It’s interesting that algebra can be useful in arithmetic.

Since arithmetic is a prerequsite to learning algebra, most algebra courses don’t focus on how to apply algebra to do arithmetic.

However, you can apply your algebra skills to arithmetic problems.

Following is an example.

Suppose that you would like to multiply 58 times 58, but don’t have a calculator (or cell phone) handy.

One way to do this, which is often taught in elementary school, is basically to multiply 8 times 58, then multiply 50 times 58, and add these together.

Algebraically, what your elementary school teacher taught you is that 58 × 58 = (50 + 8) × 58 = 50 × 58 + 8 × 58.

You may not have realized it at the time, but your teacher was applying the distributive property of mathematics.

When you solve the problem this way, first you multiply 8 × 58 to get 464. Then you multiply 50 × 58 to get 2900. Finally, you add these together to get 3364.

Now let’s see how algebra can supply an alternative solution.

I’d prefer to work with nice round numbers. Hey, 60 is a round number close to 58. If I could work with a round number, like 60, and a small number, like 2, that would make the arithmetic much simpler to do without a calculator.

So let’s write 58 as 60 minus 2.

Then the problem is 58 × 58 = (60 – 2) × (60 – 2).

Use the f.o.i.l. method from algebra. Recall that the f.o.i.l. method stands for “first,” “outside,” “inside,” “last.”

One example of the f.o.i.l. method from algebra is (x + y)(x + y) = x² + xy + xy +y² = x² + 2xy +y².

Let’s apply this to the arithmetic problem. We can think of (60 – 2) as (x + y), where x = 60 and y = –2.

58 × 58 = (60 – 2) × (60 – 2) = 60 × 60 + 2 (60)(–2) + (–2) × (–2).

Look what we did. We wrote the original problem, 58 × 58, in terms of three simple multiplication problems: 60 × 60, 2 × 60 × (–2), and (–2) × (–2).

58 × 58 = 3600 – 240 + 4 = 3364.

Note that (–2) × (–2) = +4 because the two minus signs cancel out.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest release:

- 300+ Mathematical Pattern Puzzles

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing

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Here is an exercise in pattern recognition.

It’s not a linear pattern.

This is an array, so there is a slight geometric element to the pattern.

See if you can figure out the missing numbers in the above puzzle.

Study the four arrays.

See if you can recognize the pattern.

Once you identify the pattern, apply it to the fifth array.

Spoiler alert.

If you scroll down too far…

You will run into the answer.

So stop scrolling down…

If you would like more time to solve the puzzle.

Ready or not.

Here it comes.

First the answer:

Now the solution.

Begin with the top left number.

Double the top left number. That makes the top right number. 5 doubled = 10.

Now multiply the top two numbers. That makes the bottom left number. 5 times 10 = 50.

Now add the bottom left number to the top right number. That makes the bottom right number. 50 plus 10 equals 60.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest releases:

- 300+ Mathematical Pattern Puzzles
- Basic Linear Graphing Skills Practice Workbook
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing

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Here is a math puzzle challenge.

**Hint**: Each of the 5 patterns below has something in common.

**Directions**: See if you can figure out which numbers go in the blanks.

- 1, 2, 4, 6, 10, 12, 16, 18, 22, _, _
- 4, 6, 10, 14, 22, 26, 34, 38, _, _
- 3, 7, 13, 19, 29, 37, _, _
- 4, 9, 25, 49, 121, 169, 289, _, _
- 5, 8, 12, 18, 24, 30, 36, 42, 52, 60, 68, _, _

If you need help, you can find hints below.

But don’t scroll too far or you’ll run into the answers and explanations.

Each pattern above has something in common.

They all involve prime numbers.

A prime number is only evenly divisible by two integers: 1 and itself.

For example, 7 is a prime number because the only integers that can multiply together to make 7 are 1 and 7.

In contrast, 6 isn’t a prime number because 2 x 3 = 6 (in addition to 1 x 6).

Here are the first several prime numbers.

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37

Each of the puzzles above relates to these prime numbers.

When you’re ready, you can find answers and explanations below.

Here are the answers and explanations to the math puzzles:

- 28, 30. Explanation: Subtract 1 from each prime number: 2 – 1 = 1, 3 – 1 = 2, 5 – 1 = 4, 7 – 1 = 6, 11 – 1 = 10, etc.
- 46, 58. Explanation: Double each prime number: 2 x 2 = 4, 3 x 2 = 6, 5 x 2 = 10, 7 x 2 = 14, 11 x 2 = 22, etc.
- 43, 53. Explanation: Every other prime number: 3 (skip 5) 7 (skip 11) 13 (skip 17) 19 (skip 23) 29 etc.
- 361, 529. Explanation: Square each prime number: 2² = 4, 3² = 9, 5² = 25, 7² = 49, 11² = 121, etc.
- 78, 84. Explanation: Add consecutive prime numbers together: 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, 8 + 13 = 21, 13 + 21 = 34, etc.

One way is to follow my blog. I will post occasional math puzzles in the future.

Another way is to check out my newest book, 300+ Mathematical Pattern Puzzles.

It starts out easy and the level of challenge grows progressively so that puzzlers of all abilities can find many puzzles to enjoy.

A wide variety of topics are covered, including:

- visual patterns
- arithmetic
- repeating patterns
- Roman numerals
- Fibonacci sequence
- prime numbers
- arrays
- analogies
- and much more

The cover was designed by Melissa Stevens at www.theillustratedauthor.net.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest releases:

- 300+ Mathematical Pattern Puzzles
- Basic Linear Graphing Skills Practice Workbook
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule

Improve Your Math Fluency. Build fluency in:

- arithmetic
- long division
- fractions
- algebra
- trigonometry
- graphing

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Slope is a measure that indicates how steep or shallow a straight line is:

- A line with greater slope is steeper.
- A line with less slope is shallower.
- A horizontal line has zero slope.

Slope can be positive or negative:

- A line with positive slope slants upward.
- A line with negative slope slants downward.
- A line with zero slope is horizontal.

The slope of a straight line equals rise over run.

- The rise between two points is vertical. It’s the change in y.
- The run between two points is horizontal. It’s the change in x.

From the graph of a straight line, determine the slope as follows:

- Mark two points on the line.
- Read the x- and y-coordinates of the two points, (x1, y1) and (x2, y2).
- Subtract y2 – y1 to get the rise.
- Subtract x2 – x1 to get the run.
- Divide the rise by the run.

Here are a few tips:

- When choosing the two points, try to find points where both x and y are easy to read without interpolating. This isn’t always possible: In that case, at least one coordinate should be easy to read without interpolating.
- Choose two points far apart. This reduces the relative error in interpolating.
- Make sure that both points lie on the straight line. Don’t choose a point that’s close to the line, but doesn’t lie on it.

**Example**: Find the slope of the straight line in the graph below.

**Solution**: First, choose two points on the line. Ideally, these points should be far apart and easy to read.

In this case, it’s easy to read both the x- and y-coordinates for the leftmost and rightmost points shown in the graph. So let’s choose those.

- The leftmost point has coordinates (0, 3).
- The rightmost point has coordinates (10, 8).

Subtract the y-values to determine the rise:

y2 – y1 = 8 – 3 = 5

Subtract the x-values to determine the run:

x2 – x1 = 10 – 0 = 10

(In coordinate graphing, recall that x is horizontal and y is vertical.)

Divide the rise by the run to find the slope:

The slope of the line is 0.5.

**Check**: You can check your answer as follows.

Look at the graph. Starting from (0, 3), the next point that’s easy to read is (2, 1).

From (0, 3) to (2, 1), the line goes one unit up (vertically) and 2 units over (horizontally).

The ratio of the rise to the run is 1 to 2. Divide the rise (of 1) by the run (of 2). The slope is 0.5. ♦

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

Newest release:

- Basic Linear
**Graphing Skills**Practice Workbook

Related books:

- Trigonometry Essentials Practice Workbook with Answers
- Learn or Review Trigonometry Essential Skills
- Algebra Essentials Practice Workbook with Answers
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule

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There is a simple way to remember the sine, cosine, and tangent of special trigonometry angles.

The special trig angles are 0º, 30º, 45º, 60º, and 90º. What makes these angles special? The 30º-60º-90º triangle is one-half of an equilateral triangle, while the 45º-45º-90º triangle is one-half of a square. In both cases, the trig functions (sine, cosine, and tangent) can be expressed as simple ratios.

Here is the trick for quickly working out the sine, cosine, and tangent of 0º, 30º, 45º, 60º, and 90º.

Write the special angles in order.

Write the integers 0 thru 4 in order.

Squareroot each number.

Divide each number by 2.

These are the sine of 0º, 30º, 45º, 60º, and 90º.

It’s that simple. Here’s a recap:

- Write the numbers 0 thru 4.
- Squareroot each number.
- Divide each number by 2.

Just write the previous numbers in reverse order.

Why does this work? Because the sine of theta equals the cosine of the complement of theta: sin(*θ*)=cos(90º–*θ*). What’s opposite to theta is adjacent to its complement.

Divide sine theta by cosine theta.

This chart shows all of the steps together.

- Write the special angles.
- Write the integers 0 thru 4.
- Squareroot each number.
- Divide each number by 2. This gives you sine of theta.
- Write the numbers in reverse order. This gives you cosine of theta.
- Divide the previous two rows (sine over cosine). This gives you tangent theta.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

- Trigonometry Essentials Practice Workbook with Answers
- Learn or Review Trigonometry Essential Skills
- Trigonometry Flash Cards (for Kindle)
- Algebra Essentials Practice Workbook with Answers
- Systems of Equations: Simultaneous, Substitution, Cramer’s Rule
- Other volumes cover fractions, long division, arithmetic, and more
- Also look for books on the fourth dimension, astronomy, conceptual chemistry, and more

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When students learn multiplication, it starts out easy.

Too easy, as 0 times anything equals 0 (e.g. 0 x 8 = 0, while anything times 1 equals itself (e.g. 4 x 1 = 4).

But it doesn’t stay easy. Most students struggle to remember multiplication facts where one of the numbers is 6, 7, 8, or 9.

However, there are some tips to aid in memorization.

For one, mirror images don’t matter. The word for this is commutative.

This means 2 x 8 = 8 x 2, for example. Both equal 16.

So if you know 6 x 8 = 48, you also know that 8 x 6 = 48. You don’t need to memorize both.

One of the tricky multiplication facts is 7 x 8. But it’s easy if you remember the trick.

Remember 5678. These are 5 thru 8 in order. 56 = 7 x 8. Piece of cake, huh?

The 9’s are easy. The answer is one decade less than multiplying by 10, then make the two digits add up to 9.

For example, 9 x 7 is in the 60’s (because 10 x 7 = 70, one decade less is 60). It’s 63 since 6 + 3 = 9.

Another example is 5 x 9. It’s in the 40’s (one decade less than 5 x 10 = 50). It’s 45 since 4 + 5 = 9.

To do the trick, after you figure out the decade (by multiplying by 10 and then subtracting 10), subtract the tens digit from 9 to get the units digit.

For example, consider 9 x 6. Multiply 10 x 6 to get 60, and subtract 10 to make 50 (one decade less). Now subtract 5 (the tens digit of 50) from 9 to get 4. Therefore, 9 x 6 = 54.

Once you get the hang of it, this makes remembering the 9’s easy. Try out all the 9’s to get some practice.

If you’re good at doubling numbers quickly, try writing 6 as 2 x 3.

Then 6 x 7 = 2 x 3 x 7. If you know 3 x 7 = 21, double 21 to get 42.

Similarly, for 6 x 4 = 2 x 3 x 4, start with 3 x 4 = 12 and double 12 to make 24.

You can use the doubling trick for the 8’s, too. Just double the number 3 times.

For example, consider 5 x 8. Double 5 three times: 10, 20, 40. So 5 x 8 = 40.

Try 8 x 6. Double 6 three times: 12, 24, 48. Therefore, 8 x 6 = 48.

It works with the 4’s, also. Just double twice.

With 4 x 9, double 9 twice: 18, 36. So 4 x 9 = 36.

That leaves 7 x 7 = 49. You should know 7 x 9 from the 9’s trick.

You can make 7 x 6 and 7 x 8 from the doubling tricks. (The latter you can also know from the 5678 trick.)

5 and under are easier. So to complete the 7’s, you really just need to memorize 7 x 7 = 49.

This covers the 6 thru 9’s, which tend to be the trickier multiplication facts.

The 10’s and 11’s are easy. For the 10’s, just add a zero, as in 8 x 10 = 80 or 6 x 10 = 60.

For the 11’s, with 1 thru 9 just double the digits, like 3 x 11 =33 or 11 x 8 = 88. Get 11 x 10 = 110 from the 10’s trick (add a zero). Then you just need to memorize that 11 x 11 = 121 to complete the 11’s.

You can get the 12’s by doubling the 6’s. For example, knowing that 6 x 5 = 30, you can find that 12 x 5 = 60 by doubling 6 x 5.

Do you know any other tips for remembering multiplication facts 0-12? If so, please share them in the comments.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

- arithmetic facts
- multi-digit arithmetic
- long division with remainders
- fractions, decimals, and percents
- algebra and trigonometry

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The Fibonacci sequence adds consecutive terms:

- 1
- 1
- 1 + 1 = 2
- 2 + 1 = 3
- 3 + 2 = 5
- 5 + 3 = 8
- 8 + 5 = 13
- 13 + 8 = 21
- 21 + 13 = 34
- 34 + 21 = 55
- 55 + 34 = 89

Since the last two terms were 55 and 89, we would add these together to get 89 + 55 = 144.

Then you would add 144 and 89 to make 233, and so on.

I saw a cool pattern involving the Fibonacci sequence recently at the Mathemagical Site:

Fibonacci Triples via Mathemagical Site

This involves Fibonacci triples.

A Fibonacci triple consists of three consecutive numbers from the Fibonacci sequence, such as:

- 1, 1, 2
- 1, 2, 3
- 2, 3, 5
- 3, 5, 8
- 5, 8, 13
- 8, 13, 21
- 13, 21, 34

As shown on the Mathemagical Site, the square of the middle number is always one less or one more than the product of the first and third numbers:

Here are a few examples:

- (2, 3, 5): 3 x 3 = 2 x 5 – 1
- (3, 5, 8): 5 x 5 = 3 x 8 + 1
- (5, 8, 13): 8 x 8 = 5 x 13 – 1

Curious about this, I’ve been working through the algebra, and finally came up with an algebraic proof, which follows.

My proof is algebraic and not necessarily obvious, but since I worked it out, I thought I would share it here.

We begin with two facts about the Fibonacci sequence:

These are two different ways of saying that if you add two consecutive numbers from the Fibonacci sequence, you get the next number in the sequence.

Now solve for x_{n} in each sequence:

Multiply these together:

Foil this out:

Recall that

Plug this into the first term on the right-hand side of the previous equation:

Distribute:

Two of these terms cancel (the remaining terms are rearranged):

Believe it or not, this basically concludes the proof. The remainder is basically interpreting this result.

This is a recursion relation that relates the square of the n-th term to the square of the previous term (x_{n-1} times itself).

Following is the Fibonacci sequence, labeling values of n:

- n = 1 is 1.
- n = 2 is 1.
- n = 3 is 2.
- n = 4 is 3.
- n = 5 is 5.
- n = 6 is 8.
- n = 7 is 13.
- n = 8 is 21.

Let’s plug in n = 3 and see what happens:

If instead you plug in n = 4, you get:

Now just plug in these two expressions (x_{3}x_{3} – x_{4}x_{2} and x_{4}x_{4} – x_{5}x_{3}) into the previous recursion relation and you can prove that all of the Fibonacci triples satisfy one of the following relations:

That is, if x_{3}x_{3} – x_{4}x_{2} = 1 and x_{4}x_{4} – x_{5}x_{3} = -1, the previous recursion relation gives similar expressions for x_{5}x_{5} – x_{6}x_{4}, x_{6}x_{6} – x_{7}x_{5}, and so on.

Copyright © 2015 Chris McMullen, author of the *Improve Your Math Fluency* series of math workbooks

- Algebra Essentials Practice Workbook with Answers
- Trigonometry Essentials Practice Workbook with Answers
- Learn or Review Trigonometry: Essential Skills

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