This question was previously asked in

DSSSB PRT General Section Official Paper - 14 Nov 2019 Shift 2

Option 1 : 7

**Calculation:**

\({3^{11}} + {3^{12}} + {3^{13}} + {3^{14}}\)

⇒ 3^{11} + 3^{11} × 3^{1} + 3^{11} × 3^{2} × 3^{11} × 3^{3}

⇒ 3^{11}(1 + 3^{1} + 3^{2} + 3^{3})

⇒ 40 × 3^{11}

**Clearly, 40 × 311 is not divisible by 7.**

__Hint__

The last digit of the given number should be multiplied by 2 and then subtracted with the rest of the number leaving the last digit. If the difference is 0 or a multiple of 7, then it is divisible

by 7.

So,

40 × 311

Multiply the last digit by 2,

0 × 2 = 0

Subtract the product (0) from the rest of the number(4 × 3^{11}).

⇒ 4 × 311 - 0 = 4 × 311

Is 4 × 3^{11} a multiple of 7? No, hence, 40 × 311 is NOT divisible by 7.