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Archive for the ‘Fractions’ Category

Fraction Hack #2: The Size of the Smaller Number

I received an interesting question from alert reader Ivasallay a couple of days ago … about fractions.

Responding to my post about the fraction “hack” of using the gap between fraction numbers, Ivasallay wrote: “What if the numerator is smaller than the gap?”

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Good question, and thanks for sharing it. My answer: Yes, the numerator could be smaller than the gap, and if it is, that can help us simplify fractions, too.

Now we could have a fraction like 15/6, in which the lesser of the two numbers is the denominator, so to keep our discussion general I’m going to talk, not about the numerator, but rather about the “smaller fraction number,” whether numerator or denominator.

The way this matters is as follows: like the gap, the smaller fraction number provides an upper limit, a greatest possible value, for the GCF of the fraction’s two numbers. So if the fraction is 12/90 (smaller number being 12), that means that the GCF can be no larger than 12. If the fraction is 3/1011, with lesser number 3, the GCF can be no larger than 3.

The reason should be obvious, and when I say this I really mean it. Take the fraction 6/792, for example. Could a number larger than 6 go into both 6 and 792? Well there may be a number larger than 6 that goes into 792 evenly, but nothing larger than 6 can go into 6 itself, right? A large peg can’t go through a tiny hole, right? So there you go. Nothing larger than 6 can go into both 6 AND 792. QED.

So what does this mean for you, the math student, or parent of a math student, or the teacher of math students? … I means you want to keep in mind that in actuality two different numbers will help you nail down the size of the GCF. One is the gap between the fraction numbers, and the other is our “new friend,” the smaller of the two fraction numbers.

And here’s another … hack fact. (Whenever I say that, you know we’re heading into ‘nerd-land,’ right?) For both limiting numbers, the gap and the smaller fraction number, the only numbers that can possibly go into both fraction numbers are the FACTORS of those limiting numbers. So for example, if your fraction is 6/50, with the smaller number of 6, the only numbers that can possibly go into 6 and 50 are the factors of 6: i.e., 6, 3, or 2.

A nice rule of thumb:  see which is smaller, the gap or the smaller fraction number. Then use that smaller number as your largest possible GCF. To nail this down, let’s do two example problems.

Example 1:  8/44. What’s smaller? 8 or the gap, 36. Obviously 8! So use 8. Test the factors of 8, which are 8, 4, 2. Notice that 8 doesn’t go into both 8 and 44. But 4 does, so 4 is the GCF, and using 4, the fraction simplifies down to 2/11.

Example 2:  22/36. What’s smaller? 22 or the gap, 14. Here the gap is smaller. So test the gap’s factors: 14, 7, 2. 14 doesn’t go into 22 and 36; nor does 7. But 2 does. So 2 is the GCF, and using 2, the fraction simplifies to 11/18.

Time for you all to try your hands at this fun practice, which catapults your “number sense” to new heights.

For each problem, 1) identify the fraction’s smaller number and the gap. 2) Say which of those two numbers is smaller. 3) Using that number’s factors, find the GCF. 4) Finally, using the GCF, simplify the fraction. Answers follow.


a)   8/42

b)  12/20

c)  36/60

d)  18/96

e)  21/91


a)   8/42:  1)  smaller # = 8; gap = 34.  2)  8 < 34. 3)  GCF = 2. 4)  4/21

b)  12/20:  1)  smaller # = 12; gap = 8.  2)  8 < 12. 3)  GCF = 4. 4)  3/5

c)  36/60:  1)  smaller # = 36; gap = 24.  2)  24 < 36. 3)  GCF = 12. 4)  3/5

d)  18/96:  1)  smaller # = 18; gap = 78.  2)  18 < 78. 3)  GCF = 6. 4)  3/16

e)  21/91:  1)  smaller # = 21; gap = 70.  2)  21 < 70. 3)  GCF = 7. 4)  3/13

Josh Rappaport is the author of five math books, including the wildly popular Algebra Survival Guide and its trusty sidekick, the Algebra Survival Workbook. And FYI:  the 2nd Edition of the Survival Guide was just released in March, so get it while it’s hot off the press! If you’d like to get tutored by Josh, you can. Josh and his remarkably helpful wife, Kathy, use Skype to tutor students in the U.S. and Canada in a wide range of subjects. They also prep students for the “semi-evil” ACT and SAT college entrance tests. If you’d be interested in seeing your ACT or SAT scores soar, shoot an email to Josh, sending it to:  We’ll keep an eye out for your email, and in our office, our tutoring is always ON … except on Saturdays.

The “Ladder of Primes”

Remember the best teachers you had? Remember how they made their classes come alive? How one of the ways they made things exciting was by using analogies — little stories that connected new concepts to things you already knew and understood?

Educational researchers today are studying what makes analogies such an effective teaching tool. They have found that the use of analogies is one of the best techniques for making concepts “stick.” By relating that which students need to know to that which they already do know, teachers create bridges in understanding, and those bridges give students a way to grasp a new and difficult concepts.

The same holds true in math class. If we teachers use powerful analogies that make concepts more memorable, students are more likely to enjoy the lesson, and as a result, they’ll be more likely to remember what was taught.

I would like to present a quick-and-easy analogy that helps students learn about our number system, on the one hand, and which also helps students work with fractions, on the other hand.

The analogy is to something I call the “Ladder of Primes.” (more…)

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) Practice Makes Perfect

As the saying goes, practice makes perfect.

And boy is that true in math! Of the standard school subjects, math requires the most practice, if you want to excel at it.

That being the case, this strikes me as a great time to practice the divisibility tricks we’ve just learned.

There are many skill areas where divisibility tricks are useful — solving proportions, factoring polynomials, multiplying fractions — but one of the most obvious is the critical skill of reducing fractions.

So now I’m offering you a chance to practice your divisibility skills for 2, 3, 4, 5 and 6. We will save the trick for 7 till we have a few more tricks “up our sleeves.”

For the following problems, answer these four questions:

1)  Which of these numbers — 2, 3, 4, 5 or 6 — divides evenly into the numerator (NM)?

2)  Do the same for the denominator (DNM).

3)  Then choose the largest number that divides into both NM and DNM. For these problems, this number will be the GCF.

4)  Finally, reduce the fraction by dividing both NM and DNM by this number.

Here’s an example that shows what you’d write:

ex)  24/42

1)  NM:  2, 3, 4, 6
2)  DNM:  2, 3, 6
3)  GCF = 6
4)  Answer:   4/7


a)  20/24
b)  25/40
c)   18/48
d)  26/60
e)  21/72
f)  30/85
g)  36/66
h)  56/92
i)  84/102
j)  99/141


a)  20/24
1)   NM:  2, 4, 5
2)  DNM:  2, 3, 4, 6
3)  GCF =  4
4)  Answer:   5/6

b)  25/40
1)   NM:  5
2)  DNM:  2, 4, 5
3)  GCF =   5
4)  Answer:  5/8

c)   18/48
1)   NM:  2, 3, 6
2)  DNM:  2, 3, 4, 6
3)  GCF =  6
4)  Answer:  3/8

d)  26/60
1)   NM:  2
2)  DNM:  2, 3, 4, 5, 6
3)  GCF =  2
4)  Answer:  13/30

e)  21/72
1)   NM:  3
2)  DNM:   2, 3, 4, 6
3)  GCF =   3
4)  Answer:   7/24

f)  30/85
1)   NM:  2, 3, 5, 6
2)  DNM:  5
3)  GCF =  5
4)  Answer:  6/17

g)  36/66
1)   NM:  2, 3, 4, 6
2)  DNM:  2, 3, 6
3)  GCF =  6
4)  Answer:  6/11

h)  56/92
1)   NM:  2, 4
2)  DNM:  2, 4
3)  GCF =  4
4)  Answer:  14/23

i)  84/102
1)   NM:  2, 3, 4, 6
2)  DNM:  2, 3, 6
3)  GCF =   6
4)  Answer:   14/17

j)  99/141
1)   NM:   3
2)  DNM:  3
3)  GCF =  3
4)  Answer:   33/47

How to Tell if 7 Divides into a Number: Is there a divisibility trick for 7?

When I talk to people about divisibility, one thing I often hear is:

“There’s not really a trick for 7, is there?”

The question makes a lot of sense. For, of all the one-digit numbers, 7 is in a sense the weirdest digit. It’s odd — and in more ways than one. It does not neatly fit into our base-10 system; it is not half of 10, as is 5. And it is not even next to 0, 1, 5 or 10. It is kind of floating in the midst of the pack, but with no position that makes it distinctive. 7 is just a very weird number. And if you don’t believe me, just look at the decimal expansion of fractions that have 7 as the denominator. 2/7, for example, equals 0.2857142857 …  No other digit from 1 – 9, acting as the denominator, creates such a weird decimal tail (7 digits before it starts to repeat!).

Be that as it may, it’s pretty amazing that there is a “bona fide” trick for figuring out if 7 divides into another number, and I’m going to share it here. But like everything else about 7, even the divisibility trick is weird, so it’s hard to explain it without an example. This being so, I’ll demonstrate this trick using 154 as an example, so you can follow the trick more easily.

The three steps to this divisibility trick:

1st)  Break the number in question into two parts:  a)  the ones digit, and b) the rest of the number, meaning the digits to the left of the ones digit. Just to have a handy way of talking, we’ll call the rest of the number the “leftover.” You read the “leftover” as a number in its own right, reading the digits the standard way, left to right.
For 154, the ones digit is 4. The “leftover” is the number 15.

2nd)   Double the ones digit, and subtract the result from the leftover.
In our example, we double 4, getting 8. Then we subtract 8 from the leftover.  15 – 8 = 7.

3rd)  If the result of the subtraction is either 0 or a multiple of 7 (positive or negative), then the original number IS divisible by 7.
In our example, we see — with amazement (haha) — that the result is 7, which is of course  divisible by 7. So hurray, the original number, 154, is divisible by 7. (Did you really think I’d give an example in which the number is NOT divisible by 7? Have you no faith?)

After all that, you may be forgiven for thinking:  Geez, that trick was so “easy” (haha) … should we really call it a trick?

Of course it is a trick! It’s just not a super-duper-cinchy trick. But actually, if you use it a few times, I think you’ll find it fun and helpful, at least from time to time.

But, just in case you don’t believe me, I’ll offer a more intuitive way to think about divisibility by 7. Just use the two principles I talked about in my post on divisibility by 4.

The two principles are the DPP and the DPS, Divisibility Principle of Products, and the Divisibility Principle of Sums.

Using these tricks means that you do the following thought-steps to test divisibility by 7.

Take a number like 371.

Bear in mind that 7 goes into 35, so it will go into 35o. Subtract 350 from 371. Check out the difference. It is, totally coincidentally (haha), 21, a multiple of 7! So 7 DOES go into 371.

Similarly, to test a number like 529, think about the nearest big multiple of 7;  7 x 7 = 49, so 7 x 70 = 490. Subtract 490 from 529, and you get 39. 7 does NOT go into 39. So fuhgeddaboudit!  7 will NOT go into 529!

Essentially you find the nearest “big” multiple of 7 just below the number in question. You subtract that from the number, and look at the difference to see if 7 goes in. If SO, the original number IS divisible by 7. If NOT, the original number is NOT divisible by 7. That simple.

By now, you might be suspecting that using these DPP and DPS principles will work for any number. At least I am hoping you’re suspecting this. Are you?

If so, good. Because they will work. You can always use this technique — finding a large multiple just below the number in question and subtracting it out — to test for divisibility by any number. True, it is not a super-neat, fast trick, like the trick for 3. But it’s reliable, and with just a little practice, you’ll get quick with it.

In any case, now’s the time to test your new skills with divisibility by 7. See if 7 goes into the following numbers, and if you use the main trick described here, show how you got your answers.


a)  91
b)  92
c)  85
d)  84
e)  188
f)  189
g)  336
h)  437
i)   672
j)  763
k)  916
l)   1,491


a)  91:  9 – 2 = 7, DIVISIBLE by 7
b)  92:  9 – 4 = 5, NOT divisible by 7
c)  85:  8 – 10 = – 2, NOT divisible by 7
d)  84:  8 – 8 = 0, DIVISIBLE by 7
e)  188:  18 – 16  =  2, NOT divisible by 7
f)  189:  18 – 18 = 0, DIVISIBLE by 7
g)  336:  33 – 12 = 21, DIVISIBLE BY 7
h)  437:  43 – 14 = 29, NOT divisible by 7
i)   672:  67 – 4 = 63, DIVISIBLE by 7
j)  763:  76 – 6 = 70, DIVISIBLE by 7
k)  916:  91 – 12 = 79, NOT divisible by 7
l)   1,491:  149 – 2 = 147, DIVISIBLE by 7

How to tell if 4 Goes into a Number — Divisibility by 4

My last post offered a neat trick for seeing if 3 divides evenly into a number.

In this post, I’ll do the same thing for the number 4.

But my approach will be a bit different in this post. Instead of just presenting the “trick,” I will help us grasp the logic behind the trick by looking at two principles of divisibility. I’m doing this because learning the principles should boost your ability to work — or should I say, play? — with numbers.

First, a question: If a number divides evenly into one number, will it divide evenly into all multiples of that number? Example, given that 6 divides evenly into 30, will 6 divide evenly into the multiples of 30, such as 60, 90, 120, 150, etc. The answer is YES. This is a basic principle of divisibility, and we’ll call it the Divisibility Principle of Multiples, or just DPM, for short.

Second, related question:  if a number divides into two other numbers evenly, will it also divide evenly into the sum of those numbers? Check this out with an example, and see if it agree with your mathematical “common sense,” aka “number sense.”

4 goes into both 20 and 8, right? So does that mean that 4 goes into the sum of 20 and 8, namely 28? Well, yes, 4 does go into 28 evenly, seven times in fact.

Test one more example with  larger numbers. 9 goes into 90 and 36, right? So does that mean that 9 also must go into 90 + 36, which is 126? Yes again. This idea harmonizes with “number sense,” and it is in fact true. And we will use this soon. We’ll call this the Divisibility Principle of Sums, or just DPS.

To get started thinking about divisibility by 4, let’s consider one nice thing about 4:  it divides evenly into a number that ends in 0,  the number 20! This is helpful because in our base-10 number system, numbers that end in 0 are “friendly” — they fit into the system neatly.

Using DPM, then, since 4 goes into 20, it goes into all the multiples of 20:  20, 40, 60, 80, and  yes, 100! Why is this a big deal? Since 4 goes into 100, we can use DPM again to say that 4 goes into all multiples of 100:  200; 300; 400;  … 700; 1,300;  2,300, … we can even be certain that 4 goes into 6,235,700 since this is a multiple of 100 [100 x 62,357  =  6,235,700]

The implication of this is major:  if we want to figure out if 4 goes into any whole number, we can ignore all but the last two digits. In other words, to figure out if 4 goes into 5,296 we need only ask: does 4 go into 96. The reason is that we already know that 4 goes into 5,200, and using DPS, if 4 goes into both 5,200 and 96, we can be certain that 4 will go into 5,296.

So we now have the first part of our trick for 4:  To find out if 4 goes into any number, look only at the last two digits.

That’s a great start. But we can get even more precise.

First ask:  before 4 goes into 20, what other numbers does 4 divide into? Simple, 4 goes into 4, 8, 12, and 16.

DPS, we recall, says that  if a number, let’s call it n, goes into two other numbers — call them a and b — then n goes into their sum:   a + b.

We can use this idea right here. Since 4 divides into 20, and it also divides into 4, 8, 12 and 16, DPS guarantees that 4 also goes into the bold numbers below:
20 + 4 = 24
20 + 8 = 28
20 + 12 = 32
20 + 16 = 36

Big deal, you say, since you already knew this from the times tables.  True, but  going up one multiple of 20, you can start to see the power of this idea.

Since 4 divides into 40, and into 4, 8, 12 and 16, 4 also goes into the bold numbers:
40 + 4 = 44
40 + 8 = 48
40 + 12 = 52
40 + 16 = 56

Once again, since 4 divides into 60, and into 4, 8, 12 and 16, 4 also goes into:
60 + 4 = 64
60 + 8 = 68
60 + 12 = 72
60 + 16 = 76

Using the same pattern, we see that 4 goes into:  80, 84, 88, 92 and 96.

Great, you might say, this shows us a pattern, but not a “trick.”
Where is this long-promised trick?

What we need to realize is that the pattern leads to a trick.

For the trick, here’s what you do:

1st) Take the two digits at the end of any whole number.

2nd) Find the lesser but nearest multiple of 20, and subtract it from the two-digit number.

3rd) Look at the number you get by subtracting. If it’s a multiple of 4, then 4 DOES got into the original number. If it is NOT a multiple of 4, then 4 does NOT go into the original number.

Words, words, words, right? Let’s see some examples to give the words some life!

Does 4 divide into 58?

—  Nearest multiple of 20 to 58 is 40.
—  58 – 40 = 18
—  18 is NOT a multiple of 4, so 4 does NOT divide evenly into 58.


Does 4 divide into 376?

—  Focus on the last two digits:  76
—  Nearest multiple of 20 to 76 is 60.
—  76 – 60 = 16
—  16 IS a multiple of 4, so 4 DOES divide evenly into 376.

Does 4 divide into 57,794?

—  Focus on the last two digits:  94.
—  Nearest multiple of 20 to 94 is 80.
—  94 – 80 = 14
—  14 is NOT a multiple of 4, so 4 does NOT divide evenly into 57,794.

Make sense? If so, then you are ready to do some serious divisibility work with 4. Here are some practice problems, and their answers.

PROBLEMS:  Tell if 4 divides evenly into the following numbers.

a)   74
b)  92
c)   354
d)   768
e)  1,596
f)   3,390
g)  52,472
h)  831,062
i)  973,236
j)   17,531,958


a)   74:  74 – 60 = 14.  4  does NOT divide evenly into 74.
b)  92:  92 – 80 = 12.  4 DOES divide evenly into 92.
c)   354:  54 – 40 = 14.  4 does NOT divide evenly into 354.
d)   768:  68 – 60 = 8.  4 DOES divide evenly into 768.
e)  1,596:  96 – 80 = 16.  4 DOES divide evenly into 1,596.
f)   3,390:  90 – 80  = 10.  4 does NOT divide evenly into 3,390.
g)  52,472:  72 – 60 = 12.  4 DOES divide evenly into 52,472.
h)  831,062:  62 – 60 = 2.  4 does NOT divide evenly into 831, 062.
i)  973,236:  36 – 20 = 16.  4 DOES divide evenly into 973,236.
j)   17,531,958:  58 – 40  =  18.  4 does NOT divide evenly int0 7,531,958.

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


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


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


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.

“Hacks” for Slaying Proportions, Part 1: the Amazing Horizontal Canceling Trick

Proportions can seem intimidating, but they’re actually one of the easiest types of word problems to master. In this series I’ll offer a number of tips that help you conquer algebraic proportion problems. 

But first, a cool shortcut you can use whenever you’re facing down an algebraic proportion …

High-Octane Boost for MathIn working with proportions, I’m amazed that few students know how a canceling process that would help them find the solution more quickly and efficiently.

So I want to share the trick, for all who’ve never seen it.

Of course, given a problem like:  6/x  =  24/32,

most of us know that we can cancel vertically with the two numbers in the fraction on the right, to get:

6/x  =  3/4

Then we just cross-multiply to get:

3x  =  24, and see that x  =  8.

In other words, we know we can cancel vertically given a proportion, just as we can cancel vertically with any fraction.

What many people don’t know though, is that there’s another way we can cancel when solving proportions — horizontally!

— What? you say.

Horizontally, I say. And no, I’m not joshing.

For example,  given the proportion:  7/4  =  21/x

you can cancel horizontally with the two numbers in the numerator:  the 7 and the 21. These reduce to 1 and 3.

The proportion then becomes:

1/4  =  3/x  [I’m really not kidding.]

Cross-multiplying, you get the answer in one quick step:   x = 12.

What’s really convenient is that you can also cancel both vertically and horizontally in the same problem. For example, in

6/x  =  42/28,

you could first cancel horizontally, to get:

1/x  = 7/28

Then you can cancel vertically, to get:

1/x  =  1/4

Cross-multiplying, you get the answer in just a step:  x = 4

I find that when students cancel before cross-multiplying, they’re more apt to get the right answer, and to get less frustrated, for the numbers they deal with remain small.

For example, in the last problem, if the student had not canceled at all, he would have a cross-multiplication mess of:

6 x 28 = 42x

That sort of problem just opens up the door to arithmetic mistakes. But canceling before cross-multiplying shuts that door since it makes the numbers smaller and easier to manage.

So now you get a chance to practice horizontal cancelling!

First use horizontal cancelling to get the answer to these
proportions. Those who’d like an added challenge might like to try them in their head:

a)   x/12  =  3/4

b)  3/7  =  x/35

c)   z/48  =  7/12

d)  y/56  =  7/8

Now go really wild! Use both horizontal and vertical canceling to make quick work of these proportions:

e)  x/9  =  16/36

f)   x/22  =  30/66

g)  32/56  =  y/14

h)  13/q  =  65/35

And here are the answers to all of these problems:

 a)  x  =  9

b)  x  =  15

c)   z  =  28

d)   y  =  49

e)   x  =  4

f)   x  =  10

g)   y  =  8

h)  q  =  7

How to Practice Dividing Fractions

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:

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)


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:

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:


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.”


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.


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:


or, or what? …  Here’s what …


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.


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.


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  Just click the links in the sidebar for more information! 

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