Monday, 5 October 2015

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How to find remainder



How to find remainder 
                                                      BY- abhijat srivastava
hello philic's,
today we learn how to find the remainder. This is very important chapter with the view to ssc examinations, so you must have mug all this questions 





let's have a look

We know that,

dividend = quotient*divisor + remainder

i.e. a = qd + r

r is always positive and less than divisor
i.e. 0 \le r < d

1. When remainder, r = 0

then, divisor completely divide the dividend i.e. a = qd

so, if we write any number 'a', of the multiple of divisor 'd' then the remainder 'r' will be zero

e.g. If we have to find the remainder of 12*X when divided by 12.

clearly number is in the form a = qd = X*12 i.e. 12 completely divided the number and remainder 'r' will be zero.

Note the following:

1. 12*X + 5 , Remainder 'r' = 5 when divided by 12 ( since Number = 12*X + 5 compare it with a = dq + r)

2. 12*X + 17 = 12*X + 12 + 5 = 12(X + 1) + 5, remainder = 5

3. 12*X + 109 = 12*X + 108 + 1 = 12*(X + 9) + 1, remainder = 1


e.g. 
1. Remainder of  {(5)^{24}} divided by 25 

{(5)^{24}} = {5^2}{.5^{22}} = {25.5^{22}}

Therefore, it is a multiple of 25, 
so, remainder = 0

2. Remainder of {(5)^{13}} divided by 25

{(5)^{13}} = {5^2}{.5^9} = {25.5^9}


Multiple of 25
so, remainder = 0

3. Remainder of {(60)^{13}} divided by 25


{(60)^{13}} = {60.60.60^{11}} = {5.12.5.12.60^{11}} = {25.144.60^{11}}

multiple of 25
so, remainder = 0




2. If 15 is divided by 13, remainder = 2

when 15*15 divided by 13 i.e. 225 divided by 13, remainder = 2*2 = 4

How this is possible ?

15*15 = (13 + 2)(13 + 2) = 13*13 + 13*2 + 13*2 + 2*2 = 13*(13 + 2 + 2) + 2*2

If we divide 13*(13+2+2) + 2*2 by 13, Remainder = 2*2 = 4 ( a = dq + r)

Since first term i.e. 13*(13+2+2) is completely divided by 13

It is true for any multiple,

Consider, remainder of {15^7} divided by 13

If you divide 15 by 13, remainder = 2

and to find the remainder, you have to multiply all the remainders

i.e. remainder = {2^7} = 128


repeat the process,
128 can further divided by 13
so, remainder = 11 (since 0 \le r < d )

another method,

or, 2^7 = (2^4)*(2^3) = 16*8 

REPEAT THE PROCESS

if you divide first term 16 by 13, remainder = 3

so, now we get 3*8 = 24

and further if you divide 24 by 13, remainder = 11

YOU CAN FREELY MULTIPLY REMAINDERS AND REPEAT PROCESS, unless you get the remainder which is less than divisor.



Ques: Find the remainder when 15*13*17*24*19 is divided by 11 ?

Approach:

15*13*17*24*19  -R-  4*2*6*2*8  -R- 48*16 -R- 4*5 -R- 20 -R- 9 


( multiply there respective remainders and multiply then repeat the process unless you will not get the remainder which is less than divisor)












1. We know that, Remainder is always positive or zero


i.e. 0 \le r < d 


where, r is remainder and 'd' is divisor



2. Consider,


1st Case: 15 = 16*0 + 15 


2nd Case: 15 = 16*1 - 1


in 1st Case when 15 is divided by 16, remainder = 15 


in 2nd Case when 15 is divided by 16, remainder = -1




3. What is the remainder of  - 17 when divided by 5 ?


Clearly, we get  -2


but since remainder cannot be negative, we have to make it positive as 5 - 2 = 3


so, the remainder is 3




4. We can use the negative remainder concept to simplify the solution and find out the remainder.





e.g. 
1.  Remainder of {(24)^{25}} when divided by 5.

since, {5^2} = 25 and when 25 divided 24 we will get the remainder -1 

so, we can write,

{(24)^{25}}  (- R -) {( - 1)^{25}} (- R  \to ) -1  (- R -)  4

( we finally get the remainder which is -1, but remainder cannot be negative. So, remainder = 5-1 = 4)




2. Remainder of 16 \times 17 \times 13 \times 11 \times 14 when divided by 15



16 \times 17 \times 13 \times 11 \times 14  (- R -) 1 * 2 * (-2) * (-4) * (-1)   \to  R \to  (-16)   \to  R  \to  -1  \to  R  \to  14




3. Remainder when {(2)^{2004}} is divided by 7


we know that, {2^3} = 8


(*Always try to find a number which is when divided by divisor gives remainder '1' or -1 or any small number)


so we can write,


{(2)^{2004}} = {(8)^{668}}


when we divide 8 by 7 we get the remainder 1


therefore, {(8)^{668}} - R - {(1)^{668}} - R -     1 



ONE MORE IMPORTANT CONCEPT:



Consider an example, remainder when 52 is divided by 8



if you directly divide 52 by 8, we get the remainder = 4


because, 52 = 8*6 + 4


but you can also write, 52/8 = 13/2


so, now we have 13 divided by 2 


and we get the remainder = 1


as, 13 = 2*6 + 1



what happen ?



Actually when we divide 52 by 4 the remainder also get divided by 4.


so, if we find the actual remainder we have to multiply 1 by 4


i.e. Remainder = 1*4 = 4



e.g. What is the remainder when {(2)^{96}} is divided by 96


since, {2^5} = 32 which is a common multiple with 96

so we can write,

\frac{{{2^5} \times {{(2)}^{91}}}}{{96}} = \frac{{{{(2)}^{91}}}}{3}  ( after dividing by 32 )


so, now we have to find the remainder of {(2)^{91}} when divided by 3

so, {(2)^{91}} \to R \to  {( - 1)^{91}} \to R \to  -1  \to R \to  2

so, the Remainder = 2

but since we divide the dividend and divisor by 32


so, now we have to multiply 32 to the outcome


Therefore, remainder = 32*2 = 64






THEOREMS

1. {a^n} + {b^n}  is divided by (a + b) when n is ODD

2. {a^n} - {b^n} is divided by (a + b) when n is EVEN 

3. {a^n} - {b^n} is always divided by (a - b)




Ques 1: What will be the remainder when {3^{444}} + {4^{333}} is divided by 5?

Sol:

Since, {3^{444}} + {4^{333}} = {({3^4})^{111}} + {({4^3})^{111}}

since, {a^n} + {b^n} is divided by (a + b) when n is ODD

Therefore, it is divided by {3^4} + {4^3} = 81 + 64 = 145 

since, it is divided by 145 

Therefore, it will certainly be divided by 5 because 5 divides 145

Therefore, Remainder = 0





Ques 2: What will be the remainder when {5555^{2222}} + {2222^{5555}} is divided by 7 ?

Sol:

When 5555 and 2222 are divided by 7 we get the remainders 4 and 3 respectively.

so, problem reduces to find the remainder when {4^{2222}} + {3^{5555}} is divided by 7

so, {4^{2222}} + {3^{5555}} = {({4^2})^{1111}} + {({3^5})^{1111}}

since,  {a^n} + {b^n} is divided by (a + b) when n is ODD

therefore, it is divided by {4^2} + {3^5} = 16 + 243 = 259

since, 259 is divided by 7,

therefore, 7 will divide the {5555^{2222}} + {2222^{5555}}

so, Remainder = 0



IF YOU WANT STUDY ALGEBRA CLICK HERE, ALONG WITH IMPORTANT QUESTIONS ASKED IN SSC


THANKS 
KP FAMILY.





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