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Archive for the ‘Division’ Category

How to tell if a Number is Divisible by 8


I’ve explained a number of divisibility rules lately, offering tricks to tell if numbers are divisible by 2, 3, 4, 5, 6 and 7.

There is also a trick for divisibility by 8, and that’s what I’d like to explain in this post.

Essentially the trick for 8 is a lot like the trick for 4. If you’d like to refresh your memory on how that trick works, just go here. (more…)

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How to See Why the Divisibility Trick for 3 Works


One of my subscribers asked why the trick for divisibility for 3 actually works. [If you missed the post on that trick, go here:]

But the gist of the trick is this:  3 divides evenly into a number if 3 divides evenly into the sum of the digits of the number.

I’ll prove this trick for a three-digit number, but you’ll see why the proof applies to numbers with as many digits as you’d like.

Let’s call our three-digit number cde, where c is the digit in the 100s place, d is the digit in the 10s place, and e is the digit in the 1s place.

We can state the value of our cde number like this:

cde =  (100 x c)  +  (10 x d)  +  e

This shows that the c-part of the number is made up of 100 groups of c. If you think about it, there’s no reason we can’t re-write the value of this digit as 99 groups of c plus 1c, or just:  (99c + c).

In the same way, the d-part of cde is made up of 10 groups of d, which we can re-write as (9d + d). And of course the e-part of the number is just e.

So now we have this:

cde =  (99c +  c)  +  (9d + d)  +  e

Using the rules of jolly old algebra, we can shuffle the terms around a bit to get this:

cde =   99c + 9d + (c + d + e)

Then, adding a set of parentheses for clarity, we get this:

cde =   (99c + 9d) + (cde)

So far, so good. But what’s the point? Well, we’re just getting to that.

Let’s think a bit more about the 99c term. Factoring out a 3, we see that 99c =  3 x 33c. Since 3 is a factor of this expression, 99c must be divisible by 3. Aha, progress, right?

Factoring the d-term, we see that 9d = 3 x 3d. Again, since 3 is a factor of this expression, 9d is  also divisible by 3.

So we now know that both 99c and 9d are divisible by 3.

We can never forget the Divisibility Principle of Sums (DPS), which says:

If a number, x, divides evenly into both a and b, then it divides evenly into their sum, (a + b).

What does that mean here? It means that since 3 divides evenly into both 99c and 9d, it must divide evenly into their sum:   (99c + 9d).

And remember that our number, cde, equals nothing more than:  (99c + 9d) + (cde)

So, we’ve just found out that 3 divides evenly into the quantity in the first parentheses:  99c + 9d

So to find out if 3 divides evenly into the whole number cde, all that’s left is to find out whether or not 3 divides into what remains, the quantity in the second parentheses:  c + d + e

But guess what? c + d + e is the sum of the digits for our number, cde. So this idea right here is the trick for divisibility for 3:  To find out if 3 divides evenly into a number, just add up the number’s digits and see if 3 goes into that sum.  If it does, then 3 does go in; if not, 3 does not go in.

So this proves the divisibility trick for a three-digit number like cde.

To see why the same trick works for numbers with four or more digits, keep in mind that the larger digits can similarly all be broken up as we broke up the digits of c and d. For example, if we have a four-digit number, bcde, then the value of the leading digit b can be viewed, first, as (1,000 x b). And then this can be split apart once more, into 999b + b. That way, this four-digit number fits into the pattern of the trick. And this same kind of split up can be done for any digit whatsoever.

So if you follow the logic of this proof, you now see that the divisibility trick rests on solid logical/mathematical ground.

Divisibility: Find out if 3 divides evenly into an integer


Quick:  What English word has 12 letters, almost half of which are are the letter “i” — well, 5 of the 12, to be exact?

Why it’s the word “D-I-V-I-S-I-B-I-L-I-T-Y” — a great thing to understand if you’re going to spend any amount of time doing math. And guess what:  virtually ALL students do a fair amount of math, so everyone would do well to master the tricks of divisibility.

With the tricks for divisibility in your command, you will have a much easier time:

—  reducing fractions
—  multiplying fractions
—  dividing fractions
—  adding and subtracting fractions
—  finding the GCF and LCM
—  simplifying ratios
—  solving proportions
—  factoring algebraic expressions
—  factoring quadratic trinomials
—  Need I say more?

I’m sure  you get the point — divisibility tricks are handy to know.

Since the tricks of divisibility are fun and interesting, too, I’ll share as many as I can think of. If, after I’m done, you know tricks I have not mentioned, feel free to share them as comments. Or, if you know any additional tricks for the numbers I’m covering, share those! It’s always fun to learn ways to get faster at math.

Today, I’ll share the trick that tells us whether or not a number is divisible by 3. Now many of you probably know  the basic trick. But even if you do, don’t skip this blogpost. For after I show how this trick is usually presented, I’ll share a few extra tricks that most people don’t know, tricks that make the basic trick even easier to use.

Here’s how the trick is usually presented.

Take any whole number and add up its digits. If the digits add up to a multiple of 3 (3, 6, 9, 12, etc.), then 3 divides into the original number. And if the digits add up to a number that is not a multiple of 3 (5, 7, 8, 10, 11, etc.), then 3 does not divide into the original number.

Example A:  Consider 311.

Add the digits:  3 + 1 + 1  = 5

Since 5 is NOT a multiple of 3, 3 does NOT divide into 311 evenly.

Example B:  Consider 411.

Add the digits:  4 + 1 + 1  =  6

Since 6 IS a multiple of 3, 3 DOES divide into 411 evenly.

Check for yourself:

311 ÷ 3 = 103.666 … So 3 does NOT divide in evenly.

But 411 ÷ 3  =  137 exactly. So 3 DOES divide in evenly.

Isn’t it great how reliable math rules are? I mean, they ALWAYS work, if the rule is correct. In what other field do we get that level of certainty?!

Corollary #1:

Now, to make the rule work even faster, consider this trick. If the number in question has any 0s, 3s, 6s, or 9s, you can disregard those digits. For example, let’s say you need to know if 6,203 is divisible by 3. When adding up the digits, you DON’T need to add the 6, 0 or 3. All you need to do is look at the 2. Since 2 is NOT  a multiple of 3, 3 does NOT go into 6,203.

So now try this … what digits do you need to add up in the following numbers? And, based on that, is the number divisible by 3, or not?

a)  5,391
b)  16,037
c)   972,132

Answers:

a)  5,391: Consider only the 5 & the 1. DIVISIBLE by 3.
b)  16,037: Consider only the 1 & 7. NOT divisible by 3.
c)   972,132: Consider only the 7, 2, 1 & 2. DIVISIBLE by 3.

Corollary #2:

Just as you can disregard any digits that are 0, 3, 6, and 9, we can also disregard pairs of numbers that add up to a sum that’s divisible by 3. For example, if a number has a 5 and a 4, we can disregard those two digits, since they add up to 9. And if a number has an 8 and a 4, we can disregard them, since they add up to 12, a multiple of 3.

Try this. See which digits you need to consider for these numbers. Then tell whether or not the number is divisible by 3.

a)  51,954
b)  62,497
c)  102,386

Answers:

a)  51,954: Disregard 5 & 1 (since they add up to 6); disregard the 9; disregard the 5 &4 (since they add up to 9). So number is DIVISIBLE by 3. [NOTE:  If you can disregard all digits, then the number IS divisible by 3.]
b)  62,497: Disregard 6; disregard 2 & 4 (Why?); disregard 9. Consider only the 7. Number is NOT divisible by 3.
c)  102,386: Disregard 0, 3, 6. Disregard 1 & 2 (Why?). Consider only the 8. Number is NOT divisible by 3.

See how you can save time using these corollaries?

Using the trick and the corollaries, determine which numbers you need to consider, then decide whether or not 3 divides into these numbers.

a)  47
b)  915
c)  4,316
d)  84,063
e)  25,172
f)  367,492
g)  5,648
h)  12,039
i)  79
j)  617
k)  924

ANSWERS:

a)  47:  Consider the 4 and 7. Number NOT divisible by 3.
b)  915:  Consider no digits. Number IS divisible by 3.
c)  4,316:  Consider the 4, 1. Number NOT divisible by 3.
d)  84,563:  Consider only the 5. Number NOT divisible by 3.
e)  71,031:  Consider the 7, 1, 1. Number IS divisible by 3.
f)  367,492:  Consider only the 7. Number NOT divisible by 3.
g)  5,648:  Consider only the 5. Number NOT divisible by 3.
h)  12,039:  Consider no digits. Number IS divisible by 3.
i)  79:  Consider only the 7. Number NOT divisible by 3.
j)  617:  Consider the 1, 7. Number NOT divisible by 3.
k)  927:  Consider no digits. Number IS divisible by 3.

Abbreviating the Order of Operations


My recent posts about “Dear Aunt Sally” have, I hope, shown how dangerous it is to teach the memory trick of Please Excuse My Dear Aunt Sally — at least without some additional explanation.

Today I propose a way to save Dear Aunt Sally, for those of you who still like her.

As you know, the memory sentence is often abbreviated PEMDAS, which stands for:  PARENTHESES, the EXPONENTS, then MULTIPLICATION, then DIVISION, then ADDITION, then SUBTRACTION.

The problem with PEMDAS is that it makes kids think they always multiply before dividing, and that they always add before subtracting.

For students attached to PEMDAS, I let them use it, but I have them write it a novel way, so they realize they must pay attention to the left-right orientation of the operation symbols.

In the new way of writing PEMDAS, I put M and D in the same place, separate by the word “or.” Then I do the same for the A and S. So the whole memory device looks like this:

Alternative for PEMDAS

My suggestion is that teachers who like PEMDAS try this and see if your students start making fewer mistakes with the order of operations.

For those of you who never liked PEMDAS on the first place, I recommend that you check out the order of operations as presently in a clearer way, as it is in my Algebra Survival Guide. To get a feel for that, you can download the chapter of the book on Positive and Negative numbers here. Just click where it says:  See Sample Chapters of the Algebra Survival Guide and Workbook.

Then you’ll want to get the Survival Guide for the chapter on Order of Operations, which you can do through Amazon.com, here. Sorry, I can’t make everything available for free … I do have a business to run, with new products I’d like to create and make available.

Best way to write PEMDAS

“Fraction Sandwich” Practice Problems: Dividing a Fraction by a Fraction (without choking!)


Last week I sent out a blog describing my FRACTION SANDWICH approach to dividing a fraction by a fraction. The title of the blog was: Dividing Fractions:  From Annoying to FUN!

In today’s post I offer ten solid problems that help students practice the FRACTION SANDWICH, with answers just below.

Note:  if your students struggle with these problems, it’s very possible that some of them could brush up on the rules of divisibility. For a quick review of those rules, go to:

http://mathforum.org/dr.math/faq/faq.divisibility.html

 

Try these:

a)  (2/9) / (8/12)

b)  (12/16) / (15/40)

c)  (36/21) / 63/35)

d)  (44/24) / (77/48)

e)  (15/49) / (21/56)

f)  (39/52) / (24/56)

g)  (27/45) / (33/65)

h)  (12/18) / (28/45)

i)  (17/51) / (99/121)

j)  (28/18) / (42/54)

 

Answers: 

 

a)  (2/9) / (8/12)  =  1/3

b)  (12/16) / (15/40)  = 2 

c)  (36/21) / 63/35)  =  20/21

d)  (44/24) / (77/48)  =  1 and 1/7

e)  (15/49) / (21/56)  =  40/49

f)  (39/52) / (24/56)  =  1 and 3/4 

g)  (27/45) / (33/65)  =  1 and 2/11

h)  (12/18) / (28/45)  =  1 and 1/14

i)  (17/51) / (99/121)  =  1/27  

j)  (28/18) / (42/54)  =   2

How to Divide Fractions: from annoying to FUN!


O.K., I’m ready to share my amazing approach to dividing a fraction by another fraction. Well, maybe not breathtaking … like Andrew Wiles’ proof of Fermat’s Last Theorem … but at least interesting. And best of all, fun and student-friendly!

Last week I asked if anyone had any tricks up their sleeves that make it easier for students to divide fractions. And I said that I would share a trick after I heard from you.

I got a nice response from Michelle, who said that she has used the mnemonic “KFC” (like the fried chicken), which in her class stands for Keep-Change-Flip. The idea being that you KEEP the first fraction, and next you CHANGE the sign from multiplication to division. Finally you FLIP the second fraction, the fraction on the right. We have similar mnemonic where I live, which goes by the phrase: Copy-Dot-Flip, with the “dot” meaning the dot of multiplication.

But what I want to share with you is a completely different approach to dividing one fraction by another, an approach that saves time, and makes it both easier and more fun — in my humble opinion — than the standard approach.

The approach I’m going to show you works for any complex fraction situation you might encounter, such as these:

fraction-images2
For this blog post, I’m going to limit my chat to complex fractions of the arithmetic type, meaning those with numbers only, and no variables. And if it seems important, I’ll do another post later on using this very same process for algebraic fractions.

So what is this amazing approach, anyway? Well, it’s based on something I discovered on day when I was just messing around with fractions divided by fractions. I realized that after you do the KFC or the Copy-Dot-Flip, what you get — in general — is actually something really easy to grasp, as this next image will show you, along with a Quick Proof:

fraction-c3b7-1

If you take a moment to think about it, the terms in the numerator of the result — terms a and d — have something in common; they were on the outside of the original complex fraction, so I call these terms the “outers.” In the same way, the terms in the denominator of the result — terms b and c — were both on the inside of the complex fraction, so I call them the “inners.”

So when you divide fractions in this vertical format, the answer is simply the outers, multiplying each other divided by the inners, multiplying each other.

I find that students find this easy to remember and a cinch to do. This next sheet summarizes the idea, and also provides a fun way of remembering the concept, thinking about the stack of terms as a fraction “sandwich.”

fraction-c3b7-2

So, to put this in words, the four-level complex fraction that you start out with can be thought of as a sandwich, with two pieces of bread at top and bottom, and slices of bologna and cheese in the middle.

The main point is that to simplify the fraction sandwich, all you need to do is put the two slices of bread together in the numerator and multiply them, And then put the bologna and cheese together in the denominator, and multiply them.

Using this idea it becomes a lot easier to simplify these complex fractions. Here’s an image that shows how it is done, and how this approach saves time over the way we were taught to do it, using reciprocals.

fraction-c3b7-a2

And there’s more good news. This new way of looking at complex fractions also gives students a cool, new way to simplify the fractions before they get the answer. And when you do simplify fully, the answer you get will be a fraction that’s already completely reduced, so you won’t have to stress about that part.

The next two pages show you this fun and easy new way to simplify:

fraction-c3b7-31

or, or what? …  Here’s what …

fraction-c3b7-4

So now you might like to see the whole process from start to finish, so you can decide for yourself if this technique is for you. Well that’s exactly what we’re showing next. As you can see I consistently highlight the outers with pink, the inners with yellow.

fraction-c3b7-51

And finally, a “harder” problem, you might say. But check it out. Is it really any harder than the one we’ve just done? You decide.

fraction-c3b7-6

In my next blog I’ll give you a few problems like these, so you can get used to this trick, and start shaving precious seconds and nano-seconds off the time it take you to do your homework, so you spend more time doing all of those things that you want to do more:  texting, watching You-Tube, taking hikes, skating (roller and ice), etc. etc. , etc. You know better than me.

Happy Teaching and Learning!

—  Josh

 

Josh Rappaport is the author of five books on math, including the Parents Choice-award winning Algebra Survival Guide. If you like how Josh explains these problems, you’ll certainly  like the Algebra Survival Guide and companion Workbook, both of which are available on Amazon.com  Just click the links in the sidebar for more information! 

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Fraction divided by a fraction


Hi,

This one is going out to all of you … on station MCHT.

Does anyone know any good tricks for the situation where a fraction divides another fraction. In other words, for a problem like [(2/3)÷(4/5}, does anyone know of a col way to make this much easer than the way most people learn this in a school?

If so, send me your thoughts In any case, after I get a bunch of your ideas, I’ll share mine. Then we can vote on which approach we like the most.

If you’ve got an idea, send it to:

into@SingingTurtle.com

Make the subject line: Dividing a Fraction by a Fraction.

Have fun!

— Josh


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