how to find remainder of any number by simple logic

Saturday, 10 May 2014

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

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Sum of all the Numbers of any number of Digits with Tricks

Tuesday, 6 May 2014

Sum of digits of 2 or 3 digits commonly ask in the aptitude section. This questions we can do without going for large calculations. With a simple formula we can do it in a single step. Please focus on how to solve it with minimum time. All we should try to minimize time and complexity to solve any problem.


-> Find out four digits distinct numbers sum which is formed by using 2,7,9,5,1 digits ?

Ans  Four digits numbers can be written as

4 digit number=1000a+100b+10c+d

Number formed with these digits with repetition = ⁵P₄   

=120

As all digits contributes equally to hundred's, ten's and one's digit. So

-> Contribution to Thousand's Digit = 120/5 = 24

-> Contribution to Hundred's Digit = 120/5 = 24

-> Contribution to Ten's Digit = 120/5 = 24

-> Contribution to One's Digit = 120/5 = 24

Remember all digits r non-zero means 2,7,9,5,1 does n't consists 0.

If it consists zero then slightly change in contribution to Thousand's digit.

That's 120/4 = 30

Note : 0 not contribute to first digit of any number or starting digit of number

Can you assume 4 digit number as 0234 ? Really not



Now Sum of All 4 Digits Numbers = 1000 * 24 * (2+7+9+5+1) + 100 * 24 * (2+7+9+5+1) + 10 * 24 * (2+7+9+5+1) + 1 * 24 * (2+7+9+5+1).

=  24000 * 24 + 2400 * 24 + 240 * 24 + 24 * 24

= 24 (24000+2400+240+24)

= 24 (26664)

= 639936

Or simply  (n)! (sum of digits) X (1111 upto n times) 


In this manner we can do the sum of any digit number. Remember digits r not repeated in this Case.



Digital Sum of Any number = sum of the digits of particular number

for ex -  23109

digital sum = 2+3+1+0+9 = 15 = 1+5 = 6



Verfiy multiplication of a number

Suppose we have two numbers

34 X 45 = 1234

Now we want to check if this is correct multiplication

(3+4) * (4+5) = (1+2+3+4)

7 * 9 = 10

63 =10

(6+3) = 10

9=10 . InCorrect Multiplication


-> 12 * 12 =144

(1+2) * (1+2) = 4+4+1

3*3 = 9

9 = 9 , Correct Multiplication


Some questions can do only with this technique as


find value of A+B ?

A45092233B * 10 = 230000010

Don't count 9 or sum 9 in finding digital sum.

A+B+1=6

A+B=5



Unit Digit of a number


->34 * 33 * 1000 * 787888 * 9999991222

Unit digit = ?

-> As above multiplication consists 1000 or a number ending with zero . So result contains zero at its end.

Unit's digit =0

Same 5 * 2 =10

5 * 4 =20

5 * 8 =40

5 * 6 =20

Any even number multiplied with Any number ends with unit's digit 5 = Unit's digit 0

->34 * 77 * 99 * 11111111 * 345 * 1002

As 5 * 2 =10

Therefore unit's digit = 0


->Unit's digit of any number ending with 1

34599999999991222331^{889988388838828}

As number end with digit 1

Unit's digit of above number = 1


-> Unit digit of any multiplication except above case (0 or 5)

89 * 877 * 76 = 9 * 7 * 6 = 63 * 6 = 3 * 6 = 18

So 8




Last Two Digits of Any Number


->Last two digit of 76ⁿ is 76

-> Last two digit from multiplication

12342 * 998 * 1007 = 42 * 98 * 07 = 28812 = 12

-> Last two digits of number end up with 1.

-> 12341³⁵²⁹ = Last two digits is multiplication of unit's digit of exponent and ten's digit of number.

In this case  9 * 4 = 36  and unit's digit 1

= So last two digit is 61


-> Last two digits of any even number

346⁹⁶= Last two digits is break the number into two parts even and odd dividing by 2.

(173)⁹⁶*2⁹⁶=

First make 173 into number end with 1

last digit of 173 is 3 and 3 raise to power 4 = 81.

So 173^4*24=(173⁴)²⁴=

Find last 2 digits of 173 raise to the power 4 becoz we need to find only last two digits.

73 raise to power 4 = ?


-> 73 raise to power 2 * 73 raise to power 2

-> 73 raise to power 2 is

73   -27   (The Method is Below to Calculate Square)
73   -27
------------
46 729

46 + 7 29

5329 is Ans

So 29 * 29 = 29(30-1) =870-29=841

Last two digit of 173⁴ is 41

Therefore 41²⁴*2²⁴= (61) * 2^10*2*2⁴= 16 * 76 * 16 =61 * 76 * 16 =  76

Last two digits is 76.



How to find square of a number easily ?

As we know we can express a number into sum of odd numbers.

1*1 = 1
2*2 = 1 + 3
3*3 = 1 + 3 + 5
4*4= 16 = 1 + 3 + 5 + 7
5*5 = 25 = 1 + 3 + 5 + 7 +.....................(2n-1)

Hence we can say that

26 * 26 = 25 * 25 + 26th Odd Number

Or

Square of number less than 50

= 43²=

Find difference from 50 = 50-43 = 7

Substract this difference from 25 = 25 - 7 = 18

Square first difference 7 * 7 =49

Required No. is  1849


Square of a number above 50

= 57²=

Add last digit to 25 = 25+7=32

Square last digit 7*7=49

Required Number is 3249


Square of any number closer to 100 or multiplication of numbers closer to 100.

By Using this Method We can solve complicated problems with ease. Thanks

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techniques to solve remainder based questions

Mostly questions in higher level aptitude consists remainder based questions. Not easy to tackle. This types of questions need a specific technique for solving. FirstofAll lets do revision of some of famous algorithm to find remainder.


1) Euler 's theorem

This theorem applies only on some condition M & N must be coprime to each other.

What is coprime numbers ?
Those numbers which have common hcf 1 is defined as coprime numbers. Therefore HCF(n1,n2,n3....) needs to be 1.

Here HCF(M,N)=1

->Take a look on problem how to solve it.

Lets say we have to find remainder of   11³⁴/12.

First, check the condition HCF(8,13)=1. If yes, then proceed

Prime Factorize 12= 2²* 3

Calculate Euler Totient Function ∅(12)= 12 (1-1/2) * (1-1/3)
                                                            = 12 * 1/2 * 2/3 = 4

So 11⁴ⁿ /12=1 (n can be any whole number)

and (11^4*8 * 11²⁾/12 =1

11³² /12=1 therefore we left with 11² /12=121/12=1

So eventually we get remainder as 1. Check your result by calculator also.



2) Fermet Theorem - Its very easy extended version of above. N must be prime and M & N should be co-prime.

->Remainder [M^N /N]= M

->Remainer [M^N-1 /N]= 1

-> M^N-M is divisible by N.

Very easy to implement, direct questions comes so apply the theorem directly.


3)  Wilson Theorem - M must be prime number

-> Remainder [(M-1)!/M]=M-1

-> Remainder[(M-2)!/M]=1


-> 11!/13 = (11-2)!/13 = 1



4) x = 1! + 2*2! + 3*3!..............10*10! 

Remainder when x+2 divided by 11!

1*1! can be written as 2! - 1!
2*2! can be written as 3!-2!
.
.
.
.
10*10! can be written as 11!-10!

After adding them we get only 11! - 1

x + 2 = 11! - 1 + 2 = 11! + 1 

(11!+1)/11! = 1

finally 1 as remainder.


Important Properties on Remainders

 5) Remainder [a*b / x ] = Remainder [a / x] * Remainder [b / x].

6) Remainder [a / x ] = 1 , then  Remainder [ a^{any power} /x ]=1

IT's all about the remainder theorems time to wrap up.


Ques - (1! + 2! + 3! + 4! + 5!)/24 what is the remainder ?

4! and 5! is divisible by 24.

so 1!+2!+3!=9/24
=9

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