# Make Maths Easy

Now let us see some easy way to find divisiblity of the numbers and some basic algebraic formulas.

**Divisibility of the numbers**

Divisible by 2: End with even number.

Divisible by 3: Sum of the digits should be multiply of 3

For example: 45=4+5=9(which is divisible by 3)

Divisible by 4: The last 2 digits have to be divisible by 4

For example: 248, here last two digit is 48 which is perfectly divisible by 4

Divisible by 5: The numbers should end with 5 or 0

Divisible by 6: If a number is divisible by 2 and 3, then its divisible by 6

Divisible by 7: Take the last digit from a number, multiply by 2. Then Subtract the remaining digits with the multiplied number. If the resultant number is divisible by 7, Then the whole number is divisible by 7.

For Example: Consider the number 5964. Now let us check the divisiblity for 7

Last digit is 4, so 4*2=8. The remaining digits are 596.

596-8=588. Since we have 3 digits once again take the last digit and multiply by 2.

So 8*2=16, The remaining digits are 58

58-16= 42. Where 42 is divisible by 7.

Divisible by 8: If the last three digits are divisible by 8 then the number is divisible by 8.

Divisible by 9: The sum of the digits should be divisible by 9

Divisible by 10: The number should end with 0

Divisible by 11: From right side of the number sum up the alternative number. That is odd place and even place numbers. If both sum are equal then its divisible by 11

For example: 6248

Sum of odd digits from right is 8+2=10

Sum of even digits from right is 4+6=10, so this is divisible by 11.

Divisible by 12: A number which is divisible by both 3 and 4.

Divisible by 15: The number divisible by both 3 and 5

**Multiplication of a number 5 ^{n}**

If 1254*25 is given

25=5^{2}, here n=2

The formula is add n number of 0’s to the right of multiplicand

Here add 2 0’s to 1254=125400

Then divide this number by 2^{n}, since n=2, 2^{2}=4

125400/4=31350.

**Division algorithm**

Dividend= (divisor*quotient)+reminder.

**Some of the Algebra Formulas**

(a+b)^{ 2}=a^{2}+b^{2}+2ab

(a-b)^{ 2}=a^{2}+b^{2}-2ab

(a+b)^{2}-(a-b)^{ 2}=4ab

(a+b)^{2}+(a-b)^{ 2}=2(a^{2}+b^{2})

a^{2}-b^{2}=(a+b)(a-b)

a^{3}+b^{3}=(a+b)( a^{2}+b^{2}-ab)

a^{3}-b^{3}=(a-b)( a^{2}+b^{2}+ab)

1+2+3+……+n=n(n+1)/2

I hope this is useful to recollect your past maths. Share some of the easy way for maths calculation which you might have known in the comments below 🙂

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