# 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

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

## Examples on Integers

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) \(\ \ \frac{n}{2} \)

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.

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) \(\ \ n=5x+18\)

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,

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

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

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).

Incorrect Choices:

(A),(C),(D),(E)

See the Solution for why we can eliminate these choices.

## Examples on Even and Odd Numbers

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

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.

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) \(\ \ 20\)

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 \(y\). \(y = x+2 = 16+2=18\).

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) \(\ \ 54\)

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 \(100\).

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.

## Review

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

\(a^{2}-b^{2}=(a+b)(a-b)\)

## SAT Tips for Numbers

- 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