the above expansion by representing the up to

504

as shown then we get

⇒ 7100 = (10000 × k) − 50 C 47 (50) 3 + 50 C 48 (50) 2 − 50 C 49 (50) + 50 C 50

We know that the formula of combinations that is

n C r = n! r! (n − r)!

By using this formula in above equation we get
Question Image

Here, we can see that the term of

503

where we again get last four digits as zeros because the total
Multiplication gets

504

.
Now, by rewriting the above equation by joining the first and second terms in the above equation we get

⇒ 7100 = (10000 × p) + 50 × 49 2! (50) 2 − 50 1! (50) + 1

Now, by multiplying the remaining terms in above equation we get

⇒ 7100 = (10000 × p) + (25 × 49 × 2500) − (2500) + 1 ⇒ 7100 = (10000 × p) + 3062500 − 2500 + 1 ⇒ 7100 = (10000 × p) + 3060001

Now, we know that adding any number to zeros we get the same number.
Now, we know that adding any number to zeros we get the same number.
Here, we can see that the last four digits in the first term are zeros.
So, adding the last four digits of the second term to the first term we get the same digits as in the second term.
Therefore we can conclude that the last four digits of