# SAT Factors, Divisibility, and Remainders

To solve problems involving factors, divisibility, and remainders on the SAT, you need to know how to:

- divide one integer by another

An integer \(a\) is

divisibleby an integer \(b\) (or is amultipleof \(b\)) if \(a\) can be written as \(b\) times another integer:\[ a = b \times \mbox{(integer)} .\]

For example, \(10 = 5 \times 2 = 2 \times 5,\) so \(10\) is divisible by \(5\) and \(2\).

- compute the remainder upon division

Dividing one integer by another integer does not always produce an integer result. For example, \(8\) divided by \(3\) is not an integer because the positive multiples of \(3\) are \(3, 6, 9, 12, \ldots,\) which does not include \(8\). Note that \(8\) lies between \(6 = 2 \times 3\) and \(9 = 3 \times 3,\) which means the largest number of threes that go into \(8\) is \(2\). We use this to find the remainder upon division:

\[ 8 = 2 \times 3+2,\]

and say 8 divided by 3 equals 2 remainder 2. Note that the remainder will always be less than the divisor, in this case dividing by 3 will always leave a remainder that is either 0, 1, or 2.

- factorize integers
- simplify expressions using algebraic manipulation

## Examples

What is the largest integer \(m \) such that \(m\) is divisible by \(6\) and \(m < 60\)?

(A) \(\ \ 52 \)

(B) \(\ \ 54 \)

(C) \(\ \ 56 \)

(D) \(\ \ 60 \)

(E) \(\ \ 66 \)

Correct Answer: B

Solution 1:Listing the positive numbers divisible by \(6,\) we have:

\[ \begin{align} 6 \times 1 &= 6\\ 6 \times 2 &= 12\\ 6 \times 3 &= 18\\ 6 \times 4 &= 24\\ 6 \times 5 &= 30\\ 6 \times 6 &= 36\\ 6 \times 7 &= 42\\ 6 \times 8 &= 48\\ 6 \times 9 &= 54\\ 6 \times 10 &= 60. \end{align}\]

Now, we can evaluate the answers as follows.

(A) Since \(52 = (6 \times 8) + 4,\) \(52\) leaves a remainder of \(4\) upon division by \(6,\) which implies \(52\) is not divisible by \(6.\) This shows choice (A) may be eliminated.

(B) Since \(54 = 6 \times 9,\) \(54\) is divisible by \(6\) and is a possible answer.

(C) Since \(56 = (6 \times 9) + 2,\) \(56\) leaves a remainder of \(2\) upon division by \(6,\) which implies \(56\) is not divisible by \(6.\) This shows choice (C) may be eliminated.

(D) Since \(60 = 6 \times 10,\) \(60\) is divisible by \(6.\) However, we are asked to find an integer strictly smaller than \(60,\) so choice (D) may be eliminated.

(E) Since \(66 = 6 \times 11,\) \(66\) is divisible by \(6.\) However, \(66\) is not strictly smaller than \(60,\) so choice (E) may be eliminated.Since all of the answers except choice (B) have been eliminated, the correct answer is (B).

Solution 2:From Solution 1, the multiples of \(6\) are

\[6, 12, 18, 24, 30, 36, 42, 48, 54, 60, \ldots\]

Since the question asks for the largest integer \(m\) divisible by \(6\) that is strictly smaller than \(60\), we look for the largest integer \(m\) in this list that is strictly smaller than \(60\). This is satisfied for the integer appearing just before \(60\) in the list, which is the integer \(54.\) Then \(m=54,\) which is choice (B).

Incorrect Answers:

(A)

If you thought \(6\) divides \(52,\) you may have arrived at this incorrect answer.

(C)

If you thought \(6\) divides \(56,\) you may have arrived at this incorrect answer.

(D)

Tip: Read the entire question carefully.

If you were looking for an integer \(m\) with \(m \leq 60,\) instead of \(m < 60,\) then you may have arrived at this incorrect answer.

(E)

Tip: Read the entire question carefully.

If you were looking for the smallest integer \(m\) with \(m \fbox{>} 60,\) then you may have arrived at this incorrect answer.

If \(p\) is the greatest prime factor of \(98\) and \(q\) is the greatest prime factor of \(77\), what is the value of \(p+q\)?

(A) \(\ \ 7\)

(B) \(\ \ 13\)

(C) \(\ \ 14\)

(D) \(\ \ 18\)

(E) \(\ \ 56\)

Correct Answer: D

Solution:The prime factorization of \(98\) is

\[98 = 49 \times 2 = 7 \times 7 \times 2 = 7^2 \times 2,\]

so the largest prime factor of \(98\) is \(p=7\). The prime factorization of \(77\) is

\[77 = 11 \times 7,\]

so the largest prime factor of \(77\) is \(q=11.\) Then \(p+q = 7 + 11 = 18,\) which is answer (D).

Incorrect Answers:

(A)

Tip: Read the entire question carefully.

If you solved for \(p\) instead of \(p+q,\) you may have arrived at this incorrect answer.

(B)

Tip: Read the entire question carefully.

If you let \(p=2\) and \(q=11\), you may have arrived at this incorrect answer. \(p\) should be thelargestprime factor of \(98\), but \(2\) is the smallest prime factor of \(98\).

(C)

Tip: Read the entire question carefully.

If you let \(p=7\) and \(q=7,\) you may have arrived at this incorrect answer. \(q\) should be thelargestprime factor of \(77\), but \(7\) is the smallest prime factor of \(77\).

(E)

Tip: Read the entire question carefully.

If you let \(p=49\) and \(q=7,\) you may have arrived at this incorrect answer. \(p\) should be a prime number, but \(49=7\times 7\) is not a prime number.

Given five positive consecutive integers, which of the following is a list of possible remainders when the integers are divided by \(4\)?

(A) \(\ \ 0, 1, 2, 0, 1\)

(B) \(\ \ 0, 1, 2, 3, 4\)

(C) \(\ \ 0, 1, 2, 3, 0\)

(D) \(\ \ 1, 2, 3, 4, 1\)

(E) \(\ \ 2, 3, 4, 0, 1\)

Correct Answer: C

Solution 1:

Tip: Replace the variables with numbers.

Tip: Look for a counter-example.

Let the five consecutive integers be \(4, 5, 6, 7,\) and \(8\). When each is divided by \(4,\) we get the list of remainders \(0, 1, 2, 3,\) and \(0\). Only answer choice (C) follows the same pattern. Note also that we can use the list \(4, 5, 6, 7,\) and \(8\) as a counter-example for choices (A), (B), (D), and (E).

Solution 2:When an integer is divided by another integer \(k\), the remainder is an integer in the range \(0,1, 2, \ldots k-1.\) Furthermore, when non-negative consecutive integers are divided by \(k\), the remainders follow the pattern

\[ \ldots 0, 1, 2, \ldots k-1, 0, 1, 2 \ldots k-1, 0 \ldots.\]

In this case, the possible remainders after division by \(4\) are \(0, 1, 2, 3\) and the remainders follow the pattern

\[\ldots 0, 1, 2, 3, 0, \ldots.\]

Therefore, any list with remainders outside of this range may be eliminated. Since \(4\) cannot be a remainder upon division by \(4\), choices (B), (D), and (E) may be eliminated.

Since choice (A) does not include the remainder value \(3\), it does not fit this pattern and can also be eliminated. Choice (C) is the only list of remainders following the pattern, and is therefore the correct solution.

Incorrect Answers:

(A), (B), (D), and (E)

See Solution 1 for how to eliminate these choices by using a counter-example. Solution 2 uses a more abstract method to eliminate these choices.

## Review

If you thought these examples difficult and you need to review the material, these links will help:

- Factorization of integers
- Remainders upon integer division
- Prime numbers

## SAT Tips for Factors, Divisibility, and Remainders

- If you can, verify your choice.
- Plug and check.
- Eliminate obviously wrong answers.
- Replace variables with numbers.
- SAT General Tips

**Cite as:**SAT Factors, Divisibility, and Remainders.

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