If X is a positive integer, is X2 + 1 an odd number?(1)X is the smallest integer that is divisible by all integers from 11 to 15, inclusive.(2)3X is an odd number.a)Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.b)Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.c)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.d)EACH statement ALONE is sufficient.e)Statements (1) and (2) TOGETHER are NOT sufficient.Correct answer is 'A'. Can you explain this answer?

Question

Answer

Steps 1 & 2: Understand Question and Draw Inferences

For X² + 1 to be odd

--> X² must be even

--> X must be even

We need to find if X is even or not

Step 3: Analyze Statement 1

X is the smallest integer that is divisible by all integers from 11 to 15, inclusive

--> X is divisible by 12 and 14

--> X is even

Sufficient.

Step 4: Analyze Statement 2

3^X is an odd number

An odd number multiplied with itself any number of times will give an odd product

--> 3^X will always be odd, whether X is even or odd

Not Sufficient.

Step 5: Analyze Both Statements Together (if needed)

We get a unique answer in step 3, so this step is not required

Answer: Option (A)

Answer

If x^2+1is odd then x^2and x are even means divisble by 2 If we multiply an odd or even number with a even number the result will be always even statment 1 expresses that x=11×12×13×14×15×I where I is an integer. Surely x/2 is an integer so x is even x is even so x^2 is even andx^2+1 is odd. SO statment is sufficient If x is an integer then values of 3^x will be 1 3 9 27 81 ... that are always odd no matters x is odd or even but for the condition x should be even SO statement 2 is not sufficient.

Answer

Statement (1): X is the smallest integer that is divisible by all integers from 11 to 15, inclusive.
If X is the smallest integer that is divisible by all integers from 11 to 15, then X must be a multiple of their least common multiple (LCM). The LCM of 11, 12, 13, 14, and 15 is 2^2 * 3 * 5 * 7 * 11 = 4620. So, X must be a multiple of 4620.
Since X is a positive integer, X can be written as X = 4620k, where k is a positive integer.

Statement (2): 3X is an odd number.
If 3X is an odd number, then X must be an odd number.

Combining the statements:
From statement (1), we know that X = 4620k, where k is a positive integer.
From statement (2), we know that X is odd.

Is X^2 - 1 an odd number?
To determine whether X^2 - 1 is odd, we need to consider the parity of X^2.

Since X is odd (from statement 2), X can be written as X = 2n + 1, where n is an integer.

Substituting this into X^2, we get (2n + 1)^2 = 4n^2 + 4n + 1 = 2(2n^2 + 2n) + 1.

We can see that X^2 is odd, since it has the form 2k + 1, where k = 2n^2 + 2n is an integer.

Therefore, X^2 - 1 = (2k + 1) - 1 = 2k, which is an even number.

Conclusion:
From the above analysis, we can conclude that X^2 - 1 is always an even number, regardless of the value of X. Therefore, statement (1) alone is sufficient to answer the question, and statement (2) alone is not necessary. Hence, the correct answer is (A).

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