# Number Series Questions for TNPSC Exams - 1

Question: 1

The numbers from1 to 29 are written side by side as follows:

1234567891011121314 …….. 2829

If this number is divided by 9, then what is the remainder?

(A) 1

(B) 2

(C) 3

(D) 4

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.

Question: 2

3897 x 999 = ?

(A) 3639403

(B) 3883203

(C) 3893103

(D) 3791203

Ans: C

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

= 3897000 - 3897 = 3893103.

Question: 3

(80)^{2} - (65)^{2} + 81 = ?

(A) 2094

(B) 2175

(C) 2256

(D) 2357

Ans: C

(80)^{2} - (65)^{2} + 81 = (80 + 65) (80 - 65) + 81

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

Question: 4

If and b are positive integers, a > b and

(a + b)^{2} - (a – b)^{2} > 29, then the smallest value of a is

(A) 3

(B) 4

(C) 6

(D) 7

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.

Question: 5

The unit’s digit of 13^{2003} is

(A) 1

(B) 3

(C) 7

(D) 9

Ans: C

3^{4} gives unit digit 1.

So, (3^{4})^{500} gives unit digit 1.

And, 3^{3} gives unit digit 7.

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

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