Guest

Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.?


Find the sum of all those integers between 100 and 800 each of which on division by 16 leaves the remainder 7.?


Grade:11

2 Answers

GAURAV AGRAWAL
29 Points
11 years ago

SUCH NUMBERS ARE 106,119,135,------------------------,791.

THEN, Tn=a+(n-1)d gives n=44

so,Sn=n/2*(a+Tn) gives Sn=19668

so,answer is 19668

Sohini Paul
31 Points
11 years ago

Any number that leaves a remainder of 7 on division with 16 can be represented as 16k+7. In this case the number lies between 100 and 800. So 6≤k≤49. Now the sequence 16k+7 is an AP with common difference 16 starting from 103 and ending with 791, containing (49-6)+1=44 terms.

So the required sum is (n/2)(a+l)=(44/2)(103+791)= 19668

 

Please like the answer if you understood the solution.

SOHINI PAUL

Dept of Electrical engg

IIT Kharagpur

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free