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 divisible by an integer $b$ (or is a multiple of $b$) 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

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:

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

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 the largest prime 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 the largest prime 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.

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

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