Number Series Questions for TNPSC Exams - 1

Ans: C

Sum of digits of numbers from 1 to 10 = 46.

Sum of digits of numbers from 11 to 20 = 56.

Sum of digits of numbers from 21 to 29 = 63.

Sum of digits of the given number = 46 + 56 + 63 = 165.

So, the required remainder is the remainder obtained on dividing 165 by 9, which is 3.

Ans: C

3897 × 999 = 3897 × (1000 - 1) = (3897 × 1000) - (3897 × 1)

                 = 3897000 - 3897 = 3893103.

Ans: C

(80)² - (65)² + 81 = (80 + 65) (80 - 65) + 81

= (145 × 15) + 81 = (2175 + 81) = 2256.

Ans: B

Since (a + b)² - (a – b)² = 4ab,

so the given expression should be a multiple of 4.

So, the least value of 4ab is 32 and so the least value of ab is 8.

Hence, the smallest value of a is 4 and that of b is 2.

Hence, a = 4.

Ans: C

3⁴ gives unit digit 1.

So, (3⁴)⁵⁰⁰ gives unit digit 1.

And, 3³ gives unit digit 7.

∴ (13)²⁰⁰³ gives unit digit = (1 × 7) = 7.