N is divisible by 4 if the last two digist form a number divisible by 4.
Let's do an example: Is the number 23894723985729316 divisible by 4? Well the last two digits is 16 and 16 is divisible by 4, so YES!

N is divisible by 5 if it ends in 0 or 5.

For 6, we just combine the rules for 2 and 3:
N is divisible by 6 if it is divisible by both 2 and 3.


For the rest, we will stick with prime divisors p. Consider multiples M of p until: We want the smallest such M. Take Consider n and p-n, and usually we just pick the lowest.

Now, to find out if a number is divisible by p, take the last digit of the number, multiply it by n, and add it to the rest of the number (OR: multiply it by (p - n) and subtract it from the rest of the number).

If you get an answer divisible by p (note that this includes 0), then the original number is divisible by p. Repeat the rule if you don't know the new number's divisibility.

Now try to see if you can come up with the rule for 7! One thing you might find interesting is the following post that discusses using a 'divisibility graph' for 7.

This trick does require some memorization though, and also requires the number given to be a perfect cube. You need to memorize the cubes of the numbers 0 through 9 (or be able to figure them out on the spot). This information is contained below:



Note that the last digits of the cubes on the right have all the numbers 1 to 9, but no number is repeated. Here is how to find the two-digit cube root of a perfect cube.

Take a number, such as 658,503 which is grouped into two parts.

1. Looking at the number we see it ends in a 3, and according to the table only 7^3 ends in a 3, thus the last digit of our number is 7.

2. Next, ignore the last 3 digits of the cube, so consider 658. Compare these digits with the table above. Note that 658 fits between 512 and 729. You always choose the smaller one, in this case 512 which happens to correspond to 8^3.

Thus, the last digit is 7 and the first digit is 8, giving an answer of 87.

Normally this trick is used for six digit perfect cubes. To help understand how this works, ask yourself - What is the last digit of (10x+y)^3? Clearly it is y^3 mod 10 (how does this relate to #1?).

Another Example:
In 474,552 we have that 343 is the immediate smallest number from 474 so the first digit is 7.
The last digit in 474,552 is 2 and only 8^3 ends in a 2, so the last digit is 8. Hence, 78^3=474,552.

The first one is basically the multiplication algorithm by 11 discovered independently by Trachtenberg.

Let us look at the second one, which is used quite a bit in Vedic Mathematics: All from nine and the last from ten.

When subtracting from a large power of ten with many columns of zeros, it is not necessary to write the notation for borrowing from the column on the left. One can instead subtract the last (rightmost) digit from 10 and each other digit from 9. For example:



This method is also used when finding the deficit from the next larger power of ten when setting up a multiplication problem using the cross-subtraction method.

The third one is Vertically and crosswise (multiplications). One use for this is for multiplying numbers close to 100.

Suppose you want to multiply 88 by 98. Both 88 and 98 are close to 100. Note that 88 is 12 below 100 and 98 is 2 below 100. This can be pictured as follows:



We subtract crosswise to get the first two digits of the answer. It doesn't matter if we do 88-2=86 or 98-12=86, both give the same number. To get the last two digits we multiply vertically: 12 x 2=24. Therefore, the answer is 8624.

The same strategy works for multiplying two numbers above 100. For example, 107 times 111. Quickly we add the surplus from 107 (which is 7) to 111 to get 118, the first 3 digits of the answer. To get the last two digits, we multiply the surplus of 107 from 100 by the surplus of 111 from 100: namely, 7 x 11=77. Thus, the answer is 11,877.

Vedic Mathematics is all about using different formulas in a variety of ways.
In the above rule we are using:

The above notation is short for:

and is often used since it's easier to see what the number actually is.

The above generalizes for numbers close to a base of 1000. Note that the second su-tra becomes quite useful for when you are computing the deficit from the base.

Even better, it gives a way to help memorize them, by allowing one to work out the answer by rule if one cannot remember it by rote. We perform each rule starting at the far right. The `number' is the digit of the multiplicand just above the place that we are currently computing. The `neighbour' is the digit immediately to the right of the `number'. When there is no neighbour, we assume it is zero. We also write a zero in front of the multiplicand.

Note that the following rules only use the operations of addition, subtraction, doubling, and `halving'.