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.

[Acme]

There are a couple refinements on the 3, 6, & 9 divisibility.

 

With this simple short cuts you can find out a number is divisible by a given number ...

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 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

Units digit is 6 so divisible by 2

3+5+2+5+6 = 21 so divisible by 3

So 35256 divisible by 6

...

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

...

If you continue the summing of digits in the 9 example until you arrive at a single digit (called taking the digital root) you can know divisibility by 3, 6, and 9. If the digital root is 9 the number divides by 3 & 9 and if the number is even it also divides by 6.

 

So with the example 50832, 5+0+8+3+2=18 and 1+8=9. So 50832 divides by 3 & 9 and because it is even it also divides by 6.

[ewmon]

Poor 7, we've forgotten divisible by 7. Here's something fascinating I just found —

 

 

Remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7. Example: 1603 → 160-2(3)=154 → 15-2(4)=7, so 1603 is divisible by 7.

 

I found it here.

[Acme]

Poor 7, we've forgotten divisible by 7. Here's something fascinating I just found

Remove the last digit, double it, subtract it from the truncated original number and continue doing this until only one digit remains. If this is 0 or 7, then the original number is divisible by 7. Example: 1603 → 160-2(3)=154 → 15-2(4)=7, so 1603 is divisible by 7.

I found it here.

 

Alas while that works, I wouldn't call it simple. If you were to take some 'large' number it would be simpler to perform long division. Is 39,084,351,522 evenly divisible by 7?

[burgess]

Thank you all for your replies and suggestions.

[Sensei]

2,4,8,10 cases can be much easier:

 

We use binary and operator to get rid of upper part of value, and check remaining by comparing it with 0:

if( ( value & 0x1 ) == 0 ) // dividable by 2

 

if( ( value & 0x3 ) == 0 ) // dividable by 4

 

if( ( value & 0x7 ) == 0 ) // dividable by 8

(this can be extended to any 2^x)

 

Value is dividable by 10, if modulo 10 is returning 0:

if( ( value % 10 ) == 0 ) // dividable by 10

if( ( value % 100 ) == 0 ) // dividable by 100

if( ( value % 1000 ) == 0 ) // dividable by 1000