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Question 1 of 5
1. Question
If the unit’s digit of 1^{2} + 2^{2} + 3^{2} +….+n^{2} is 4, then the unit’s digit of 1^{10} + 2^{10} + 3^{10 }+….+n^{10} is
Correct
Any natural number ‘n’ and n^{5} have the same unit’s digit. Therefore, n^{2} and n^{10} have the same unit’s digit. All the numbers in the second series individually have the same respective unit’s digit as the corresponding number in the previous series. The overall unit’s digit will also be the same.
Incorrect
Any natural number ‘n’ and n^{5} have the same unit’s digit. Therefore, n^{2} and n^{10} have the same unit’s digit. All the numbers in the second series individually have the same respective unit’s digit as the corresponding number in the previous series. The overall unit’s digit will also be the same.

Question 2 of 5
2. Question
The number of natural numbers which are smaller than 2 x 10^{8}, and which can be written only by means of the digits 1 and 2 is
Correct
The required numbers are 1, 2, 11, 12, 21, 22,……..122222222.
Let us calculate how many number are these.
There are 2 one digit such numbers. There are 2^{2} two digit such numbers. And so on.
There are 2^{8 }eight digit such numbers. All the nine digit numbers begging with 1 and written by means of 1 and 2 are smaller than 2.10^{8}. Thus, there are 2^{8} such nine digit numbers.
Hence, the required number of numbers is
2 + 2^{8} + 2^{3} + …. + 2^{8} + 2^{8}
= 2(2^{8}1)/(21) + 2^{8} = 2^{9} – 2 + 2^{8} = 766.Incorrect
The required numbers are 1, 2, 11, 12, 21, 22,……..122222222.
Let us calculate how many number are these.
There are 2 one digit such numbers. There are 2^{2} two digit such numbers. And so on.
There are 2^{8 }eight digit such numbers. All the nine digit numbers begging with 1 and written by means of 1 and 2 are smaller than 2.10^{8}. Thus, there are 2^{8} such nine digit numbers.
Hence, the required number of numbers is
2 + 2^{8} + 2^{3} + …. + 2^{8} + 2^{8}
= 2(2^{8}1)/(21) + 2^{8} = 2^{9} – 2 + 2^{8} = 766. 
Question 3 of 5
3. Question
Consider the sequence 1, 3, 3, 3, 5, 5, 5, 5, 5, 7, 7, 7, 7, 7, 7, 7, 9…What will be the 4800^{th} term?
Correct
Incorrect

Question 4 of 5
4. Question
Find the number of threedigit numbers which when divided by 11 leave a remainder of 9, and when divided by 7 leave a remainder of 2?
Correct
Incorrect

Question 5 of 5
5. Question
A two digit number is divided by the sum of its digits. What is the maximum possible remainder?
Correct
The maximum possible remainder must be less than 18 as the sum of any two digits cannot be greater than 18. So we check when the sum of digits is 18. If the sum of digits is 18 the only possible remainder is 9 in case of 99. Similarly if the sum of digits is 17 the maximum possible remainder is 14 in case of 98. Similarly if the sum of digits is 16 the maximum possible remainder is 15 in case of 79. The remainder we have already got is 15 and all other sums of digits will be 15 or less than 15. So 15 has to be the answer
Incorrect
The maximum possible remainder must be less than 18 as the sum of any two digits cannot be greater than 18. So we check when the sum of digits is 18. If the sum of digits is 18 the only possible remainder is 9 in case of 99. Similarly if the sum of digits is 17 the maximum possible remainder is 14 in case of 98. Similarly if the sum of digits is 16 the maximum possible remainder is 15 in case of 79. The remainder we have already got is 15 and all other sums of digits will be 15 or less than 15. So 15 has to be the answer