Is x^2 + 9 prime? (1) x is odd (2) 3 x 7­

Official solution from Veritas Prep.

I have a question about something..this is something I have had trouble with. Let me explain, in this question..there has nothing been stated about what "x" is..they have provided us simply with the equation. Are questions given in the same format on the GMAT? or is it explicitly going to be specified that "x is a real number". Please help as it will clear a lot for me. Request help from Bunuel

By default on the GMAT all numbers are real.

In the 1st statement if x=1 then the answer is 10 which is not prime so shouldnt that make 1) insufficient? What am I doing wrong?

This is a n YES/NO data sufficiency question. A statement is sufficient if you can get a definite YES or definite NO to the question asked. Here we need to find out whether x^2 + 9 is a prime number. (1) says that x is an odd number. For an odd number x^2 + 9 = odd + odd = even. The only even prime is 2 but x^2 + 9 is more than 2, therefore it's NOT a prime. We have a definite NO answer to the question, which means that it's sufficient.

Hope it's clear.

What a tricky question!!! The lesson is to never assume that x is an integer unless it is stated.

Statement 1: (odd*odd) + odd = odd + odd = even. All even numbers are divisible by 2, so the statement is not prime. Statement 1 is sufficient.

Statement 2: 3<=x<=7. If you plug in x={3,4,5,6,7} you will see that all the results are divisible. However, x is not limited to integers. x can be pi, which will make this statement insufficient.

Since A is sufficient and B is insufficient, the answer is A.

Just to be totally clear, in Statement 2, you still have to find at least one \(x\) that produces a prime, or else you won't have "insufficient"; \(\pi\) won't work because \(x^2+9\) still will not be prime in that case. But try \(\sqrt{10}\)...

Option A- the sum of 2 odd numbers is always an even never hence A is enough to establish that the answer will NOT be prime.
Option B can have numbers that are not integers too. Hence insufficient.
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\({x^2} + 9\,\,\,\mathop = \limits^? \,\,\,{\rm{prime}}\)

\(\left( 1 \right)\,\,\,x\,\,{\rm{odd}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^2} + 9\,\, \ge \,\,\,10\,\,\,{\rm{and}}\,\,{\rm{even}}\,\,\,\,\,\left( {{\rm{also}}\,\,{\rm{when}}\,\,x \le - 1} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\,\,\)

\(\left( 2 \right)\,\,\,3 \le x \le 7\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,x = 3\,\,\,\, \Rightarrow \,\,\,{x^2} + 9 = 18\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,x = \,\,\sqrt {10} \,\,\,\left( {\sqrt 9 < \,\,\sqrt {10} < \sqrt {49} } \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^2} + 9 = 19\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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