[satanica]

Originally Posted by crushkiller

What is the units digit of 1^1 + 2^2 + 3^3 + … + 2008^2008?
1. 8
2. 6
3. 4
4. 2
5. 0

The answer is 8.

We get the following 2008 last digits.
1 - 201 times
3 - 201 times
4 - 201 times
5 - 201 times
6 - 603 times
7 - 201 times
9 - 200 times
0 - 200 times

Hence, the last digit is same as last digit of 1+3+4+5+8+7 i.e. 8.

[jha16june]

Originally Posted by crushkiller

puyssssss solve this out
What is the units digit of 1^1 + 2^2 + 3^3 + … + 2008^2008?
1. 8
2. 6
3. 4
4. 2
5. 0

we need to see from 1 to 10.
1+4+7+6+5+6+3+6+9+0 =>7
Then we have 2000/10 = 200 cycles.
So we need to see for next 8 only. which gives 8 as the last digit. Which sud be the answer.

[deepakraam]

My take for the above prblm is 8.

When we add first 9 numbers,we get 47 which will repeat for 2000 times.So the final 8 terms will contribute to the unit's digit.The ans is 8.

-Deepak.

[atul011]

Originally Posted by crushkiller

puyssssss solve this out
What is the units digit of 1^1 + 2^2 + 3^3 + … + 2008^2008?
1. 8
2. 6
3. 4
4. 2
5. 0

i am getting the units digit as 8.
Plz post the correct answer.

[Flameboy007]

Originally Posted by deepakraam

My take for the above prblm is 8.

When we add first 9 numbers,we get 47 which will repeat for 2000 times.So the final 8 terms will contribute to the unit's digit.The ans is 8.

-Deepak.

Hey, how r u gettin 47?
Isn it 1+4+9+6+5+6+9+4+1 = 45?

I am getting 3 as the answer.

[Flameboy007]

Originally Posted by Flameboy007

Hey, how r u gettin 47?
Isn it 1+4+9+6+5+6+9+4+1 = 45?

I am getting 3 as the answer.

Ohhhhhhh. Got it. Guess masoom n I did the same mistake.
All apologies.

Now I am gettin 8.

[masoom]

Originally Posted by Flameboy007

Ohhhhhhh. Got it. Guess masoom n I did the same mistake.
All apologies.

Now I am gettin 8.

yups dude.... as most of the colleagues went home in long weekend felling a bit aloof in office.. and hence commiting blunders..

[prateek7563]

Find the greatest number of 4 digits which when divided by 10,11,15 and 22 gives remainder 3,4,5,8,15 respectively?

[prateek7563]

n! has 13 zeroes. Whats the least and highest value of n?

[holy_luv]

Originally Posted by prateek7563

n! has 13 zeroes. Whats the least and highest value of n?

least: 55
highest:59

all we have to do is find 2 nos. havin 13 fives in its factorial
number no. of 5s
5 1
10 1
15 1
20 1
25 2
30 1
35 1
40 1
45 1
50 2
55 1
total 13

same for 56,57,58,59...hence highst no.: 59