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In this article ,We would learn the basic fundamental approach & applications of Remainder Theorem to solve out numerous problems of power based system asked in various competitive exams including CAT,CDS, SSC CGL, SSC CHSL, SSC CPO, BANK PO, BANK CLERK, NTSL,RAILWAY, LDC,CSAT etc. like any one day exam.
Try to Memorise Powers of certain numbers:-
Q. Find the remainder when 37100 is divided by 7 .
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In this article ,We would learn the basic fundamental approach & applications of Remainder Theorem to solve out numerous problems of power based system asked in various competitive exams including CAT,CDS, SSC CGL, SSC CHSL, SSC CPO, BANK PO, BANK CLERK, NTSL,RAILWAY, LDC,CSAT etc. like any one day exam.
Concept:-
v When you are to find remainder of a power based system that get divided by a divisor then, you should try to convert that system in such a way that the main(actual)base of the power will leave +1 or -1 as remainder on dividing by the divisor given in the question; which is explained through following examples.
v
(-1)odd = -1 And
(-1)Even = +1
e.g. R[35/9] = -1 =
9-1=8
R[3521/9] = (-1)21
= -1 = 9 -1 =8
R[3520/9] = (-1)20 = +1
R[9999/100]= (-1)99 = -1= -1+100
= 99
Q. What will be the remainder (719 + 2) is divided
by 6 ?
Sol:- (719 + 2)/6 = 19 + 2 = 1+2 = 3
Try to Memorise Powers of certain numbers:-
21 =2
|
22 = 4
|
23 = 8
|
24 = 16
|
25 = 32
|
26 = 64
|
27 = 128
|
28 = 256
|
29 = 512
|
210=1024
|
31 =3
|
32 =9
|
33 =27
|
34 = 81
|
35 = 243
|
36 =729
|
71 = 7
|
72 = 49
|
73 = 343
|
74 = 2401
|
Q. Find the remainder when 263 is divided by 9 .
Sol:-263/9 = (23)21/9 [we should find power of 2 that gives +1/-1 as remainder]
= 821/9 (As 23 = 8 & 8 gives remainder -1)
= (-1)21
= -1 = -1+9 = 8
Q. Find the remainder when 369 is divided by 82 .
Sol:-369/82 =3 x (34)17/82
= 3 x (81)17/82 [As 34
= 81& 82-1=81]
= 3
x (-1)17
= -3 = -3+82 = 79
Q. Find the remainder when 34321 is divided by 80 .
Sol:- 34321/80 = 3 x (34)1080/80
= 3 x (81)1080/80 [As 34
= 81& 80+1=81]
= 3
x (+1)1080
= 3
Alter:- 34321/80 {R[4321/4]=
1;so
3 remains only}
= 31/80 [As 34
= 81& 80+1=81]
= 3
Sol:- (37)100/7
[Here 37>7;so 1st divide this & obtain remainder]
= 2100/7
= 2 x (23)33/7 [As
23 = 8& 7+1=8]
= 2 x 833/7
= 2 x 133
= 2 x 1 = 2
Q. Find the remainder when 740 is divided by 400 .
Sol:- 740/400
= (74)10/400 {74=2401& 2401 gives -1
remainder by dividing 400}
= (2401)10/400
= 110 =1
Q. Find the remainder when 235 is divided by 10 .
Sol:- 235/10 {Here we can cancel by 2 to simplify}
= 135/10
= 1
Hence, the remainder
will be 1 x 2 = 2
In the following article of cyclicity or pattern theorem you will solve these questions easily and more difficult problem quickly but this fundamental ideas will help you a lot to create your basic concepts more strong.
If you have any doubt, please let me know.