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“Algebra Survival” Program, v. 2.0, has just arrived!


The Second Edition of both the Algebra Survival Guide and its companion Workbook are officially here!

Check out this video for a full run-down on the new books, and see how — for a limited time — you can get them for a great discount at the Singing Turtle website.

 

Here’s the PDF with sample pages from the books: SAMPLER ASG2, ASW2.

And here’s the website where you can check out the books more fully and purchase the books.

 

 

 

 

 

 

 

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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…)

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

NOW TRY THESE PROBLEMS:

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

ANSWERS:

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

Divisibility by 7: Is there really a trick?


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.

PRACTICE:

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

ANSWERS:

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

Divisibility by 4 — Principles & Trick


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!

EXAMPLE 1:
Does 4 divide into 58?

PROCESS:
—  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.


EXAMPLE 2:

Does 4 divide into 376?

PROCESS:
—  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.

EXAMPLE 3:
Does 4 divide into 57,794?

PROCESS:
—  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

ANSWERS:

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.

Multiplication Trick — Multiplying by 5


Time for a math trick …

Q:  How do you multiply an even number by 5 in lightning speed?

A:  Divide the number by 2, then tack on a “0.”

Example:   5 x 24

Divide 24 by 2 to get 12.

Tack a “0” onto 12 to get 120. Presto, nothing up your sleeve. It’s that easy.

Why does it work? Hint: Think about how we multiply by 10. Then think about how multiplying by 5 compares to multiplying by 10.

Rotated version of File:Symbol support2 vote.svg.

Image via Wikipedia


Try these for fun (answers at bottom of post):

a)  5  x  16

b)  5  x  8

c)  5  x  28

d)  5  x  64

e)  5  x  142

f)  5  x  2,468

g)  5  x  6,042

h)  5  x  86,432

j)  5  x  888,888


Answers:

a)  5  x  16 = 80

b)  5  x  8 = 40

c)  5  x  28 = 140

d)  5  x  64 = 320

e)  5  x  142 = 710

f)  5  x  2,468 = 12,340

g)  5  x  6,042 = 30,210

h)  5  x  86,432 = 432,160

j)  5 x 888,888 = 4,444,440

k)  5  x  2,486,248 = 12,431,240