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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
We are told that, for any natural number n1 let an be the largest integer not exceeding √n
So for n = 1, the largest integer not exceeding √1 will be 1
For n = 2, the largest integer not exceeding √2 will be 1
For n = 3, the largest integer not exceeding √3 will be 1
For n = 4, the largest integer not exceeding √4 will be 2
We see a pattern here regarding the squares of the numbers,
Listing down all the perfect squares,
1, 4, 9, 16, 25, 36, 49, 64, ...
We see that the difference between 4 and 1 is 3 and there were three natural numbers in the given pattern with the value as 1,
So we can write for the rest of the numbers as well,
3 numbers will have value 1, giving a total value of 3
5 numbers will have value 2, giving a total value of 10
7 numbers will have value 3, giving a total value of 21
9 numbers will have value 4, giving a total value of 36
11 numbers will have value 5, giving a total value of 55
13 numbers will have value 6, giving a total value of 78
Now, only the values of a49, a50 will have the value of 7, total value of 14.
Adding these values, we get the total sum as 217, which is the answer.