**Hello students, try solving these questions. The questions might seem easy but accuracy is the differentiating factor in such questions.**

- What are the last two digits of 7^2008 ?
- 01
- 21
- 61
- 71

- If n*n = 12345678987654321, what is the value of n?
- 12344321
- 1235789
- 111111111
- 11111111

- The sum of all the natural numbers from 200 to 600 (both inclusive) which are neither divisible by 8 nor by 12 is
- 123968
- 133068
- 133268
- 187332

**Solution: **

- = 2401^502

Last two digits = 01

- n = 111111111
- Number of numbers divisible by 8 from 200 to 600 = 51

Sum of numbers divisible by 8 = (51/2) (200 + 600) = 20400

Number of numbers divisible by 12 from 200 to 600 = 34

Sum of numbers divisible by 12 = (34/2) (204 + 600) = 13668

Number of numbers divisible by both 8 and 12 = 17

Sum of numbers divisible by both 8 and 12 = (17/2) (216 + 600) = 6936

Sum of all the numbers from 200 to 600 = (401/2) (200 + 600) = 160400

Therefore, required answer = 160400 – (20400 + 13668 – 6936) = 133268

Level: Medium

Happy learning!!!

Signing off,

Binit Gourav,

Academic Research Associate, BFactory