Factors, LCM, and HCF Practice Questions

Since \(n\) must be a non-negative integer, \(n\) must be either a positive integer or zero. Also, note that the base of the exponent \(12^n\) is even and that raising \(12\) to the \(n\)th exponent is equivalent to multiplying \(12\) by itself \(n\) number of times. Since the product of even integers is always even, the value of \(12^n\) will always be even as long as \(n\) is a positive integer. For example, if \(n = 1\), then \(12^1 = 12\); if \(n = 2\), then \(12^2 = 144\), etc.



Since integer \(3,176,793\) is odd, it cannot be divisible by an even number. As a result, if \(n\) is a positive integer, then \(12^n\) (an even number) will never be a divisor of \(3,176,793\).



However, if \(n\) is equal to zero, then \(12^n = 12^0 = 1\). Since \(1\) is the only possible divisor of \(3,176,793\) that will result from raising \(12\) to a non-negative integer exponent (recall that all other outcomes will be even and thus will not be divisors of an odd integer), the value of \(n\) must be \(0\).



\(0^{12} - 12^0 = 0 - 1 = -1\)



The correct answer is B.

If the square root of \(p^2\) is an integer, \(p^2\) is a perfect square. Let's take a look at \(36\), an example of a perfect square to extrapolate some general rules about the properties of perfect squares.




Statement I:\(36\)'s factors can be listed by considering pairs of factors \((1, 36)\) \((2, 18)\) \((3,12)\) \((4, 9)\) \((6, 6)\). We can see that they are \(9\) in number. In fact, for any perfect square, the number of factors will always be odd. This stems from the fact that factors can always be listed in pairs, as we have done above. For perfect squares, however, one of the pairs of factors will have an identical pair, such as the \((6,6)\) for \(36\). The existence of this "identical pair" will always make the number of factors odd for any perfect square. Any number that is not a perfect square will automatically have an even number of factors. Statement I must be true.



Statement II: \(36\) can be expressed as \(2 \times 2 \times 3 \times 3\), the product of \(4\) prime numbers. A perfect square will always be able to be expressed as the product of an even number of prime factors because a perfect square is formed by taking some integer, in this case \(6\), and squaring it. \(6\) is comprised of one two and one three. What happens when we square this number? \((2 \times 3)^2 = 2^2 \times 3^2\). Notice that each prime element of \(6\) will show up twice in \(6^2\). In this way, the prime factors of a perfect square will always appear in pairs, so there must be an even number of them. Statement II must be true.



Statement III: \(p\), the square root of the perfect square \(p^2\) will have an odd number of factors if \(p\) itself is a perfect square as well and an even number of factors if \(p\) is not a perfect square. Statement III is not necessarily true.



The correct answer is D.

\(\gcd(16,n)=4\) implies \(n\) is divisible by \(4\) but not by \(8\) (i.e., \(n\) has exactly two factors of 2).

\(\gcd(n,45)=3\) implies \(n\) is divisible by \(3\) but not by \(5\), and also not by \(9\).

Check the choices:
\(12=2^2\cdot 3\) gives \(\gcd(16,12)=4\) and \(\gcd(45,12)=3\). So \(n=12\) works.