Question:

The least value of x, for which the expression \( x^2+x+17 \) will not give a prime number, is

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Substitute the constant term itself. If x equals 17, every term of x squared plus x plus 17 is a multiple of 17, so the value cannot be prime. Check that 7, 11 and 13 all give primes.
Updated On: Jul 17, 2026
  • 7
  • 11
  • 13
  • 17
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The Correct Option is D

Solution and Explanation

Step 1: Understand what the question wants.
A prime number is a whole number bigger than 1 whose only factors are 1 and itself. The expression \( x^2+x+17 \) is a well-known prime-generating polynomial: it turns out prime for many small whole values of \( x \). We have to find the smallest listed \( x \) that breaks the pattern, that is, the one that makes the value composite.
Since only four values are offered, the safe method is to plug each one in, starting from the smallest, and test the result for primality.

Step 2: Test x = 7.
\[ 7^2+7+17 = 49+7+17 = 73 \]
To test 73, we only need to try prime divisors up to \( \sqrt{73} \approx 8.5 \), that is 2, 3, 5 and 7. It is odd, its digits add to 10 so it is not divisible by 3, it does not end in 0 or 5, and \( 73 = 7 \times 10 + 3 \) so 7 does not divide it. So 73 is prime and option (A) fails.

Step 3: Test x = 11.
\[ 11^2+11+17 = 121+11+17 = 149 \]
Check divisors up to \( \sqrt{149} \approx 12.2 \): 2, 3, 5, 7 and 11. It is odd; digit sum 14 rules out 3; it does not end in 0 or 5; \( 149 = 7 \times 21 + 2 \); \( 149 = 11 \times 13 + 6 \). So 149 is prime and option (B) fails.

Step 4: Test x = 13.
\[ 13^2+13+17 = 169+13+17 = 199 \]
Check divisors up to \( \sqrt{199} \approx 14.1 \): 2, 3, 5, 7, 11 and 13. It is odd; digit sum 19 rules out 3; it does not end in 0 or 5; \( 199 = 7 \times 28 + 3 \); \( 199 = 11 \times 18 + 1 \); \( 199 = 13 \times 15 + 4 \). So 199 is prime and option (C) fails.

Step 5: Test x = 17.
\[ 17^2+17+17 = 289+17+17 = 323 \]
Here a shortcut helps. Put \( x = 17 \) into the expression symbolically:
\[ 17^2+17+17 = 17(17+1+1) = 17 \times 19 = 323 \]
Every term carries a factor of 17, so the whole value must be a multiple of 17. Since \( 323 = 17 \times 19 \) has factors other than 1 and itself, it is composite, not prime.

Step 6: Pick the option.
Among the four choices, 7, 11 and 13 all give primes, and only 17 gives a composite number. So the least listed value that fails is 17.

Final Answer:
At \( x = 17 \) the expression equals \( 17 \times 19 = 323 \), which is not prime.
\[ \boxed{17} \]
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