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```

```
9 years ago

```
1) 2137^754=(2130+7)^754
on  binomial expansion,we find that the first teram is not divisible by 10  while others are divisible by 10..
SO only  first term will contribute to last digit..
the first  term is 7^754
the last  digit of this will be the last digit of given question.'
now  observe the last digit in following arguments.
7^1=7,
7^2=49,
7^3=513
7^4=last digit being 1
and then the last digits  repeat
the last digits are  7,9,3,1,7,9,3,1,
SO 7^(4n+1)---last digit is 7
7^(4n+2)---- last digit is 9
..
so on dividing 754 by 4 we  get 2 as remainder.........
so the last digit will be  9.......

```
9 years ago
```							2) there are 5P4 ways of forming a 4 digit number with each digit appearing at most once.Consider first the one's place.  In those 120 numbers, 1,2,3,4, and 5  appear exactly the same number of times, that is, each digit appears  120/5 = 24 times.
For the tens place, same  thing:  1,2,3,4,and 5 appear equally the same number of times.  Each  digit appears 24 times.
The argument for the hundreds and thousands  places are the same.
Thus  the sum of all those distinct 120 numbers is    1,000*24*(1+2+3+4+5)+ 100*24*(1+2+3+4+5) + 10*24*(1+2+3+4+5) + 1*24*(1+2+3+4+5)
= 1,111*24*(1+2+3+4+5)
= 1,111*24*15
= 3,99,960 i.e option (B)

```
9 years ago
```							3) HINT : Use binomial expansion of (bc + a (b + c))6.
answer is 3 . 6!/2!3! = 720/12 = 60

```
9 years ago
```							hey sudeesh what about the other question , you have attempted the easiest one; i have posted this questions thinkinkg that some one will answer the tough questions but ......
```
9 years ago
``` dear sudheesh,
see this method you should solve like this,well your attempt was good.i'm enclosing you with solution of 19th question also.
```
9 years ago
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