"There is a simple rule how to determine if a number is divisible by three.
Add up the digits and see if the sum of those digits (a much smaller number) is divisible by three.
This rule simplifies finding out if a big number is divisible by three, by reducing it to a small number.
Why is this proposition true?
Consider, for simplicity, a 3 digit number 'abc", such as 321.
Suppose that a+b+c is divisible by 3.
Our number, written as 'abc' is actually
100a+10b+c.
For instance, 321 = 100*3+10*2+1.
Let's rewrite 100a+10b+c as 99a+a + 9b+b + c.
99a+a + 9b+b + c = (99a + 9b) + (a + b + c)
The first part is always divisible by 3 since numbers with all nines are always divisible by 3 (9 = 3*3, 99=33*3, 999=333*3 etc).
That the second part (a+b+c) is divisible by 3, is a given.
So, we have a sum of two parts, both of which are divisible by 3. This sum is, therefore, also divisible by three. Theorem proven."
~http://www.algebra.com/algebra/homework/divisibility/lessons/divisibility-by-3-three.lesson~
By tautology. If it did not work, it would not be a divisibility rule!
there is a divisibility for 24 the rule is you can divide 24 as 6 and 4 i think
If the sum of the digits is divisible by 3, the number is also.
What is the divisblity rule by 8
There is no easy rule for divisibility by 34.
By tautology. If it did not work, it would not be a divisibility rule!
A number is a multiple of 4 if the last 2 digits are a multiple of 4 The 10s digit is even and the last digit is 0, 4 or 8 The 10s digit is odd and the last digit is 2 or 6 A number is a multiple of 8 if the last 3 digits are a multiple of 8 The 100s digit is even and the last 2 digits are a multiple of 8 The 100s digit is odd and the last 2 digits are 4 times an odd number
If the rules for 2 and 3 work, the number is divisibale by 5.
It is 3 6 9
117 is a composite. I know because suming the digits is 9. A number is a multiple of 117 if it's a multiple of 9 and 13 at the same time
A number is a multiple of 741 if it's a multiple of 3, 13 and 19 at the same time 741 = 3 x 13 x 19
A number is a multiple of 2,345,678,910 is it's a multiple of 9, 10, 109, 157 and 1523 at the same time if you mean this number 2,345,678,910 = 2 x 3^2 x 5 x 109 x 159 x 1523
3 - 31
You have to use the rules of 4 and 9 Using the rules of 2 and 18 won't work because the smallest common multiple of 2 and 18 is 18 not 36. 3 and 12 won't work either because the smallest common multiple of 3 and 12 is 12 not 36. However 4 and 9 does work because their biggest common divisor is 1 so multiplying them works. The biggest common divisor of 2 and 18 is 2 and the biggest common divisor of 3 and 12 is 3
There is no divisibility rule for 13 because it is a prime number. If you are thinking: why is there a divsibility rule for 2 and 3 then. Well, i don't know so go look it up on google.
there is a divisibility for 24 the rule is you can divide 24 as 6 and 4 i think
A number has to be even and also divisible by 3.