Question

Consider the following statements. For any integer n.
I. ${{\text{n}}^{\text{2}}}$+3 is never divisible by 17.
II. ${{\text{n}}^{\text{2}}}$+4 is never divisible by 17. Then
A. Both I and II are true
B. Both I and II are false
C. I is false and II is true
D. I is true and II is false

Answer 1

Hint-In this question, we need to check the divisibility of a number with 17. As we know, a product of two numbers a and b is divisible by another number c only if a or b is divisible by third number i.e. c in this case.

Complete step-by-step solution -
As given, n is an integer.
Now, ${{\text{n}}^{\text{2}}}$+4 can be written as
$ \Rightarrow $${{\text{(k + 8)}}^{\text{2}}}{\text{ + 4 = }}{{\text{k}}^{\text{2}}}{\text{ + 16k + 68}}$
(∵ we know that 68 is divisible by 7)
K(k+7) is divisible by 17 if k or (k+16) is divisible by 17.
But ${{\text{n}}^{\text{2}}}$+3 can’t be written in the form of 17’s multiple.
Thus, we say ${{\text{n}}^{\text{2}}}$+3 is divisible by 17 but ${{\text{n}}^{\text{2}}}$+4 is never divisible by 17.
So, in this question, we conclude that
Statement I is true but statement II is false.
Therefore, in this question
Option (D) is the correct answer.

Note- Here, we mention a short trick to find divisibility by 17. A solution to the problem is to extract the last digit and subtract 5 times the last digit from the remaining number and repeat this process until a two-digit number is obtained. If the obtained two-digit number is divisible by 17, then the given number is divisible by 17.