x is a positive integer less than

Question

x is a positive integer less than 30, which of the following option correctly indicates the number of values of x for which x^2/18 has 2 factors?

A. 0
B. 1
C. 4
D. 5
E. 7

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IanStewart · Accepted Answer

A positive integer with two factors is prime, and if we call that prime 'p' here, we know x^2/18 = p, so x^2 = 18p, and

x^2 = (3^2)(2)(p)

The left side, x^2, is the square of an integer, so the right side must be too (they're equal, so they're the same number). But in the prime factorization of a square, every exponent must be even. The "3^2" is fine, but we only have the single '2' on the right side. We need another '2' on the right to get up to at least 2^2, so p must contain a '2'. But p is prime, so p must be exactly equal to 2. So x^2 = 3^2*2^2, and if x is positive, x = 3*2 = 6, and we have only one value of x.

Ananabanana · Answer

At first I approached this question by tabulating all the scenarios, but it took quite a lot of time. So, I decided to solve this question purely with inferences to see if it would save me some time. And yes it did! So, here is my approach (not sure if I missed anything in this series of inferences

, pls let me know if you find gaps

) Constraints: x: integer; 1<=x <30 say A =x^2/18 has 2 factors means A = prime number (property/definition of Prime) 1. For A to be a Prime: A must be an integer first. So x^2 must be divisible by 18 = 3^2 * 2. Thus, x must be divisible by 2. 2. Say x = 2k --> x^2 = (2k)^2 = 4*k^2 ; A = \frac{4k^2}{3^2 *2} = \frac{2 * k^2}{3^2} 3. Because A must be an integer, k^2 must be divisible by 3^2; thus k^2 = 3^2 * n then, we have A = \frac{2 * 3^2 * n}{3^2} <=> A=2n always even 4. Meanwhile A is a prime and there is only 1 even prime which is 2 --> thus A=2 and there is only 1 value of x resulting in A=2 (there could be 2 values of x= 6 or -6 to have A=2; but don't forget we have a constraint that x is positive integer) Hence, answer is B: 1 value. Hope this helps.

rak08 · Answer

Bunuel   IanStewart

is it wrong if it thought so

x can be 1 which is apparent
2^5 = 32 
so cant be
if 2 cant be the case no other integer can above 2

we need x to be (prime)^9 hence the above.

thereby B

Bunuel · Answer

Yes, it’s wrong. Here’s why:

If x = 1, then x^2/18 = 1/18, not even an integer, so it cannot have 2 factors. 
 The condition is not about forcing x itself to be of the form prime^9.  
 The requirement is that x^2/18 must equal a prime. 
 Setting x^2/18 = p gives x^2 = 18p. For this to be a perfect square, p can only be 2, which makes x^2 = 36, so x = 6.

Therefore the only solution is x = 6.

Mehta17 · Answer

For a number to have 2 factors it should be a Prime number 
therefore, x^2/18 = Prime number

X^2 is divisible by 18
=> x^2 is divisible by 2*3^2
 since x^2 is divisible by 2*3^2 
X should be divisible by 2*3 = 6

Given x should be less than 30 
Therefore x can be 6,12,18,24

At x=6 
X^2/18 = 36/18 = 2 (prime)
At x=12,18 and 24, not prime

Therefore only 1 value of x where x^2/18 is prime 
Hence answer is B

x is a positive integer less than

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x is a positive integer less than

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