Remainders, Divisibility Practice Questions

Question 1 of 3 free Easy

When the positive integer x is divided by 9, the remainder is 5. What is the remainder when 3x is divided by 9?

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Correct answer: E

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.

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 $3$).

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 $6$.

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 $5$).

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

The correct answer is E.

Question 2 of 3 free Medium

If (x # y) represents the remainder that results when the positive integer x is divided by the positive integer y, what is the sum of all the possible values of y such that (16 # y) = 1?

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Correct answer: D

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.

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.

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.

Therefore, the possible values of y are 3, 5, and 15. The sum of these integers is 3 + 5 + 15 = 23.

The correct answer is D.

Question 3 of 3 free Medium

If k and x are positive integers and x is divisible by 6, which of the following CANNOT be the value of $ \sqrt{288kx} $?

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Correct answer: B

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} $.

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:

$$\begin{array}{|c|c|}\hline y & \text{Solution} \\\hline 1 & 24\sqrt{3k} \\\hline 2 & 24\sqrt{6k} \\\hline 3 & 72\sqrt{k} \\\hline k & 24k\sqrt{3} \\\hline \end{array}$$

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.

The correct answer is B.

Question 4 of 5 preview Hard

If n is an integer and $ n^4 $ is divisible by 32, which of the following could be the remainder when n is divided by 32?

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Question 5 of 5 preview Medium

$ 10^{25} $ – 560 is divisible by all of the following EXCEPT:

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