Question

What is the last two digits of 2005+2005^2........2005^2005

Answer 1

Answer:

I think last two digits are 25.

Step-by-step explanation:

When the number is with unit digit 5...

1. If the digit in the tens place is even and the exponent y is odd, then the number ends in 25.
2. If the digit in the tens place is even and the exponent y is even, then the number ends in 25.

In our example... We have to find

${2005}^{1} + {2005}^{2} + ... + {2005}^{2005}$

Now, in 2005 the digit in unit place is 5 and the tenth place digit is even but exponent is alternately odd and even.

Hence, both the time in odd and even exponent,

the number ends with 25 that is ends with 5

Then in our example

${2005}^{1} + {2005}^{2} + ... + {2005}^{2005}$

The summation of exponential value is 2005 times....

Hence in unit place of summation is 5 with 2005 times.

$2005 \times 5 = 10025$

Therefore unit place of summation is 5.

Now, by another condition tenth place of power is 2.....hence the tenth place in summation is 1002 + (2005 times 2)

$1002 + (2 \times 2005) = \\ 1002 + 4010 = 5012$

Hence the tenth place of summation is 2.

So the last two digits of the summation are 25.

Plz mark the answer as brainliest...

Hope it will help...

Thanks.

#VIVEK9090