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**simplyjasper****Member**- Registered: 2009-11-15
- Posts: 24

First the slightly easier ones...

Definition:

2|10 means 2 divides 10 or 10 is divisible by 2 and leave no remainder

**Test for Divisibility by 5**

- Check if the last digit is 0 or 5

Example: 123512341325

Since the light digit is 5, the number is divisible by 5

**Test for Divisibility by 2**

- Check if the last digit is even (0, 2, 4, 6, 8)

Example: 12312348

Since the last digit is 8, the number is divisible by 2

**Test for Divisibility by 4**

- Check if the last 2 digits is divisible by 4

Example: 123124124

Since 4|24, the number is divisible by 4

**Test for Divisibility by 8**

- Check if the last 3 digits is divisible by 8

Example: 9798743648

Since 8|648, the number is divisible by 8

**Test for Divisibility by 16**

- Check if the last 4 digits is divisible by 16

Example: 7849758432

Since 16|8432, the number is divisible by 16

**Test for Divisibility by 11**

- By subtracting the sum of all the even position of the digit by the sum of all the odd position of the digit

Example: 19203182

Odd sum: (1 + 2 + 3 + 6) = 12

Even sum: (9 + 0 + 1 + 2) = 12

Difference is 0 and 0 is divisible by 11. Hence the digit is divisible by 11...

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**simplyjasper****Member**- Registered: 2009-11-15
- Posts: 24

**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, so will be the number

Example: 4991

Steps:

499 - 2(1) = 497

49 - 2(7) = 35

Since 7|35, the number is divisible by 7

**Test for Divisibility by 13**

- Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, so will the number

Example: 50661

Steps:

5066 + 4(1) = 5070

507 + 4(0) = 507

50 + 4(7) = 78

7 + 4(8) = 39

Since 13|39, the number is divisible by 13

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**iamfriendly****Member**- Registered: 2009-07-03
- Posts: 72

thanks for the divisibility rule of 7. was always trying to find that ...

BUt u could just use long division in the mean time.....

any way,

*Last edited by iamfriendly (2010-02-20 07:30:00)*

EYE AM FRIENDLY, THAT'S O U NEED 2 NO,

psst,

Don't trust strangers, EYE AM FRIENDLY, THAT'S O U NEED 2 NO ..........psst, Don't trust strangers......

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

simplyjasper wrote:

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, so will be the number

Example: 4991

Steps:

499 - 2(1) = 497

49 - 2(7) = 35

Since 7|35, the number is divisible by 7

So you take a number whose last digit is , and you form the new number

Rearranging gives

, i.e.Very ingenious. (In fact

.)I suppose the divisibility test for 13 uses the same trick.

*Last edited by JaneFairfax (2010-03-31 01:31:17)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Right, now we know the trick, lets use it to devise more divisibility tests!

**Test for Divisibility by 17**

Subtract five times the last digit from the remaining truncated number. If the result is divisible by 17, so will be the number.

Example: 12342

Steps:

1234 − 5(2) = 1224

122 − 5(4) = 102

Since 17∣102, the number is divisible by 17.

**Test for Divisibility by 19**

Add twice the last digit to the remaining leading truncated number. If the result is divisible by 19, so will the number.

Example: 10203

Steps:

1020 + 2(3) = 1026

102 + 2(6) = 114

11 + 2(4) = 19

Since 19∣19, the number is divisible by 19.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**Test for Divisibility by 31**

Subtract three times the last digit from the remaining truncated number. If the result is divisible by 31, so will be the number.

Example: 111011

Steps:

11101 − 3(1) = 11098

1109 − 3(8) = 1085

108 − 3(5) = 93

Since 31∣93, the number is divisible by 19.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

IN GENERAL:

**Test for Divisibility by a number of the form n = 10b+a, where b is a positive integer and a ∈ {1,3,7,9}**

NB: *n* need not be prime for this to work. It only needs to be coprime with 10 (which such a number necessarily is). We consider the cases for *a*.

*a* = 1:

Subtract *b* times the last digit from the remaining truncated number.

*a* = 3:

Add 3*b*+1 times the last digit to the remaining truncated number.

*a* = 7:

Subtract 3*b*+2 times the last digit from the remaining truncated number.

*a* = 9:

Add *b*+1 times the last digit to the remaining truncated number.

Examples:

To test for divisibility by 53, add 3(5)+1 = 16 times the last digit to the truncated number. Thus, for 44944: 4494 + 16(4) = 4558; 455 + 16(8) = 583; 58 + 16(3) = 106; as 53∣106, 44944 is divisible by 53.

To test for divisibility by 91, subtract 9 times the last digit from the the truncated number. Thus for 55601: 5560 − 9(1) = 5551; 555 − 9(1) = 546; 54 − 9(6) = 0; as 91∣0, 55601 is divisible by 91.

Note that 91 = 7×13 is not prime. However, 91 is coprime with 10 and so the method works, enabling us to save time by not having to test for 7 and 13 separately.

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**ZHero****Real Member**- Registered: 2008-06-08
- Posts: 1,889

Thx Jane for such a nice Explanation!

An Interesting Collection of Mathematical Things

*Last edited by ZHero (2010-03-31 05:44:28)*

If two or more thoughts intersect with each other, then there has to be a point.

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**cool_jessica****Member**- Registered: 2012-06-26
- Posts: 15

We also have a divisibility test for 6 and 9.

For a number to be divisible by 6, it should be divisible by both 2 and 3.

Example:

like 18 is divisible by both 3 and 2 so it is divisible by 6.

For a number to be divisible by 9, its sum of digits must be equal to 9.

Example:

like 18 , here sum of the digit is 1 + 8 = 9, so it is divisible.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi all!

Here's a test for divisibility by 7, 11 and 13 together. The product 7*11*13 = 1001 so subtract

multiples of 1001 from the number N until a number less than 1001 occurs; that is, calculate

C := N-k*(1001) < 1001 for an appropriate k. Then N divides by 7 or 11 or 13 if and only if

C does also.

Example: N=5148 Subtract 5005 to get 143. 143 divides by both 11 and 13 but not by 7 so

5148 divides by both 11 and 13 but not by 7.

Of course if C=0 then it and N divide by 7, 11, 13 all three.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi cool-jessica!

The test you mention in post #9 for divisibility by 9 is also the basis for the checking method called

"CASTING OUT NINES." In the 1920's when my dad went to school an arithmetic problem was not

considered finished until they checked it by casting out nines. The method is seldom taught these

days, probably because calculators are so handy for checking arithmetic problems.

HOWEVER try checking addition, subtraction, multiplication, division and factoring POLYNOMIALS

on a calculator. Can you find a calculator (not computer application like Maple, etc.) that can do this?

Even computer applications are not so handy for this due to the cumbersome way of entering

polynomials like 5*x^4-3*x^3+7*x^2+12*x-22. Not only does it take a while to enter such, we

are prone to "hit wrong keys" and mess up the check. Using the short form, the polynomial in

the preceding sentence can be written 5 -3 7 12 -22 and the cast out digit (COD) 8, can be

obtained by adding all the columns to get -25+24=-1. Then add 9 to get the 8. Adding 9 to any

single negative digit produces the corresponding positive digit. Either can be used to check.

Also there are a few tricks for making the "casting out" process faster and easier.

Using CASTING OUT NINES, such polynomial problems can be checked in SECONDS in our heads.

Just entering the polynomials in these computer applications takes more time than it would to

check using casting out nines. Most folks are not aware that CO9's can be used on polynomials.

By the way, the reason CO9's works on polynomials is that the coefficients of the terms are

typically written in base ten. If these coefficients were written in a different base, say b, then

we could "cast out (b-1)'s", for example in base 16 we would cast out 15's. The larger the base,

the more accurate the check. Small bases don't do so well, especially base 2.

If a person is really bad at arithmetic and makes lots of mistakes, then CO9's will catch the error

about 89% of the time since it usually takes TWO mistakes to cause a 1 in 9 chance of the problem

"checking out OK" by coincidence. Usually making only ONE mistake in a problem will cause CO9's

to indicate an error in the problem. So if a person is highly accurate in their calculations, the

accuracy rate is more like 99%+. I have found in my experience that CO9's is the fastest and

easiest method of checking arithmetic and polynomial operations. It has also worked out to be

generally as accurate as any other method (all methods being subject to 'OPERATOR ERRORS').

Be Blessed!!!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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