SAT Numbers

To solve problems about numbers on the SAT, you need to know:

• how to manipulate algebraic expressions
• the definition of consecutive integers

Consecutive integers: $\ldots, n-1, n, n+1, n+2, n+3 \ldots$, where $n$ is an integer.
Consecutive integers differ by $1$.

• the definition of even and odd numbers

Even numbers: $..., -6, -4, -2, 0, 2, 4, 6, ... , 2n, ...$, where $n$ is an integer.
Note: $0$ is an even integer.
Consecutive even integers differ by $2$.

Odd numbers: $..., -7, -5, -3, -1, 1, 3, 5, 7, ..., 2n+ 1, ...$, where $n$ is an integer.
Consecutive odd integers differ by $2$.

• the properties of even and odd integers

$$\begin{array}{l l l l l} \text{even} & \pm & \text{even} &=& \text{even}\\ \text{odd} & \pm & \text{odd}&=&\text{even}\\ \text{even} &\pm& \text{odd}&=& \text{odd}\\ \text{even} & \times& \text{even}&=& \text{even}\\ \text{even} &\times& \text{odd}&=& \text{even}\\ \text{odd} &\times& \text{odd}&=& \text{odd}\\ \end{array}$$

If $n$ is an integer, which of the following need NOT be an integer?

(A) $\ \ n^2$
(B) $\ \ 2n$
(C) $\ \ n+2$
(D) $\ \ n-2$
(E) $\ \ \frac{n}{2}$

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Correct Answer: E

Solution:

Tip: Replace variables with numbers.
If $n = 1$, then

(A) $\ \ n^2 = 1^2=1.$
(B) $\ \ 2n = 2\cdot 1 = 2.$
(C) $\ n+2 = 1 +2 =3.$
(D) $\ \ n-2 = 1-2 = -1.$
(E) $\ \ \frac{n}{2} = \frac{1}{2}.$

$\frac{1}{2}$ is not an integer. Hence the answer is (E).

In fact, for an integer $n$, $\frac{n}{2}$ is an integer if and only if $n$ is even.

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Incorrect Choices:

(A)
The product of two integers is an integer. Hence $n^2$ is an integer.

(B)
The product of two integers is an integer. Hence $2n$ is an integer.

(C)
The sum of two integers is an integer. Hence $n + 2$ is an integer.

(D)
The difference of two integers is an integer. Hence $n - 2$ is an integer.

In a set of six consecutive integers, let the sum of the three smallest numbers be $m$ and the sum of the three greatest numbers be $n$. Which of the following represents $n$ in terms of $m$?

(A) $\ \ n=m+9$
(B) $\ \ n=x+12$
(C) $\ \ n=3m+12$
(D) $\ \ n=3x+12$
(E) $\ \ n=5x+18$

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Correct Answer: A

Solution 1:

Tip: Replace variables with numbers.
Let the six consecutive integers be $1, 2, 3, 4, 5$, and $6$. Then, $m=1+2+3=6$ and $n=4+5+6=15$. $n$ is $9$ greater than $m$, or $n=m+9.$

Only answer choice (A) has this form; we eliminate the other choices.

Solution 2:

Let $x$ be the smallest of the six integers. Since consecutive integers differ by $1,$ the six integers are: $x, x+1, x+2, x+3, x+4$,and $x+5.$

$m$ is the sum of the three smallest integers. Therefore,

$$\begin{array}{l c l l l} m &=& x+x+1+x+2 &\quad \text{sum of three smallest integers} &(1)\\ m &=& 3x +3 &\quad \text{combine like terms} &(2)\\ m-3 &=& 3x &\quad \text{express}\ 3x\ \text{in terms of}\ m &(3)\\ \end{array}$$

$n$ is the sum of the three greatest integers. So,

$$\begin{array}{l c l l l} n &=& x+3+x+4+x+5 &\quad \text{sum of three greatest integers} &(4)\\ n &=& 3x + 12 &\quad \text{combine like terms} &(5)\\ n &=& m-3 +12 &\quad \text{substitute}\ 3x\ \text{with}\ m-3 &(6)\\ n &=& m+9 &\quad -3+12 = 9 &(7)\\ \end{array}$$

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Incorrect Choices:

(B)
Tip: Eliminate obviously wrong answers.
Offered to confuse you, this wrong answer expresses $n$ in terms of $x$. But the problem asks for $n$ in terms of $m$.

(C)
If in step $(6)$ of Solution 2 you substitute $x$ with $m$, instead of with $m-3$, you will get this wrong answer.

$$\begin{array}{l c l l l} n &=& 3\fbox{x} + 12 &\quad \text{combine like terms} &(5)\\ n &=& 3\fbox{m} +12 &\quad \text{mistake: substituted}\ x\ \text{with}\ m &(6)\\ \end{array}$$

Also, notice that $m$ is the sum of six consecutive integers, the smallest of which is $x$. Therefore, $m>x$. In other words, $m$ and $x$ can't be equal.

(D)
$n=3x+12$, assuming that $x$ is the smallest integer. But, the question asks for $n$ expressed in terms of $m$, not expressed in terms of $x$. Eliminate this choice.

(E)
Tip: Read the entire question carefully.
If you assume that $x$ is the smallest of the numbers, and if you solve for the sum of the six consecutive integers instead of $n$ in terms of $m$, you may select this wrong answer. A good reason to eliminate this answer immediately is that $n$ is expressed in terms of $x$, and not in terms of $m$.

The integer $66$ can be expressed as a sum of $m$ consecutive integers. Which of the following could be the value of $m$?

\(\begin{array}{r r l} &\text{I}.&2\\
&\text{II}.&4\\
&\text{III}.&6\\
\end{array}\)

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) II and III only

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Correct Answer: B

Solution:

Tip: Consecutive integers: $\ldots, n-1, n, n+1, n+2, n+3 \ldots$, where $n$ is an integer.
We analyze each of the possibilities, as follows.

I. We check if $66$ can be expressed as the sum of two consecutive integers. Let these integers be $x$ and $x+1$. Then,

$$\begin{array}{r c l l l} x+x+1&=&66 &\quad \text{sum of two consecutive integers}\\ 2x+1&=&66 &\quad \text{combine like terms}\\ 2x&=&65 &\quad \text{subtract}\ 1\ \text{from both sides}\\ x&=&32.5 &\quad \text{divide both sides by}\ 2\\ \end{array}$$

But, $x=32.5$ is not an integer. We have reached a contradiction. $66$ cannot be expressed as the sum of two consecutive integers. We can eliminate answer choices (A), (C), and (D).

II. We check if $66$ can be expressed as the sum of four consecutive integers. Let these integers be $x, x+1, x+2,$ and $x+3$. Then,

$$\begin{array}{r c l l l} x+x+1+x+2+x+3&=&66&\quad \text{sum of four consecutive integers}\\ 4x+6&=&66 &\quad \text{combine like terms}\\ 4x&=&60 &\quad \text{subtract}\ 6\ \text{from both sides}\\ x&=&15 &\quad \text{divide both sides by}\ 4\\ \end{array}$$

We can express $66$ as the sum of $15, 16, 17$, and $18$.

III. Again, we check if $66$ can be expressed as the sum of six consecutive integers. Let these integers be $x, x+1, x+2, x+3, x+4$and $x+5$. Then,

$$\begin{array}{r c l l l} x+x+1+x+2+x+3+x+4+x+5&=&66 &\quad \text{sum of six consecutive integers}\\ 6x+15&=&66 &\quad \text{combine like terms}\\ 6x&=&51 &\quad \text{subtract}\ 1\ \text{from both sides}\\ x&=&8.5 &\quad \text{divide both sides by}\ 6\\ \end{array}$$

But, $x=8.5$ is not an integer. We have reached a contradiction. $66$ cannot be expressed as the sum of six consecutive integers. We can eliminate answer choice (E).

Since the only option that works is II, the answer is (B).

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Incorrect Choices:

(A), (C), (D), (E)
See the Solution for why we can eliminate these choices.

Let $x$ and $\frac{x+7}{2}$ both be integers. Then $x$ must be:

(A) a multiple of $7$
(B) even
(C) odd
(D) positive
(E) negative

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Correct Answer: C

Solution 1:

Tip: Replace variables with numbers.
Try $x=1$. Then $\frac{x+7}{2}=\frac{1+7}{2}=\frac{8}{2}=4$, which is an integer. Since $1$ is not a multiple of $7$, it is not even, nor is it negative, we can eliminate answers (A), (B), and (E).

Try $x=2$. Then $\frac{x+7}{2}=\frac{2+7}{2}=\frac{9}{2}=4.5$, which is not an integer. Therefore, $x$ can't be even. Eliminate answer (B).

The only remaining option is (C).

Solution 2:

Tip: Look for short-cuts.
Tip: Know the properties of even and odd numbers.
The only way $\frac{x+7}{2}$ can be an integer is if the numerator is divisible by $2$. That is, the numerator must be even. $7$ is odd, and only odd $\pm$ odd $=$ even. Therefore, $x$ is odd.

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Incorrect Choices:

(A), (B), (D), (E)
See Solution 1 for how we can eliminate these choices by replacing the variable with a numbers.

If $x$ and $y$ are even consecutive integers, $x<y$, and $4x - 3y = 10$, what is the value of $y$?

(A) $\ \ 4$
(B) $\ \ 14$
(C) $\ \ 16$
(D) $\ \ 18$
(E) $\ \ 20$

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Correct Answer: D

Solution 1:

Tip: Plug and check.
Tip: Use a calculator.
(A) If $y=4$, then $x=2$ and $4\cdot 2 - 3\cdot 4 = -4 \neq 10$. This choice is wrong.
(B) If $y=14$, then $x=12$ and $4\cdot 12 - 3\cdot 14 = 6 \neq 10$. This choice is wrong.
(C) If $y=16$, then $x=14$ and $4\cdot 14 - 3\cdot 16 = 8 \neq 10$. This choice is wrong.
(D) If $y=18$, then $x=16$ and $4\cdot 16 - 3\cdot 18 = 10 = 10$. Correct answer.
(E) If $y=20$, then $x=18$ and $4\cdot 18 - 3\cdot 20 = 12 \neq 10$. This choice is wrong.

Solution 2:

Tip: Even numbers: $\ldots, -6, -4, -2, 0, 2, 4, 6, \ldots , 2n, \ldots$, where $n$ is an integer.
Consecutive even integers differ from each other by $2$. Since $x<y$, $x+2=y$. We substitute for $y$ in the given equation:

$$\begin{array}{l c l l l} 4x - 3y &=&10 &\quad \text{given} &(1)\\ 4x - 3(x+2) &=& 10 &\quad \text{substitute}\ y\ \text{with}\ x+2 &(2)\\ 4x-3x-6 &=& 10 &\quad \text{use distributive property} &(3)\\ x-6 &=& 10 &\quad \text{combine like terms} &(4)\\ x &=& 16 &\quad \text{add}\ 6\ \text{to both sides} &(5)\\ \end{array}$$

But, we are looking for $y$. $y = x+2 = 16+2=18$.

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Incorrect Choices:

(A)
Tip: Read the entire question carefully.
You might get this wrong answer if you assume $x>y$ and conclude that $x-2=y$. But we are told that $x<y$.

Alternatively, you might forget to distribute the negative sign in step $(3)$ of Solution 2, like this:

$$\begin{array}{l c l l l} 4x - 3(x+2) &=& 10 &\quad &(2)\\ 4x-3x\fbox{+}6 &=& 10 &\quad \text{mistake: didn't distribute the negative sign} &(3)\\ \end{array}$$

(B)
Tip: Plug and check.
If $y=14$, then $x=12$ and $4\cdot 12 - 3\cdot 14 = 6 \neq 10$. This choice is wrong.

(C)
Tip: Read the entire question carefully.
If you are solving for $x$, you will get this wrong answer.

(E)
Tip: Plug and check.
If $y=20$, then $x=18$ and $4\cdot 18 - 3\cdot 20 = 12 \neq 10$. This choice is wrong.

An evenly-odd number is defined as a positive integer which when once divided evenly by $2$, cannot be divided evenly by $2$ again. For example, $6$ is an evenly-odd number because it can be evenly divided by $2$ only once: $6/2=3$.

An evenly-even number is defined as an integer which can be divided evenly by $2$, the result can itself be divided evenly by $2$ and so on, until the result is $1$. For example, $16$ is an evenly-even number because $16/2 =8; 8/2=4; 4/2=2; 2/2=1$.

How many evenly-even and evenly-odd integers are there from $1$ to $100$?

(A) $\ \ 6$
(B) $\ \ 25$
(C) $\ \ 31$
(D) $\ \ 49$
(E) $\ \ 54$

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Correct Answer: C

Solution 1:

By definition, an evenly-odd number, when halved evenly, cannot be halved evenly again. This means that the result of the division by $2$ is an odd number. So, the evenly-odd numbers are formed when all the odd numbers are multiplied by $2$. Between $1$ and $100$, we count $25$ evenly-odd numbers:

$$\begin{array}{l c l l} 1\cdot 2&=&2 &\quad \text{note:}\ 2>1\\ 3\cdot 2&=&6\\ 5\cdot 2&=&10\\ 7\cdot 2&=&14\\ \vdots\\ 47\cdot 2&=&94\\ 49\cdot 2&=&98 &\quad \text{note:}\ 98<100\\ \end{array}$$

By definition, evenly-even numbers can be divided repeatedly by $2$ without a remainder until the result is $1$. The only numbers that can be repeatedly divided by $2$ and not leave a remainder are powers of $2$. We count $6$ evenly-even numbers between $1$ and $100$:

$$\begin{array}{l c l} 2^{1}&=&2\\ 2^{2}&=&4\\ 2^{3}&=&8\\ 2^{4}&=&16\\ 2^{5}&=&32\\ 2^{6}&=&64\\ \end{array}$$

Therefore, there are $25+6=31$ evenly-odd and evenly-even numbers between $1$ and $100$.

Solution 2:

Tip: For an arithmetic sequence: $a_{n}=a_{1}+(n-1)d$
Instead of counting the evenly-odd numbers, like we did in Solution 1, we realize that consecutive evenly-odd numbers differ by $4$, and therefore they form an arithmetic sequence: $2, 6, 10, \ldots$. We are only interested in those evenly-odd numbers that are between $1$ and $100$, so we have the finite sequence $2, 6, 10 \ldots 94, 98$. To find how many terms are in that sequence, we use the arithmetic sequence rule: $a_{n}=a_{1}+(n-1)d$, where $a_{n} = 98$ is the last term of the sequence, $a_{1=2}$ is the first term in the sequence, $n$ is the number of terms in the sequence, and $d=4$ is the difference between two consecutive terms.

$$\begin{array}{l c l l} a_{n}&=&a_{1}+(n-1)d &\quad \text {arithmetic sequence formula}\\ 98&=&2+(n-1)\cdot 4 &\quad \text{plug in known values}\\ 96&=&(n-1)\cdot 4 &\quad \text{subtract}\ 2\ \text{from both sides}\\ 24&=&n-1 &\quad \text{divide both sides by}\ 4\\ 25&=&n &\quad \text{add}\ 1\ \text{to both sides}\\ \end{array}$$

There are $6$ evenly-even numbers between $1$ and $100$: $2, 4, 8, 16, 32, 64$.

So, we have $25+6=31$ evenly-odd and evenly-even numbers between $1$ and $100$.

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Incorrect Choices:

(A)
If you only count the evenly-even numbers, instead of both evenly-odd and evenly-even numbers, you will get this wrong answer.

(B)
If you only count the evenly-odd numbers, instead of both evenly-odd and evenly-even numbers, you will get this wrong answer.

(D)
If you count the number of integers from $1$ to $49$, instead of the number of odd integers from $1$ to $49$ which, when multiplied by $2$ will form the evenly-odd numbers, then you will get this wrong answer.

(E)
If you count the number of integers from $1$ to $49$ and the number of evenly-even numbers from $1$ to $100$, you will get this wrong answer.

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

• Even and Odd Numbers
• SAT Reasoning Skills
• Combining Like Terms
• Distributive Property
  $a(b+c)=ab+ac$
• Simple Equations
• Applying the Difference of Two Squares Identity
  $a^{2}-b^{2}=(a+b)(a-b)$

• Know the properties of even and odd numbers.
• Consecutive integers: $\ldots, n-1, n, n+1, n+2, n+3 \ldots$, where $n$ is an integer.
• Even numbers: $\ldots, -6, -4, -2, 0, 2, 4, 6, \ldots , 2n, \ldots$, where $n$ is an integer.
• Odd numbers: $\ldots, -7, -5, -3, -1, 1, 3, 5, 7, \ldots, 2n+ 1, \ldots$, where $n$ is an integer.
• SAT General Tips