Bunuel wrote:
7. Is x the square of an integer?
The question basically asks whether x is a perfect square (a perfect square, is an integer that is the square of an integer. For example 16=4^2, is a perfect square).
Perfect square always has even powers of its prime factors. The reverse is also true: if a number has even powers of its prime factors then it's a perfect square. For example: \(36=2^2*3^2\), powers of prime factors 2 and 3 are even.
(1) When x is divided by 12 the remainder is 6. Given that \(x=12q+6=6(2q+1)=2*3*(2q+1)\). Now, since 2q+1 is an odd number then the power of 2 in x will be odd (1), thus x cannot be a perfect square. Sufficient.
(2) When x is divided by 14 the remainder is 2. Given that \(x=14p+2\). So, x could be 2, 16, 30, ... Thus, x may or may not be a perfect square. Not sufficient.
Answer: A.
first and foremost, thanks for this problem set. it's helped me immenselyCan you offer more explanation on St. (1) and St. (2), specifically:
St. (1): which portion of the equation represents any specific "power" that is distributed to 2 ? All I can gather is that 2 will be multiplied by 3 (odd) and then multiplied by some odd number (2q + 1). In effect you are multiplying an even (6) by some odd number. How does this tell us the power that 2 is raised to?
Example:
x = 2*3*[2*2 +1) = 18
Prime factors of 18 = 2^1 and 3^2. How does the equation (2q+1) ensure that 2 will always be raised to an odd power?
St. (2): Does it make sense/is it possible to break it out in similar fashion as you did to statement 1? x = 2(7p+1) ? The (7p+1) could yield an odd or even number, depending on what q is (as displayed in your example).
Also, what program do you use for equation writing?