CLAT-BBA-IPM Solving Article Day 17

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

  1. What are the last two digits of 7^2008 ?
    1. 01
    2. 21
    3. 61
    4. 71

 

  1. If n*n = 12345678987654321, what is the value of n?
    1. 12344321
    2. 1235789
    3. 111111111
    4. 11111111

 

  1. The sum of all the natural numbers from 200 to 600 (both inclusive) which are neither divisible by 8 nor by 12 is
    1. 123968
    2. 133068
    3. 133268
    4. 187332

 

Solution:

  1. = 2401^502

Last two digits = 01

  1. n = 111111111
  2. 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

You May Also Like

Leave a Reply

Your email address will not be published. Required fields are marked *