Free Math Tips

Mathematical trick: multiply by powers of 2

Wednesday, January 13, 2010

Students who follow courses in electronics and systems know that it is essential to know the binary system and as a result quickly calculate the powers of 2 of a number. As repeatedly explained below, the lecture method, with a little 'training will make you fast in mental calculation.
To multiply a number by 2, 4, 8, 16, 32 or any other power of 2 is sufficient to double the product as many times as is necessary for instance if you want to multiply a number by 16 is sufficient to recall that 16 is the product of 4 x 4 or in another form:

16 = 2 x 2 x 2 x 2

Let's see some examples

Example 1

We make the product 18 x 16 in

18 x 2 = 36
36 x 2 = 72
72 x 2 = 144
144 x 2 = 288

Example 2

Execute the product of 27 x 8

27 x 2 = 54
54 x 2 = 108
108 x 2 = 216

Example 3

Execute the product of 63 x 8

63 x 2 = 126
126 x 2 = 252
252 x 2 = 504

Mathematical trick: multiply a number by 5, 25 or 125

Multiply by 5

Multiplying a number by 5 and multiply it by like 10 and then divide by 2.

Remember that (obvious but better remembered) that multiplying a number by 10 means add one zero to the bottom of the number.

16 × 5 = (16 × 10) / 2 = 160 / 2 = 80

Another example:

82 × 5 = (82 × 10) / 2 = 820 / 2 = 410

Again

6840 × 5 = 68400 / 2 = 34200

Multiply by 25

Multiplying a number by 25 you multiply the number by 100 (remember that you must add two zeros to the bottom number) and then divide by 4 (derived from 100 = 25 × 4).

Remember, divide by 4 and divide two to say this because 2 times for 2 × 2 = 4

84 × 25 = 8400 / 4 = 4200 / 2 = 2100

38 × 25 = 3800 / 4 = 1900 / 2 = 950

Multiply by 125

The same method, multiply by 125 means to multiply the number by 1000 and divide by 8 (derived from 1000 = 125 × 8).

Note that 8 × 2x2 = 2 and multiply by 1000 means add 3 zeros at the end of the number.

24 × 125 = 24000 / 8 = 12000 / 4 = 6000 / 2 = 3000

58 × 125 = 58000 / 8 = 29000 / 4 = 14500 / 2 = 7250

Mathematical trick: multiply by 11

To multiply a number by 11 and add enough pairs of numbers from right within the given number except the numbers on the edges that should be repeated.

For example:

consider the following product: 324 x 11

* From the right type 4 (the last number to the right of 324) first digit of the product;
* Try the sum 4 +2 = 6 (the sum of 1 'and 2' figure), you get the second digit of the product;
* Try the sum 3 +2 = 5 (the sum of the 2 'and 3' figure), you get the second digit of the product;
* Repeat 3, last number of the product.

Summing up (*):

(3) (3 +2) (4 +2) (4) = 3564

(*) Groups are not intended parentheses multiplied together

Now let's see what happens when the sum of two numbers exceeds the internal ten.

Consider the product:

967 x 11

* Begin to write from right to left, the first digit of the product is the 7;
* Sum 6 +7 = 13 write 3 (2 'figure of the product) and over 1;
9 +6 = 15 * sum added to the carryover of the amount above 15: 15 +1 = 16 writing 6 with carryover of 1;
* Last figure 9 we add the carry of the sum above 9 +1 = 10 which is the last digit of the product;

and therefore:

967 x 11 = 10637

Simple right?
With a little 'training can become very fast and impress your classmates :-)

Mathematical trick: multiply a number by 9, 99 or 999

Very often I hear from students' math is one thing to genes and the affirmation that I hate more: "I am not inclined towards mathematics, but only for matters literary" I'm thinking that if a student says this is because believe someone has done this. In my opinion we are all potentially good at mathematics and literature.

Use the brain may be faster than using a calculator just to do some 'training and use some trick.

Here's the rule for multiplying a number by 9, 99 or 999

Multiplying a number by 9 means multiply by 10-1

Therefore, 9 × 9 and how to say 9x (10-1) from which 9 × 10-9 to 90-9 in 81 words

Here's another example: 78 × 9 = 780-78 = 702

To multiply a number by 99 to multiply the number 100-1

Therefore, 54 × 99 = 54x (100-1) = 5400-54 = 5346.

Multiply by 999 using the same method used for 9 and 99

42 × 999 = 42x (1000-1) = 42000-42 = 41958

Dividing by 14

Tuesday, January 12, 2010

Sara Heikali builds on her divisibility test for seven:

How can you know if a number with three or more digits 
is divisible by the number fourteen?

Check if the last digit of the original number is odd or 
even. If the number is odd, then the number is not 
divisible by fourteen. If the number is even, then apply 
the divisibility rule for seven. 

(Keep in mind, the odd and even test is to see if the number 
is divisible by two.) If the original even number is 
divisible by seven, then it is also divisible by fourteen. 
If the original even number is not divisible by seven, it 
is not divisible by fourteen.

Dividing by 13

Here's a straightforward method supplied by Scott Fellows:

Delete the last digit from the given number. Then subtract nine times the deleted digit from the remaining number. If what is left is divisible by 13, then so is the original number.

Rafael Ando contributes:

Instead of deleting the last digit and subtracting it ninefold from the remaining number (which works), you could also add the deleted digit fourfold. Both methods work because 91 and 39 are each multiples of 13.

For any prime p (except 2 and 5), a rule of divisibility could be "created" using this method:

Find m, such that m is the (preferably) smallest multiple of p that ends in either 1 or 9.

Delete the last digit and add (if multiple ends in 9) or subtract (if it ends in 1) the deleted digit times the integer nearest to m/10. For example, if m = 91, the integer closest to 91/10 = 9.1 is 9; and for 3.9, it's 4.

Verify if the result is a multiple of p. Use this process until it's obvious.

Example 1: Let's see if 14281581 is a multiple of 17.
In this case, m = 51 (which is 17×3), so we'll be deleting the last number and subtracting it fivefold.

1428158 - 5×1 = 1428153
142815 - 5×3 = 142800
14280 - 5×0 = 14280
1428 - 5×0 = 1428
142 - 5×8 = 102
10 - 5×2 = 0, which is a multiple of 17, so 14281581 is multiple of 17.
Example 2: Let's see if 7183186 is a multiple of 46.
First, note that 46 is not a prime number, and its factorization is 2×23. So, 7183186 needs to be divisible by both 2 and 23. Since it's an even number, it's obviously divisible by 2.
So let's verify that it is a multiple of 23:

m = 3×23 = 69, which means we'll be adding the deleted digit sevenfold.
718318 + 7×6 = 718360
71836 + 7×0 = 71836
7183 + 7×6 = 7225
722 + 7×5 = 757
75 + 7×7 = 124
12 + 7×4 = 40
4 + 7×0 = 4 (not divisible by 23), so 7183186 is not divisible by 46.
Note that you could've stopped calculating whenever you find the result to be obvious (i.e., you don't need to do it until the end). For example, in example 1 if you recognize 102 as divisible by 17, you don't need to continue (likewise, if you recognized 40 as not divisible by 23).
The idea behind this method it that you're either subtracting m×(last digit) and then dividing by 10, or adding m×(last digit) and then dividing by 10.

Jeremy Lane adds:

It may be noted that while applying these rules, it is possible to loop among numbers as results.

Example: Is 1313 divisible by 13? Using the procedure given we take 13×3 and obtain 39. This multiple ends in 9 so we add four-fold the last digit.

131 + 4×3 = 143 14 + 4×3 = 26 2 + 4×6 = 26 ...

Example: Is 1326 divisible by 13? Using the procedure given we take 13×7 = 91. This is not the smallest multiple, but it does show looping. The smaller multiple does loop at 39 as well. There are some examples where we would still need to recognize certain multiples. So we subtract nine-fold the last digit.

132 - 9×6 = 78 7 - 9×8 = -65 (factor out -1) 6 - 9×5 = -39 (again factor out -1) 3 - 9×9 = -78 (factor out -1)

This only occurs though if the number does happen to be divisible by the prime divisor. Otherwise, eventually you will have a number that is less than the prime divisor.

And here's a more complex method that can be extended to other formulas:

1 = 1 (mod 13)
10 = -3 (mod 13)    (i.e., 10 - -3 is divisible by 13)
100 = -4 (mod 13)   (i.e., 100 - -4 is divisible by 13)
1000 = -1 (mod 13)  (i.e., 1000 - -1 is divisible by 13)
10000 = 3 (mod 13)
100000 = 4 (mod 13)
1000000 = 1 (mod 13)

Call the ones digit a, the tens digit b, the hundreds digit c, ..... and you get:

a - 3×b - 4×c - d + 3×e + 4×f + g - .....

If this number is divisible by 13, then so is the original number.

You can keep using this technique to get other formulas for divisibility for prime numbers. For composite numbers just check for divisibility by divisors.

Dividing by 11-2

Any number written in our decimal system is made 
up of powers of 10. For example, 

   65,321 = 6(10^4) + 5(10^3)+3(10^2) + 2(10)+1. 

Now if you are interested in knowing whether 65,321 is a multiple of 5 
(that is, is exactly divisible by 5), you can tell just by looking 
at the last digit. That's because 10^4, 10^3, 10^2, and 10 are all 
divisible by 5, so the last digit on the end gets to cast the deciding 
vote.

Deciding whether 65,321 is a multiple of 3 is a little harder, since 
no power of 10 is a multiple of 3.  But 

         10 =  9   + 1 and
        100 = 99   + 1 and 
       1000 = 999  + 1 etc.

So if you had 65,321,

     65,321 = 6(10^4) + 5(10^3)+3(10^2) + 2(10)+1

you could rewrite this as

     65,321 = 6(9999+1) + 5(999+1) + 3(99+1) + 2(9+1)+1
            = 6(9999) + 5(999) + 3(99) + 2(9) + (6+5+3+2+1)

The first part is divisible by 3 so 65,321 is a multiple of 3 if the 
sum of its digits 6+5+3+2+1 is divisible by 3.

Now we are ready for 11.

Eleven is a little more complicated, but look at this neat pattern:

99                is a multiple of 11
9999              is a multiple of 11
999999            is a multiple of 11
...
any even number of 9's makes a multiple of 11.

If you look at how we wrote 
   
     65,321 = 6(9999+1) + 5(999+1) + 3(99+1) + 2(9+1) + 1

you can see that every other term contains a number that is a multiple 
of 11.  Unfortunately, numbers with an odd number of 9's are not 
divisible by 11.

But look at this pattern for these numbers:

10 = 10^1       = 11 - 1
1000 = 10^3     = 1001 - 1
100000 = 10^5   = 100001 - 1
10000000 = 10^7 = 10000001 - 1

and each of these numbers: 11, 1001, 100001, 10000001, ... is a 
multiple of 11. (Check it out by long division . . . it's worthwhile 
looking at the pattern you get when you divide 1 00 00 00 00 00 1 by 
11!)

Back to the example of 65,321.

    65,321 = 6(9999+1) + 5(1001 - 1) + 3(99+1) + 2(11-1) + 1

           = 6(9999) + 6 + 5(1001) - 5 + 3(99) + 3 + 2(11) - 2 + 1

All the terms with parentheses are multiples of 11. So 65,321 will 
be divisible by 11 if the remaining numbers are a multiple of 11. 
That is,
              6 - 5 + 3 - 2 + 1

Just the numbers you spoke about in your letter.

In general, the numbers in the odd positions are part of the "multiple 
of 999...999 + 1" and so are added to the total. The numbers in the 
even positions are part of the multples of "1000...0001 - 1" and so 
get subtracted.

The final test is to look at this alternating sum and difference of 
digits. But the result does not have to be 0 in order to be a multiple 
of 11...ANY multiple of 11 will do. For example,  

         8030209 gives 8 + 3 + 2 + 9 = 22.  

Since 22 is a multiple of 11, so is 8,030,209.