Wednesday 5 June 2013

Divisibility Rules

Dividing by 3
    Add up the digits: if the sum is divisible by three, then the number is as well. Examples:
    1. 111111: the digits add to 6 so the whole number is divisible by three.
    2. 87687687. The digits add up to 57, and 5 plus seven is 12, so the original number is divisible by three.
    Why does the 'divisibility by 3' rule work?
From: Dr. Math
To: Kevin Gallagher
Subject: Re: Divisibility of a number by 3

As Kevin Gallagher wrote to Dr. Math
On 5/11/96 at 21:35:40 (Eastern Time),

>I'm looking for a SIMPLE way to explain to several very bright 2nd 
>graders why the divisibility by 3 rule works, i.e. add up all the 
>digits; if the sum is evenly divisible by 3, then the number is as well.
>Thanks!
>Kevin Gallagher

The only way that I can think of to explain this would be as follows:
Look at a 2 digit number:

   10a + b = 9a + (a + b).

We know that 9a is divisible by 3, so 10a + b will be divisible by 3
if and only if a + b is.  Similarly, 

   100a + 10b + c = 99a + 9b + (a + b + c),

and 99a + 9b is divisible by 3, so the total will be iff a + b + c is.

This explanation also works to prove the divisibility by 9 test.  
It clearly originates from modular arithmetic ideas, and I'm not sure if
it's simple enough, but it's the only explanation I can think of.

    
    
    Another visitor suggests this as an easier explanation, relying on decomposition and place value:
    We know that 9 is divisible by 3
    
      10 = 9 + 1
    
    and
    
      30 = 9*3 + 3
    
    Similarly,
    
    3000 = 999*3 + 3
    
    With this kind of decomposition in mind, examine any number;
    for example, 1235:
    
    1235 = 1000
            200
             30
           +  5
    
    Now,
    
    1000 = 999*1 + 1
     200 = 99*2  + 2
      30 = 9*3   + 3
       5 =         5
    
    Add these remainders:
    
    1 + 2 + 3 + 5 = 11
    
    Eleven is not divisible by three, which tells us that 
    our initial number, 1235, is not divisible by 3.
    
    Another interesting fact about 3 is that any 3-digit
    number with sequential digits, e.g., 123, 234, 456,
    is divisible by 3.
    
Dividing by 4
    Look at the last two digits. If the number formed by its last two digits is divisible by 4, the original number is as well.
    Examples:
    1. 100 is divisible by 4.
    2. 1732782989264864826421834612 is divisible by four also, because 12 is divisible by four.
Dividing by 5
    If the last digit is a five or a zero, then the number is divisible by 5.
Dividing by 6
    Check 3 and 2. If the number is divisible by both 3 and 2, it is divisible by 6 as well.
    Robert Rusher writes in:
    Another easy way to tell if a [multi-digit] number is divisible by six . . .
    is to look at its [ones digit]: if it is even, and the sum of the [digits] is
    a multiple of 3, then the number is divisible by 6.
    
Dividing by 7
    To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.
    Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.
    Matthew Correnti describes this method:
If you do not know if a two-digit number, call it ab, is divisible 
by 7, calculate 2a + 3b. This will yield a smaller number, and if 
you do the process enough times you will eventually -- if the 
number ab is divisible by 7 -- end up with 7.

You can use a similar method if you have a three-digit number abc: 
take the digit a and multiply it by 2, then add it to the number bc, 
giving you 2a + bc; repeat and reduce until you recognize the 
result's divisibility by seven. With a four-digit number abcd, take 
the digit a and multiply by 6, then add 6a to bcd giving. This 
usually gives you a three-digit number; call it xyz. Take that x 
and, as described previously, multiply x by two and add to yz 
(i.e., 2x + yz). Again, repeat and reduce until you recognize the 
result's divisibility by seven.
    
    
    Ahmed Al Harthy writes in:
To know if a number is a multiple of seven or not, we can use also 
3 coefficients (1 , 2 , 3). We multiply the first number starting 
from the ones place by 1, then the second from the right by 3, 
the third by 2, the fourth by -1, the fifth by -3, the sixth by -2, 
and the seventh by 1, and so forth.

Example: 348967129356876. 

6 + 21 + 16 - 6 - 15 - 6 + 9 + 6 + 2 - 7 - 18 - 18 + 8 + 12 + 6 = 16 
means the number is not multiple of seven.

If the number was 348967129356874, then the number is a multiple of seven 
because instead of 16, we would find 14 as a result, which is a multiple of 7. 

So the pattern is as follows: for a number onmlkjihgfedcba, calculate

a + 3b + 2c - d - 3e - 2f + g + 3h + 2i - j - 3k - 2l + m + 3n + 2o.

Example:  348967129356874.

Below each digit let me write its respective figure.

3  4  8  9  6  7  1  2  9  3  5  6  8  7  4
2  3  1 -2 -3 -1  2  3  1 -2 -3 -1  2  3  1

(3×2) + (4×3) + (8×1) + (9×-2) + (6×-3) + (7×-1) + 
(1×2) + (2×3) + (9×1) + (3×-2) + (5×-3) + (6×-1) + 
(8×2) + (7×3) + (4×1) =  14 -- a multiple of seven.
    
    
    Another visitor observes:
Here is one formula for seven...

3X + L

L = last digit
X = everything in front of last digit.

All numbers that are divisible by seven have this in common. 
There are no exceptions.

For example, 42: 3(4) + 2 = 14.
Seven divides into 14, so it divides into 42.

Next example, 105: 3(10) + 5 = 35.
Seven divides into 35, so it divides into 105.

Here is another formula for seven:

4X - L

When using this formula, if you get zero, seven or a multiple of seven, 
the number will be divisible by seven.

For example, 56: 4(5) - 6 = 14.
Seven divides into 14, so it divides into 56.

Next example, 168: 16(4) - 8 = 56.
Seven divides into 56, so it divides into 168.

Similarly:

The formula for 2 is 2X + L
The formula for 3 is 4X + L
The formula for 4 is 6X + L
The formula for 5 is 5X + L
The formula for 6 is 2X + L 
and 4X + L -- in other words, the formulas for 2 and 3
must work before the number is divisible by 6.
The formula for 9 is X + L
The formula for 11 is X - L
The formula for 12 is 2X - L
The formula for 13 is 3X - L
The formula for 14 is 4X - L 
and 2X + L -- in other words, the formulas for 7 and 2 
            must work before the number is divisible by 14.
The formula for 17 is 7X - L
The formula for 21 is X - 2L
The formula for 23 is 3X - 2L
The formula for 31 is X - 3L
    
    
    Sara Heikali explains this way to test a number with three or more digits for divisibility by seven:
    1. Write down just the digits in the tens and ones places.
    2. Take the other numbers to the left of those last two digits, 
    and multiply them by two.
    3. Add the answer from step two to the number from step one.
    4. If the sum from step three is divisible by seven, then the 
    original number is divisible by seven, as well. If the sum is 
    not divisible by seven, then the original number is not 
    divisible by seven.
    
    For example, if the number we are testing is 112, then
    1. Write down just the digits in the tens and ones places: 12.
    2. Take the other numbers to the left of those last two digits, 
    and multiply them by two: 1 × 2 = 2.
    3. Add the answer from step two to the number from step one: 
    12 + 2 = 14.
    4. Fourteen is divisible be seven. Therefore, our original 
    number, one hundred twelve, is also divisible by seven.
    

Dividing by 8
    Check the last three digits. Since 1000 is divisible by 8, if the last three digits of a number are divisible by 8, then so is the whole number.
    Example: 33333888 is divisible by 8; 33333886 isn't. How can you tell whether the last three digits are divisible by 8? Phillip McReynolds answers:
    If the first digit is even, the number is divisible by 8 if the last two digits are. If the first digit is odd, subtract 4 from the last two digits; the number will be divisible by 8 if the resulting last two digits are. So, to continue the last example, 33333888 is divisible by 8 because the digit in the hundreds place is an even number, and the last two digits are 88, which is divisible by 8. 33333886 is not divisible by 8 because the digit in the hundreds place is an even number, but the last two digits are 86, which is not divisible by 8.
    Sara Heikali explains this test of divisibility by eight for numbers with three or more digits:
    1. Write down the units digit of the original number.
    2. Take the other numbers to the left of the last digit,
    and multiply them by two.
    3. Add the answer from step two to the number from step one.
    4. If the sum from step three is divisible by eight, then the 
    original number is divisible by eight, as well. If the sum is 
    not divisible by eight, then the original number is not 
    divisible by eight.
    
    For example, if the number we are testing is 104, then
    1. Write down just the digits in ones place: 4.
    2. Take the other numbers to the left of that last digit,
    and multiply them by two: 10 × 2 = 20.
    3. Add the answer from step two to the number from step one:
    4 + 20 = 24.
    4. Twenty-four is divisible be eight. Therefore, our original
    number, one hundred and four, is also divisible by eight.
    
Dividing by 9
    Add the digits. If that sum is divisible by nine, then the original number is as well.
Dividing by 10
    If the number ends in 0, it is divisible by 10.
Dividing by 11
    Let's look at 352, which is divisible by 11; the answer is 32. 3+2 is 5; another way to say this is that 35 -2 is 33. Now look at 3531, which is also divisible by 11. It is not a coincidence that 353-1 is 352 and 11 × 321 is 3531.
    Here is a generalization of this system. Let's look at the number 94186565.
    First we want to find whether it is divisible by 11, but on the way we are going to save the numbers that we use: in every step we will subtract the last digit from the other digits, then save the subtracted amount in order. Start with
              9418656 - 5 = 9418651     SAVE 5
         Then 941865  - 1 = 941864      SAVE 1
         Then 94186   - 4 = 94182       SAVE 4
         Then 9418    - 2 = 9416        SAVE 2
         Then 941     - 6 = 935         SAVE 6
         Then 93      - 5 = 88          SAVE 5
         Then 8       - 8 = 0           SAVE 8
    Now write the numbers we saved in reverse order, and we have 8562415, which multiplied by 11 is 94186565.

    Here's an even easier method, contributed by Chis Foren: Take any number, such as 365167484.
    Add the first, third, fifth, seventh,.., digits.....3 + 5 + 6 + 4 + 4 = 22
    Add the second, fourth, sixth, eighth,.., digits.....6 + 1 + 7 + 8 = 22
    If the difference, including 0, is divisible by 11, then so is the number.
    22 - 22 = 0 so 365167484 is evenly divisible by 11.
Dividing by 12
    Check for divisibility by 3 and 4.
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:
  1. Find m, such that m is the (preferably) smallest multiple of p that ends in either 1 or 9.
  2. 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.
  3. 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 14
    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.
    

Short-cut to find square of any number

Eg 1) 362   
Step 1: Choose the base number.36 is closer to 40. Therefore 40 is the base number.
Step 2:Find how much the given number is more or less of the base number.36 is  -4 of 40.
Step 3: Square the difference. (-4)² =16.
Step 4: Add the difference to the given number. 36-4=32.
Step 5: Multiply the above result with the base number. 32 x 40 =1280.
Step 6: Add the result of step 3 with the result of step 5.
1280 + 16=1296.
Ans: 1296.
Let’s do with another one.
Eg 2) 62²
Step 1: Base number is 60.
Step 2: 62-60=2.
Step 3: 2²=04 (Write 0 in the ten’s place if the square is a single digit)
Step 4: 62+ 2=64.
Step  5: 64 x 60=3840
Step 6: 3840 + 04=3844
Ans:3844.
Eg 3) 79²
Step 1: Base number is 80.
Step 2: 79-80 =-1.
Step 3: (-1)²=01
Step 4: 79-1=78.
Step 5: 78 x 80=6240.
Step 6: 6240 + 01=6241.
Eg 4) 84²
Step 1: 80 is the base
Step 2: 84-80 =4
Step 3: (4)²=16
Step 4:84 + 4=88
Step 5: 88 x 80=7040
Step 6: 7040+16=7056
Eg 5) 77²
Step 1: 80 is the base number.
Step 2: 77-80=-3
Step 3: (-3)²=09
Step 4: 77-3=74
Step 5: 74 x 80 =5920
Step 6:5920 + 09=5929
Eg 6) 49²
Step 1: 50 is the base number.
Step 2: 49-50=-1
Step 3: (-1)²=01
Step 4: 49-1=48
Step 5: 48 x 50=2400
Step 6: 2400 + 01=2401

A Mind-Blowing 3-Digit Number Squaring Shortcut

Now You Can Square Any Number Whose Tens Digit is 5 Without Using a Calculator

To begin with, I shall first introduce my canny friend, the Rules. It will be necessary to examine these rules in order to make your way about this method simpler. Since the tens digit has already been determined and defined to be 5, the next factor that would really affect the result of the square would be the value of the hundreds digit. Now this is where the rule comes in.
Step 1) Assuming the three-digit number to be squared is written in algebraic form,
100a +10b+c, where b is defined as 5 and both a and c are any number from 1 - 9 and 0 - 9 respectively. First, we shall deal with a as in the following:
a x (a + 1); {this will give us the first one or two digits of the answer}
Step 2) Next we shall obtain the following third digit and so on by performing the following:
25 + (2a + 1) x c; {this will give us the following digits of the answer. You may also assume that (2a +1) is the same as adding the number that comes right after a to a and multiplying that sum by the last the digit of the number to be squared}
Step 3) Finally, we determine the last two digit of the answer simply by obtaining the square of c and read the results obtained in step1 and 2 followed by the one obtained in step 3 and that is the complete answer.

c2;
{gives us the last two digits of the answer, hence, it is better to think that the last two digits are as easy in obtaining as finding the square of the last digit of the number to be squared}

Example:
a) Find the square of 253.
1) 2 x (2+1) = 6
2) 25 + (2 x 2 +1) x 3 =40
3) 32 = 09
4) Read 64009 (the answer!)
b) What is the square of 457?
1) 4 x 5 = 20 2) 25 + (4 x 2 +1) x 7 = 88
3) 72 = 49
4) Read 208849 (the answer!)
c) What is 9592?
1) 9 x 10 = 90
2) 25 + (9 + 10) x 9 = 196 {note that the result is a three digit number, hence the value in the hundreds digit is a carry digit that is added to the result in step 1}
3) 92 = 81 4) Read 919681 (the answer!)
Additional Tip: A shortcut to multiplying any single digit number by 19 would be as
simple as the following example showing 19 x 7:
Step1) Take the double of 7 and subtract it by one
That gives us, 14 - 1 = 13
Step 2) Subtract 7 from 10.
Which is, 10 - 7 = 3
Step 3) Read the result in step 1 followed by the result in step 2 and you will get 133,
which is the answer.
Hence the general rule for a x 19, would be:
(2a -1) x 10 + (10 - a) {this simply eliminates the hassle of having to deal with carries}
Or alternatively, you may also try; multiplying the number to be multiplied by 19, by 20 instead and subtract that number again from the product,
i.e: 7 x 20 - 7 = 133

 

Multiply Up to 20X20 In Your Head


In just FIVE minutes you should learn to quickly multiply up to 20x20 in your head.  With this trick, you will be able to multiply any two numbers from 11 to 19 in your head quickly, without the use of a calculator.
I will assume that you know your multiplication table reasonably well up to 10x10.
Try this:

  • Take 15 x 13 for an example.
  • Always place the larger number of the two on top in your mind.
  • First add 15 + 3 = 18
  • Add a zero behind it (multiply by 10) to get 180.
  • Multiply the covered lower 3 x the single digit above it the "5" (3x5= 15)
  • Add 180 + 15 = 195.
That is It! Wasn't that easy? Practice it on paper first! 

Take another example: 

15 x 16
Step1:  15+6 =21
Step2:  21x10=210
Step3: 5x6=30
Step4: 210+30=240 (answer) 

Divisibility by prime numbers under 50

A number is divisible by 2 if its last digit is also (i.e. 0,2,4,6 or 8).
A number is divisible by 3 if the sum of its digits is also. Example: 534: 5+3+4=12 and 1+2=3 so 534 is divisible by 3.
A number is divisible by 5 if the last digit is 5 or 0.
Most people know (only) those 3 rules. Here are my rules for divisibility by the PRIMES up to 50. Why only primes and not also composite numbers? A number is divisible by a composite if it is also divisible by all the prime factors (e.g. is divisible by 21 if divisible by 3 AND by 7). Small numbers are used in these worked examples, so you could have used a pocket calculator. But my rules apply to any number of digits, whereas you cannot test a 30 or more digit number on your pocket calculator otherwise.
Test for divisibility by 7. Double the last digit and subtract it from the remaining leading truncated number. If the result is divisible by 7, then so was the original number. Apply this rule over and over again as necessary. Example: 826. Twice 6 is 12. So take 12 from the truncated 82. Now 82-12=70. This is divisible by 7, so 826 is divisible by 7 also.
There are similar rules for the remaining primes under 50, i.e. 11,13, 17,19,23,29,31,37,41,43 and 47.
Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.
Test for divisibility by 13. Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. Apply this rule over and over again as necessary.
Example: 50661-->5066+4=5070-->507+0=507-->50+28=78 and 78 is 6*13, so 50661 is divisible by 13.
Test for divisibility by 17. Subtract five times the last digit from the remaining leading truncated number. If the result is divisible by 17, then so was the first number. Apply this rule over and over again as necessary.
Example: 3978-->397-5*8=357-->35-5*7=0. So 3978 is divisible by 17.
Test for divisibility by 19. Add two times the last digit to the remaining leading truncated number. If the result is divisible by 19, then so was the first number. Apply this rule over and over again as necessary.
EG: 101156-->10115+2*6=10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, so 101156 is divisible by 19.
My original divisibilty webpage stopped here. However, I have had a number of mails asking for divisibility tests for larger primes, so I've extended the list up to 50. Actually even with 37 most people cannot do the necessary mental arithmetic easily, because they cannot recognise even single-digit multiples of two-digit numbers on sight. People are no longer taught the multiplication table up to 20*20 as I was as a child. Nowadays we are lucky if they know it up to 10*10.
Test for divisibility by 23. 3*23=69, ends in a 9, so ADD. Add 7 times the last digit to the remaining leading truncated number. If the result is divisible by 23, then so was the first number. Apply this rule over and over again as necessary.
Example: 17043-->1704+7*3=1725-->172+7*5=207 which is 9*23, so 17043 is also divisible by 23.
Test for divisibility by 29. Add three times the last digit to the remaining leading truncated number. If the result is divisible by 29, then so was the first number. Apply this rule over and over again as necessary.
Example: 15689-->1568+3*9=1595-->159+3*5=174-->17+3*4=29, so 15689 is also divisible by 29.
Test for divisibility by 31. Subtract three times the last digit from the remaining leading truncated number. If the result is divisible by 31, then so was the first number. Apply this rule over and over again as necessary.
Example: 7998-->799-3*8=775-->77-3*5=62 which is twice 31, so 7998 is also divisible by 31.
Test for divisibility by 37. This is (slightly) more difficult, since it perforce uses a double-digit multiplier, namely eleven. People can usually do single digit multiples of 11, so we can use the same technique still. Subtract eleven times the last digit from the remaining leading truncated number. If the result is divisible by 37, then so was the first number. Apply this rule over and over again as necessary.
Example: 23384-->2338-11*4=2294-->229-11*4=185 which is five times 37, so 23384 is also divisible by 37.
Test for divisibility by 41. Subtract four times the last digit from the remaining leading truncated number. If the result is divisible by 41, then so was the first number. Apply this rule over and over again as necessary.
Example: 30873-->3087-4*3=3075-->307-4*5=287-->28-4*7=0, remainder is zero and so 30873 is also divisible by 41.
Test for divisibility by 43. Now it starts to get really difficult for most people, because the multiplier to be used is 13, and most people cannot recognise even single digit multiples of 13 at sight. You may want to make a little list of 13*N first. Nevertheless, for the sake of completeness, we will use the same method. Add thirteen times the last digit to the remaining leading truncated number. If the result is divisible by 43, then so was the first number. Apply this rule over and over again as necessary.
Example: 3182-->318+13*2=344-->34+13*4=86 which is recognisably twice 43, and so 3182 is also divisible by 43.
Update : Bill Malloy has pointed out that, since we are working to modulo43, instead of adding factor 13 times the last digit, we can subtract 30 times it, because 13+30=43. Why didn't I think of that!!! :-(
Finally, the Test for divisibility by 47. This too is difficult for most people, because the multiplier to be used is 14, and most people cannot recognise even single digit multiples of 14 at sight. You may want to make a little list of 14*N first. Nevertheless, for the sake of completeness, we will use the same method. Subtract fourteen times the last digit from the remaining leading truncated number. If the result is divisible by 47, then so was the first number. Apply this rule over and over again as necessary.
Example: 34827-->3482-14*7=3384-->338-14*4=282-->28-14*2=0 , remainder is zero and so 34827 is divisible by 47.
I've stopped here at the last prime below 50, for arbitrary but pragmatic reasons as explained above.
Other blogreaders (sadly even people from .edu domains, who should be able to do the elementary algebra themselves) have asked why I sometimes say ADD and for other primes say SUBTRACT, and ask where the apparently arbitrary factors come from. So let us do some algebra to show the method in my madness.
We have displayed the recursive divisibility test of number N as f-M*r where f are the front digits of N, r is the rear digit of N and M is some multiplier. And we want to see if N is divisible by some prime P. We need a method to work out the values of M. What you do is to calculate (mentally) the smallest multiple of P which ends in a 9 or a 1. If it's a 9 we are going to ADD, Then we will use the leading digit(s) of the multiple +1 as our multiplier M. If it's a 1 we are going to SUBTRACT later. then we will use the leading digit(s) of the multiple as our multiplier M.
Example for P=17 : three times 17 is 51 which is the smallest multiple of 17 that ends in a 1 or 9. Since it's a 1 we are going to SUBTRACT later. The leading digit is a 5, so we are going to SUBTRACT five times the remainder r. The algorithm was stated above. Now let's do the algebraic proof. Writing N=10f+r, we can multiply by -5 (as shown in the example for 17), getting -5N=-50f-5r. Now we add 51f to both sides (because 51 was the smallest multiple of P=17 to end in a 1 or a 9), giving one f (which we want), so 51f-5N=f-5r. Now if N is divisible by P (here P=17), we can substitute to get 51f-5*17*x=f-5r and rearrange the left side as 17*(3f-5x)=f-5r and therefore f-5r is a multiple of P=17 also. Q.E.D.

Thursday 30 May 2013

Squaring numbers ending in 1

Here’s a simple trick to square any number that ends in 1.
  • Subtract 1 from the number.
  • Square the difference. (Squaring of such number is easy as it ends in ‘0’)
  • Add the difference twice to its square.
  • Add 1.
Example: If the number to be squared is 61  
1.    Subtract 1=> 61 - 1 = 60.
2.    Square the difference => 60 × 60 = 3600.
3.    Add the difference twice to its square => 3600 + 60 + 60 = 3720.
4.    Add 1 => 3720 + 1 = 3721.
So, 612 = 3721
812 =?

1)    81 - 1 = 80 (Subtract 1). 
2)    802 = 6400 (square the difference). 
3)    6400 + 80 + 80 = 6560 (add the difference twice to its square).
4)    6560 + 1 = 6561 (add 1)
So, 812 = 6561
If the number to be squared is a 3-digit number, let’s say 121
1)    121 - 1 = 120 (Subtract 1). 
2)    1202 = 14400 (square the difference). 
3)    14400 + 120 + 120 = 14640 (add the difference twice to its square).
4)    14640 + 1 = 14641 (add 1)
So, 1212 = 14641
2512 =?
1)    251 - 1 = 250  
2)    2502 = 62500 
3)    62500 + 250 + 250 = 63000
4)    63000 + 1 = 63001
So, 2512 = 63001

Ref. http://faster-maths.blogspot.com/2013/05/squaring-numbers-ending-in-1.html#.UadByNjKWVo

Sunday 12 May 2013

Divisibility Rules

Dividing by 3
    Add up the digits: if the sum is divisible by three, then the number is as well. Examples:
    1. 111111: the digits add to 6 so the whole number is divisible by three.
    2. 87687687. The digits add up to 57, and 5 plus seven is 12, so the original number is divisible by three.
    Why does the 'divisibility by 3' rule work?
    Look at a 2 digit number: 10a+b=9a+(a+b). We know that 9a is divisible by 3, so 10a+b will be divisible by 3 if and only if a+b is. Similarly, 100a+10b+c=99a+9b+(a+b+c), and 99a+9b is divisible by 3, so the total will be if a+b+c is.
    This explanation also works to prove the divisibility by 9 test. It is originates from modular arithmetic ideas.
  Dividing by 4
    Look at the last two digits. If the number formed by its last two digits is divisible by 4, the original number is as well.
    Examples:
    1. 100 is divisible by 4.
    2. 1732782989264864826421834612 is divisible by four also, because 12 is divisible by four.
  Dividing by 5
    If the last digit is a five or a zero, then the number is divisible by 5.
  Dividing by 6
    Check 3 and 2. If the number is divisible by both 3 and 2, it is divisible by 6 as well.
    Another easy way to tell if a [multi-digit] number is divisible by six . . . is to look at its [ones digit]: if it is even, and the sum of the [digits] is a multiple of 3, then the number is divisible by 6.
  Dividing by 7
    To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.
    Example: If you had 203, you would double the last digit to get six, and subtract that from 20 to get 14. If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.
    Another methods is:
    To know if a number is a multiple of seven or not, we can use also 
    3 coefficients (1 , 2 , 3). We multiply the first number starting 
    from the ones place by 1, then the second from the right by 3, 
    the third by 2, the fourth by -1, the fifth by -3, the sixth by -2, 
    and the seventh by 1, and so forth.
    
    Example: 348967129356876. 
    
    6 + 21 + 16 - 6 - 15 - 6 + 9 + 6 + 2 - 7 - 18 - 18 + 8 + 12 + 6 = 16 
    means the number is not multiple of seven.
    
    If the number was 348967129356874, then the number is a multiple 
     of seven because instead of 16, we would find 14 as a result, which is a
      multiple of 7. So the pattern is as follows: for a number onmlkjihgfedcba,
      calculate a + 3b + 2c - d - 3e - 2f + g + 3h +

    2i - j - 3k - 2l + m + 3n + 2o.
    
    Example:  348967129356874.
    
    Below each digit let me write its respective figure.
    
    3  4  8  9  6  7  1  2  9  3  5  6  8  7  6 
    2  3  1 -2 -3 -1  2  3  1 -2 -3 -1  2  3  1
    
    (3×2) + (4×3) + (8×1) + (9×-2) + (6×-3) + (7×-1) + 
    (1×2) + (2×3) + (9×1) + (3×-2) + (5×-3) + (6×-1) + 
    (8×2) + (7×3) + (6×1) =  16 -- not a multiple of seven.
    
  Dividing by 8
    Check the last three digits. Since 1000 is divisible by 8, if the last three digits of a number are divisible by 8, then so is the whole number.
    Example: 33333888 is divisible by 8; 33333886 isn't. How can you tell whether the last three digits are divisible by 8?
    If the first digit is even, the number is divisible by 8 if the last two digits are. If the first digit is odd, subtract 4 from the last two digits; the number will be divisible by 8 if the resulting last two digits are. So, to continue the last example, 33333888 is divisible by 8 because the digit in the hundreds place is an even number, and the last two digits are 88, which is divisible by 8. 33333886 is not divisible by 8 because the digit in the hundreds place is an even number, but the last two digits are 86, which is not divisible by 8.
  Dividing by 9
    Add the digits. If that sum is divisible by nine, then the original number is as well. This holds for any power of three.
  Dividing by 10
    If the number ends in 0, it is divisible by 10.
  Dividing by 11
    Take any number, such as 365167484. Add the first, third, fifth, seventh,.., 
    digits.....3 + 5 + 6 + 4 + 4 = 22
    Add the second, fourth, sixth, eighth,.., digits.....6 + 1 + 7 + 8 = 22
    If the difference, including 0, is divisible by 11, then so is the number.
    22 - 22 = 0 so 365167484 is evenly divisible by 11.
  Dividing by 12
    Check for divisibility by 3 and 4.

 

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