BTGmoderatorLU wrote:Source: e-GMAT

X is a number which on squaring produces Y. If Y has 3 factors, how many such X are present in the first 20 natural numbers?

A. 2
B. 4
C. 5
D. 7
E. 8

The OA is E.

We have Y = X^2 such that Y has three factors.

Question: How many such Xs are present in the first 20 natural numbers?

Starting with X = 2, we have Y = 4; the factors of Y = 4 are 1, 2 and 4 -- three factors, so X = 2 is one such number.
Same goes with X = 3
But X cannot be 4 since 4^2 = 16 has 1, 2, 4, 8, and 16 -- a total of 4 factors

With this, we can intuit that X must be a prime number.

The number of prime numbers within 20 are 2, 3, 5, 7, 11, 13, 17, and 19 -- a total of eight prime numbers.

The correct answer: E

Hope this helps!

-Jay
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BTGmoderatorLU wrote:Source: e-GMAT

X is a number which on squaring produces Y. If Y has 3 factors, how many such X are present in the first 20 natural numbers?

A. 2
B. 4
C. 5
D. 7
E. 8

Since Y has 3 factors, Y must be of the form p^2 where p is a prime. Since Y = X^2, so X is a prime. Since the primes up to 20 are 2, 3, 5, 7, 11, 13, 17 and 19, there are 8 such values of X.

Let's illustrate with some examples. Let's assume that X is prime: X = 2. Then X^2 = Y = 4, which has exactly 3 factors, namely 1, 2, and 4. Now let's assume X is not prime: X = 6. Then X^2 = Y = 36, which has 9 factors, namely 1, 2, 3, 4, 6, 9, 12, 18, and 36. These two examples give us the hint that only if X is a prime number, then X^2 (or Y) will have exactly 3 factors, namely 1, X, and Y. Thus, the prime numbers up to 20 are the only values that will work.

Answer: E