Tests of Divisibility- Simple tricks

[burgess]

With this simple short cuts you can find out a number is divisible by a given number
Divisible by 2: A number is divisible by 2, if its unit’s digit is any of 0, 2, 4, 6, 8.
Example: 6798512

Divisible by 3: A number is divisible by 3, if sum of its digits divisible by 3.
Example : 123456
1+2+3+4+5+6 = 21
21 is divisible by 3 so 123456 is also divisible by 3

Divisible by 4: if the last two digits of a given are divisible 4, so the number can be divisible by 4.
Example : 749232
Last two digits are 32 which are divisible by 4 so the given number is also divisible by 4

Divisible by 5: If unit’s digit of a number is either ‘0’ or ‘5’ it is divisible 5.
Example : 749230

Divisible by 6: If a given number is divisible by 2 and 3 (which are factors of 6), then the number is divisible by 6.
Example : 35256
Unit’s digit is 6 so divisible by 2
3+5+2+5+6 = 21 so divisible by 3
So 35256 divisible by 6

Divisible by 8: if last 3 digits of a given number is divisible 8, then the given number is divisible 8.
Example: 953360
360 is divisible by 8, so 953360 is divisible by 8

Divisible by 9: A number is divisible by 9, if sum of its digits divisible by 9.
Example : 50832
5+0+8+3+2 = 18 divisible by 9 so 50832 divisible by 9

Divisible by 10: A number is divisible 10, if it ends with 0.
Example : 508320

Divisible by 11: A number is divisible by 11,if the difference of sum of its digits at odd places and sum of its digits at even places , is either 0 or a number divisible by 11.
Example : 4832718
(sum of digits at odd places ) – (sum of digits at even places)
=(8+7+3+4)-(1+2+8) = 11 which is divisible by 11.
So 4832718 is divisible by 11.

I hope this simple tricks, will be very helpful to solve math’s homework problems easily.

 

[burgess]

If anybody knows the divisibility test for 7..Please share it here...
Thanks

 

[Nono713]

Thanks for posting this. The test for divisibility by 7 is rather tricky apparently and is probably more work than just performing the division, I don't think many people use it in practice. Myself I only mentally check for divisibility by 2, 3, 4, 5, 6, 8, 9, other powers of 2 (up to 64) and the trivial ones e.g. 10/25/etc..

 

[Jay1]

Thanks for posting this. The test for divisibility by 7 is rather tricky apparently and is probably more work than just performing the division, I don't think many people use it in practice. Myself I only mentally check for divisibility by 2, 3, 4, 5, 6, 8, 9, other powers of 2 (up to 64) and the trivial ones e.g. 10/25/etc..

As I understand it, a specific test for divisibility by 7 has not yet been discovered. Where's Fermat when we need him?

 

[Dethrone]

The divisibility by 7 test is on wikipedia.

 

[Jay1]

The divisibility by 7 test is on wikipedia.

Interesting. Apparently I stand corrected.

A book I have shows various methods of testing divisibility, but they skip 7 and say no specific method exists other than trial and error. But that book dates back to the 1970s, when the Earth was young, we were on the verge of another ice age and household computers were a dream on the horizon.

The method shown on Wikipedia for divisibility by7 is so convoluted and hard to remember that I think it would be easier to simply do the division and check if the remainder is zero. LOL

Some are easy and you can almost do it in your head, but 7 can be a pain.

 

[I like Serena]

Interesting. Apparently I stand corrected.

A book I have shows various methods of testing divisibility, but they skip 7 and say no specific method exists other than trial and error. But that book dates back to the 1970s, when the Earth was young, we were on the verge of another ice age and household computers were a dream on the horizon.

The method shown on Wikipedia for divisibility by7 is so convoluted and hard to remember that I think it would be easier to simply do the division and check if the remainder is zero. LOL

Some are easy and you can almost do it in your head, but 7 can be a pain.

Wiki doesn't just give a divisibility method for 7 - it gives 5 methods!

As for dating back, the Arab mathematician al-Banna (1256-1321 AD) had tests for 7, 8, and 9.
I think this was when the Earth was a little bit younger than 1970. ;)

 

[Jay1]

Wiki doesn't just give a divisibility method for 7 - it gives 5 methods!

As for dating back, the Arab mathematician al-Banna (1256-1321 AD) had tests for 7, 8, and 9.
I think this was when the Earth was a little bit younger than 1970. ;)

Yes, I noticed.
It's amazing some of the math that far predates calculators and modern computation methods. I often marvel at what they could do in the ancient past and I still use concepts developed by ancient Greeks to develop my programs today.

Imagine what Newton or Einstein could have done if they only had PCs in those days!

 

[soroban]

I learned this trick in high school.

Divisibility by 7

Subtract 2 times the unit-digit from the "rest of the number".
If the result is divisible by 7, the number is divsible by 7.
Repeat, if necessaryExample: test $483$

$\quad 48 - 2(3) \,=\,42$

Since $42$ is divisibe by $7$, so is $483.$Example: test $5537$

$\quad 553 - 2(7) \,=\,539$

Is $539$ divisible by $7$?
Repeat the test.
$\quad 53 - 2(9) \,=\,35$

Since $35$ is divisible by $7$, so is $5537.$

 

[I like Serena]

I learned this trick in high school.

Divisibility by 7

Subtract 2 times the unit-digit from the "rest of the number".
If the result is divisible by 7, the number is divsible by 7.
Repeat, if necessaryExample: test $483$

$\quad 48 - 2(3) \,=\,42$

Since $42$ is divisibe by $7$, so is $483.$Example: test $5537$

$\quad 553 - 2(7) \,=\,539$

Is $539$ divisible by $7$?
Repeat the test.
$\quad 53 - 2(9) \,=\,35$

Since $35$ is divisible by $7$, so is $5537.$

Nice! That made me want to prove it. ;)

Proof

Let $x = 10a +b$, where $b$ is a single digit.

Then:
$$a -2b \equiv 0 \pmod 7\Rightarrow 3a-6b\equiv 0 \pmod 7 \Rightarrow x = 10a+b\equiv 0 \pmod 7$$
In words: if $a-2b$ is divisible by $7$, so is $x=10a+b$.

 

[velho]

If anybody knows the divisibility test for 7..Please share it here...
Thanks

I am the author of a rule of divisibility by seven that is included in my unpublished but officially registered book: "Divisibility by Seven: the End of a Myth?"
The algorithm of this rule that must be applied from right to left repetitively eliminating the last two digits each time until the last two digits or the isolated digit to the left is reached:
N = a,bcd
a' ≣ (− cd mod 7 + a) mod 7, if 7∣a'b then 7∣N
If the application reaches an isolated digit to the left, think of "a" as a zero.
Example:
N = 5,574,569
Using common language: 69 to 70 equals 1; 1 + 4 = 5; eliminating 69 we have : 55755; 55 to 56 equals 1; 1 + 5 = 6; eliminating 55 we have: 567; 67 to 70 equals 3; 3 + 0 = 3; eliminating 67 we have 35; 7|35 and 7|N
This is not a trick, it is a rule of divisibility by 7. Try this number with any other known trick, rule, test (whatever) to understand why I define my procedure as a rule.

 

[Joe1224]

Here is a simple trick. You first need to know these values.
7*1=7
7*2=14
7*3=21
7*4=28
7*5=35
7*6=42
7*7=49
7*8=56
7*9=63
7*10=70
When you have a number higher than 7 you add one of the numbers in the table above such that you get a number which ends in one or more zeroes and then remove the zeroes (trick for dividing by powers of ten). Repeat until you obtain a number you know is divisible by 7.
Example: 566349
566349+21=566370
56637+63=56700
567+63=630
After dividing by ten (by removing the zero) we get 63 which we know is divisible by 7 thus 566349 is also divisible by 7.

Note this trick works for any odd number which doesn't end with a five, you only have to know the multiplication table for that number up to 10 to use this trick.