Remainders, Divisibility Practice Questions

If there is a remainder of \(5\) when \(x\) is divided by \(9\), it must be true that \(x\) is five more than a multiple of \(9\). We can express this algebraically as \(x = 9a + 5\), where \(a\) is a positive integer.<br><br>

The question asks for the remainder when \(3x\) is divided by \(9\). If \(x = 9a + 5\), then \(3x\) can be expressed as \(3x = 27a + 15\) (we just multiply the equation by \(3\)).<br><br>

If we divide the right side of the equation by \(9\), we get \(3a + \frac{15}{9}\). \(9\) will go once into \(15\), leaving a remainder of \(6\).<br><br>

Alternatively, we can pick numbers. If we add the divisor (in this case \(9\)) to the remainder (in this case \(5\)) we get the smallest possibility for \(x\). \(9 + 5 = 14\) (and note that \(\frac{14}{9}\) leaves a remainder of \(5\)).<br><br>

\(3x\) then gives us \(3(14) = 42\). \(\frac{42}{9}\) gives us \(4\) remainder \(6\) (since \(4 \times 9 = 36\) and \(36 + 6 = 42\)).<br><br>

The correct answer is E.

The definition given tells us that when x is divided by y, a remainder of (x # y) results. Since (16 # y) = 1, when 16 is divided by y, a remainder of 1 results.<br><br>We can express this algebraically as: 16 = qy + 1, where q is the quotient. Solving for qy gives us: 15 = qy. This means y must be a factor of 15.<br><br>The factors of 15 are 1, 3, 5, and 15. However, y cannot equal 1 because the divisor must be greater than the remainder. If y = 1, then dividing 16 by 1 gives a remainder of 0, not 1.<br><br>Therefore, the possible values of y are 3, 5, and 15. The sum of these integers is 3 + 5 + 15 = 23.<br><br>The correct answer is D.

The value \( \sqrt{288kx} \) can be simplified to \( 12\sqrt{2kx} \). Given that x is divisible by 6, for the purpose of solving this problem x might be restated as 6y, where y may be any positive integer. The expression \( \sqrt{288kx} \) could then be further simplified to \( 12\sqrt{12ky} \) or \( 24\sqrt{3ky} \).<br><br>Therefore, each answer choice CAN be a solution if and only if there is an integer y such that \( 24\sqrt{3ky} \) equals that answer choice. The following table shows such an integer value of y for four of the possible answer choices:<br><br>$$\begin{array}{|c|c|}\hline y & \text{Solution} \\\hline 1 & 24\sqrt{3k} \\\hline 2 & 24\sqrt{6k} \\\hline 3 & 72\sqrt{k} \\\hline k & 24k\sqrt{3} \\\hline \end{array}$$<br><br>The answer choice that cannot be the value of \( \sqrt{288kx} \) is \( 24\sqrt{k} \). For this expression to be a possible solution, y would have to equal \( \frac{1}{3} \), which is not a positive integer. Put another way, this solution would require that \( x = 2 \), which cannot be true because x is divisible by 6.<br><br>The correct answer is B.