how to find remainder of any number by simple logic

In this post i will describe how to find remainder of any by simple logic. Do nothing just cram it up and apply it. It would definitely save your precious time in examination. First, let's do some revision on divisibility test.


Even Number - A number divisible by  2.

Odd Number - A number not divisible by 2.

Integers - ............-3,-2,-1,0,1,2,3,4........

Whole Number - 0,1,2............

Prime Number - A number which is not divisible by any number except 1 and number itself.

Composite Number - A number which has more than 2 factors.

-ve Integerals  -  ........................-3,-2,-1

Real Number - Any number which can be represented on number line. For ex - 0.002,-2,0.0003,5.

Complex Number - Number which cannot placed over number line. Always represent in form of x+iy.

Real Part - x
Imaginary Part - y


These r the revision of basic mathematics on numbers. Please see what is given in the questions then after go for approach.

Lets do revision on divisibility test.


We know all the divisibility tests upto 10. So i m explaining from 10 onwards.


Divisibility by 11 - Lets say we have a number 1221 we have to find either it is divisible by 11 or not.

 Condition must follow for divisibility of 11

11k = sum of even digits - sum of odd 

k can be any integral value.

Here in this case ,

    = 1+2 -  (2+1)
    = 0

Hence, it is divisible by 11.



Divisibility by 7 -  Lets' say we have a 8588. Double the last digit of this number, means 8*2= 16, substract 858 - 16 = 842, Do the same thing. 2 * 2 =4 , 84-4 =80, Hence this number not divisible by 7.

Divisibility by 13 - We have 4561. Four times the last digit. 1 * 4 . Substract from residual 456 + 4 = 460, Repeat it. 0 * 4 =0 , 46 + 0 , 46. 46 is not divisible by 13. Hence 4561 not divisible by 13.

Divisibility by 17 -  4561, add five times of last digit to the residual number. 456 + 5 * 1 = 461, Repeat, 46 + 5 * 1 = 51, 51 is divisible by 17. Hence 4561 also.

Divisibility by 19 - 4561, add 2 time of last digit to residual number. 456 + 2 * 1 = 458, 45 + 8 * 2 = 66 . 66 is not divisible by 19. Hence 4561 also.


Don't go beyond this otherwise it would be complicated for U as well as for me.


All prime number greater than 3 can be expressed in terms or either 6K - 1 or 6K + 1.


Cyclicity of Numbers-


Always remember  6 * 6 = 36 , 6 * 6 * 6 = 216 , ...... 6 is repeating in unit digit thats' called cyclicity.

in Case of 2 , 2⁴=16, so last digit six we obtained. 2⁴*2⁴*2⁴.....=abcd6. 6 at last digit. So eventually cyclicity in power of 2 is 2⁴ⁿ .

9 * 9 = 81, we got 1 as last digit. 9²ⁿ Or 3⁴ⁿ= 1 as last digit.

In Case of  3, 3⁴ⁿ = 1 is always at last digit.

In Case of 4, 4²ⁿ= 6 at last digit

In Case of 5 = 5ⁿ = 5 at last digit

In Case of 6 = 6²ⁿ = 6 at last digit

In Case of 7 = 7⁴ⁿ = 1 at last digit

Same for 8,9,10, do get confused.


Suppose we have to find the unit digit of 2⁴⁵*3⁸⁷*99³³

99³³=9³³=9^2*16*9=9
3⁸⁷=3^4*21*4³=1*64=4
2⁴⁵=2^4*11*2¹=6*2=12=2

So Last digit is 9*4*2=72 =2


Its very simple. Just remember cyclicity of these numbers.




Ques - A man has some marbles if he divides them into four equal groups, 2 balls left. If he divides into 7 groups, 6 balls left. 9 groups then 7 balls remained. Smallest no. of marbles so he could have.

->  Lets a man  has n no. of balls.

1st case

n=4x+2

2nd case

n=7y+6

3rd case

n=9z+7



It can be solved by hit & trial ,

4x + 2 = 7y +6 (start from zero always)

4x = 7y + 4

x = 7y/4 +1

smallest value of y is 4.

n = 7 * 4 + 6 = 28+6=34

Just make relations and do it by special equations.




Complex Remainder finding theorems and questions click here