need clear explanation

[trisa120]

1.Col A: The remainder when (10^8 + 10^9 + 10^10 + 10^11 + 10^12) divided by 11
Col B: 0
answer: A

2. What is the remainder, when (7^6 + 7^7 + 7^8 + 7^9 + 7^10) is divided by 14?
answer: 7

3. Given ‘X’ as a set of elements that has numbers between 1 and 100 inclusive and are not divisible by either 5 or 7.
Col A: Number of elements in X
Col B: 68

4. Given a series 3, 33, 333 ………. find the hundreds digit of the sum of first 10 numbers of the series?
answer:24

5. In how many maximum parts can a circular region be divided by using 3 lines which cut the circle at exactly 2 places?
answer:7

[xperience]

Originally Posted by trisa120

1.Col A: The remainder when (10^8 + 10^9 + 10^10 + 10^11 + 10^12) divided by 11
Col B: 0
answer: A

I am not really sure what answer:A refers to, but here is the explanation nevertheless

(10^8 + 10^9 + 10^10 + 10^11 + 10^12)

{= (11-1)^8 + (11-1)^9 + (11-1)^11 + (11-1)^12}
when you expand this, all terms containing the number 11 will be divisible by 11 and the only terms remaining will be as follows:

(-1)^8 + (-1)^9 + (-1)^11 + (-1)^12
= 1 + (-1) + (-1) + 1
= 2 - 2
=0

hence there is no remainder left when (10^8 + 10^9 + 10^10 + 10^11 + 10^12) is divided by 11

Cheers..!

[xperience]

Originally Posted by trisa120

2. What is the remainder, when (7^6 + 7^7 + 7^8 + 7^9 + 7^10) is divided by 14?
answer: 7

(7^6 + 7^7 + 7^8 + 7^9 + 7^10)
= [7*(7^5+7^6+7^7+7^8+7^9)]/14
=(7^5+7^6+7^7+7^8+7^9)/2
= (6+1)^5 + (6+1)^6 + (6+1)^7 + (6+1)^8 + (6+1)^9
Now, all terms containing 6 should be divisible by 2, and the only terms remaining will be

(1)^5 + (1)^6 + (1)^7 + (1)^8 + (1)^9
= 5/2
So remainder should be 1. I wonder if the OA is right

[xperience]

Originally Posted by trisa120

3. Given ‘X’ as a set of elements that has numbers between 1 and 100 inclusive and are not divisible by either 5 or 7.
Col A: Number of elements in X
Col B: 68

The total number present is 100-1+1 = 100
from 1 to 100, numbers divisible by 5 = 20
from 1 to 100, numbers divisible by 7 = 14
Hence total numbers left is 100-34 = 66

Out of this, we have subtracted two numbers twice, 35 and 70. They need to be subtracted only once, so we shall add 2 numbers to 66
Hence total numbers is 66+2 = 68

Cheers..!

[xperience]

Originally Posted by trisa120

4. Given a series 3, 33, 333 ………. find the hundreds digit of the sum of first 10 numbers of the series?
answer:24

let's dissect this the usual way:
1st number will have one 3
2nd number will have two 3's
3rd number will have three 3's
--
--
--
--
10th number will have ten 3's

Now when you add them up, the left most digit will become 3 (Unit's digit)
the digit next to it will be 6 (ten's digit)
3rd will be 9 (hundredth's digit)
and so on and so forth
So the hundredth digit will be 9.
Wonder where 24 came from..!
BTW, a digit can never be a 2-digit number. Please check

[Fiver]

Agree with the solutions for all but the 4th question.
4] We'll have to add 10 3s to arrive at the unit digit of the sum of all these 10 nos. 3*10 = 30. Hence the unit digit of the sum will be a 0 with 3 carried over to the tens place.
Now we'll need to add 9 3s to arrive at the tens digit of the sum 3*9=27 + 3 carried over. Hence the tens place of the sum will also be a 0 with again 3 carried over to the hundreds place.
Now we'll need to add 8 3s to arrive at the hundreds place of the sum 3*8 = 24 + 3carried over=27.
So if your question is 'what is the sum of the numbers (excluding the carried over no.) forming the hundreds place of the sum of these 10 nos.?' Then the answer is 24.
Or if your question is 'what is the hundreds place of the sum of these 10 nos?
then the answer is 7.