REMAINDER THEOREM
(PRACTICE SET)
1.
2.
22+24+26+28+210.....260 is divided by 7. Find the
eku D;k gS\
remainder :
(a) -4
(b) 2
(c) -2
(d) 4
(a) 1
(b) 2
(c) 3
(d) 0
2
4
6
8
10
2 +2 +2 +2 +2 .....2
60
8.
times the quotient and four times the remainder.
is divided by 5. Find the
If the remained is 12 then the dividend is :
remainder :
3.
(a) 0
(b) 0
(c) 0
(d) 0
4.
5.
Hkkx ds ,d iz'u esa] Hkktd] HkkxiQy dk 8 xquk gS rFkk 'k
4 xquk gS] ;fn 'ks"kiQy 12 gS rc HkkT; D;k gS&
x+x2+x3+x4......x6000 is divided by 4
Find the remainder If
x
R 1
7
9.
(a) 300
(b) 288
(c) 512
(d) 524
The divisor is 25 times the quotient and 5
times the remained. If the quotient is 16 the
(a) 0
(b) 1
dividend is:
(c) 2
(d) 3
3100+3200+3300+3400.....312000 is divided by 7. Find the
Hkktd] HkkxiQy dk 25 xquk rFkk 'k"kiQy dk 5 xquk gSA ;
dk eku 16 gS rks HkkT; Kkr dhft,\
remainder :
(a) 6400
(b) 6480
(c) 400
(d) 480
(a) 1
(b) 2
(c) 3
(d) 0
10. On dividing 397246 by a certain number, the
quotient is 865 and the remainder is 211. Find
2 2 +2 4 +2 6 +2 8 ....2 64 is divided by 7. Find the
the divisor.
remainder :
6.
If a problem involving division, the divisor is eight
(a) 7
(b) 6
(c) 5
(d) 8
dks fdlh fuf'pr la[;k ls Hkkx nsus ij HkkxiQy 865 vk
'ks"kiQy 211 gS] Hkktd Kkr dhft,\
397246
5100+5200+5300+5400....... 59000 is divided by 13.
Find the remainder :
(a) 10
(b) 11
(c) 12
(d) 13
(a) 456
(b) 459
(c) 458
(d) 457
11. A number when divided by 14 leaves a remainder
of 8 but when the same number is divided by 7. If
will leave the remainder.
7.
61284
If x is the remainder when 3
is divided by 5
and y is the remainder when 496 is divided by 6,
then what the value of (2x-y) ?
x cprk
;fn 361284 dks 5 ls foHkkftr fd;k tkrk gS rks
'ks"kgSS vkSj
96
4 dks 6 ls foHkkftr fd;k tkrk gS
y rks
'ks"k cprk (2x-y)
gS dk
;fn fdlh la[;k dks 14 ls Hkkx fn;k tkrk gS rks 'ks"kiQy 8 izkI
gS ;fn mlh la[;k dks 7 ls Hkkx fn;k tk;s rks 'ks"kiQy crkb,
(a) 3
(b) 2
(c) 1
(d) Can't be determinde
1
12. When a number is divided by certain divisor,
remainder is 35 by another no. is divided by the
same divisor remainder is 27. If the sum of both
number is divided by the same certain divisor,
remainder is 20. Find the certain divisor.
(x3-6x+7)
dks(x+1) ls Hkkx nsus ij izkIr 'ks"kiQy gS
(a) 2
(b) 12
(c) 0
(d) 7
18. If (x 5 -9x 2 +12x-14) is divided by (x-3) then
;fn ,d la[;k dks fdlh fuf'pr Hkktd ls Hkkx fn;k tkrk gS rks remainder is :
'ks"kiQy 35 izkIr gksrk gS ;fn ,d nwljh la[;k dks mlh Hkktd ls Hkkx
5
2
dks(x-3) ls ykx nsus ij izkIr
fn;k tkrk gS] rks 'ks"kiQy 27 vkrk gS ;fn nksuksa la[;kvksa (x
ds -9x
;ksx+12x-14)
dks
gS&
mlh fuf'pr Hkktd ls Hkkx fn;k tkrk gS rks 'ks"kiQy 20 izkIr'ks"kiQy
gksrk
gS] fuf'pr Hkktd dk eku Kkr djsa\
(a) 184
(b) 56
(a) 44
(b) 43
(c) 42
(d) 41
13. If two number are each divided by same divisor,
the remainder are respectively 3 and 4. If the sum
of the two numbers be divided by the same divisor,
The remainder is 2. The divisor is :
(c) 2
(d) 1
19. If x3-5x2+4p is divisible by (x+2), then the value of p
is :
;fn x3-5x2+4p dj
eku crkb,\
(a) 7
(x+2) ls iw.kZr% foHkkftr
p dk
gS rks
(b) -2
;fn nks la[;kvksa esa ls izR;sd dks fdlh fuf'pr Hkktd ls Hkkx fn;k
(c) 3la[;kvksa
(d) -7
tk;sA rks 'ks"kiQy Øe'k% 3 vkSj 4 izkIr gksrk gSA ;fn nksuksa
dk ;ksx dks mlh fuf'pr ls Hkkx fn;k tk;s rks 'ks"kiQy 2 izkIr
gksrk
20. If
(x-a) is a factar of (x3-3x2a+2a2x+b), then the value
gS] Hkktd Kkr dhft,\
of b is :
(a) 9
(b) 7
(c) 5
(d) 3
14. A number x when divided by 289 leaves 18 as the
remainder. The same number when divided by 17
;fn (x-a) O;atd (x3-3x2a+2a2x+b) dk ,d xq.ku[k.M
b dk eku Kkr djsa\
gSa] rc
(a) 0
(b) 2
(c) 1
(d) 3
leaves y as a remainder the value of y is:
la[;k x dks 289 ls Hkkx nsus ij 'ks"kiQy 18 izkIr gksrk gSA mlh la[;k
51 52 is divided by
21. Find the remainder when 50
y izkIr gksrkygS]
dks 17 ls Hkkx nsus ij 'ks"kiQy
dk eku crkb,\
 
(a) 5
(b) 2
(c) 3
(d) 1
15. 173×192×99×96 find the last two digits?
173×192×99×96 ds vfUre nks vad Kkr dhft,\
(a) 64
(b) 65
(c) 66
(d) None of these
16. 39×55×57×24×13872×9871. Find the last two digits.
39×55×57×24×13872×9871 ds vfUre nks vad Kkr dhft,\
52
5051
dks 11 ls Hkkx nsus ij izkIr 'ks"kiQy g
 
(a) 6
(b) 4
(c) 7
(d) 3
22. Find the remainder when (3334 )35 is divided by7.
(a) 21
(b) 20
(3334 )35 djs 7 ls foHkkftr djus ij izkIr 'ks"kiQy
(c) 19
(d) 18
gS\
17. On dividing (x3-6x+7) by (x+1), then the remainder
is :
2
11.
(a) 5
(b) 4
(c) 6
(d) 2
23. If (17)41+ (29)41 is divided by 23. Find the remainder.
25. The expression 8n-4n, where n is a natural number
(17)41+ (29)41 dks 23 ls foHkkftr djus ij izkIr 'ks"kiQyis always divisible by :
gS%
O;atd 8n-4n tgkan ,d izkÑr la[;k gS iw.kZr% foHkk
(a) 1
(b) 6
(c) 0
(d) 12
24. If mn-nm = (m+n) : (m, n) are prime numbers, then
what can be said about m and n.
gksxk&
(a) 15
(b) 18
(c) 36
(d) 48
;fn mn-nm = (m+n) tgka(m, n) vHkkT; la[;k gSamrc 26. If (x+2) and (x-1) are the factor of (x3+10x2+mx+n)
the value of m and n are :
rFkkn ds ckjs esa lR; gS&
(a) m, n are only even integer./m, n dsoy le
;fn (x+2) vkSj (x-1) O;atd (x3+10x2+mx+n) ds
iw.kk±d gSA
xq.ku[k.M gSmrks
vkSjn dk eku crkb,\
(b) m, n are only odd integer./m, n dsoy fo"ke
(a) m = 5, n = -3
(b) m = 17, n = -8
iw.kk±d gSA
(c) m = 7, n= -18
(d) m = 23, n = -19
(c) m is even and n is odd/m le rFkkn fo"ke la[;k
gSA
(d) None of these. /buesa esa ls dksbZ ugha
3
SOLUTIONS
Cyclicity of 7 is 6
22
R4
1. :
7
=
a+a 2 +a 3 +..... a 120
7
24
R2
7
Every 6 terms give remainder '0'
120 is divied by 6
So, remainder will be '0' till a120
6
=
=
2
R 1
7
Alternate
4+2+1
R 0
7
every three terms give remainder 0
a
3100
=
R 4
7
7
a2 =
16
R2
7
a3 =
64
R 1
7
Let 22 = a
2
3
4
5
6
a+a +a a +a +a ...........a30
0
0
7
The total term of this series is 120 and 120 is
divided by 6 perfectly.
R=0
2. :
22
R 4
5
So, remainder = 0
5 : Let a = 22
4
=
2
R 1
5
26
R 4
=
5
Every two terms give remainder 'o'
6
8
60
2
4
a +a + 2 +2 ...........2
0
0
remainder will be '0'
3.
Every three terms give remainder '0'
Cyclicity of 7 is '6'
a=
22
R4
7
a2 =
42
R2
7
a3 =
43
R 1
7
a+a 2 +a 3 +a 4 ....a 32
7
Till thirty terms, it will give remainder = 0
Extra terms = a31, a32
Every 6 terms gives remainder '0'
6000
divide perfectly so remainder will be '0'
6
4.
Let a = 3100
100
a
3
=
7
7
4
4
=
3
R  4 (it is not '1')
7
a 31 +a 32
7
The ............. of 7 is 6
a1 +a 2
4+2
=
R 6
7
7
6.
Let 5100 = a
Quotient =
a+a 2 +a 3 +a 4 ...a 90
13
48
6
8
Dividend= divisor × quotient + remainder
= 48×6 +12
100
a=
4
5
5
=
R 1
13 13
= 288+12
= 300.
a=1
9.
(b) Dividend = divisor × quotient + Remainder.
1+1+1....1 till 90 terms
13
=
7.
Acq.
Divident = 16×25 = 5×R
90
R  12
13
=R=
1
×16×25
5
R = 80
361284
(c)
5
dividend = [(16×25)×16] +80
dividend, 6480.
Cyclicity 4
61284
4
R
10. (b) Divisor =
11. (c)
14 N x
 0.
8
261284
1
5
397246  211
 459
865
Remainder
N = 14x+8.
According to question.
x=1
N 14x+8
=
7
7
496 4  495 2  495


6
6
3
8
1
7
Remainder =
=
2  (1)95
2×2 = 4 [We divide by 2 so multiply by 2]
3
12. (c)
y =4
2x-y = 2×1-4 = -2.
8.
(a) Remainder = 12.
Divisor = 4×12 = 48
N1
,
D
N2
,
D
N1 +N 2
D



R1
R2
R3
D = R1 + R2- R3
The divisor = 35+27-20 = 42.
5
13. (c) D = R1+R2-R3
17. (b)
x+1 = 0
 x= -1
= 3+4-2
put the value of x = -1 in equation.
D=5
R = x3-6x+7 = (-1)3 - 6 (-1) +7
= -1+6+7
Remainder 18

14. (d)
17
17
R = 12.
18. (a)
y=1
 x= 3.
x-3 = 0
Put of value of x = 3 in x5-9x2+12x-14
173  192  99  96
15. (a)
100
R = (3)5-9(3)2+ + 12(3)-14
= 243-81+36-14.
Simplifying by 4.
-2
-8
-1
-1
= 279-95




R = 184.
173 ×
192 ×
92 ×
24
19. (a)
x+2 = 0
 x = -2 (put)
25
=
2  ( 8)  (1  1)  (1)
25
16
=
= 16
25
So, Actual Remainder = 16×4 = 64.

x3-5x2+ 4p = 0

(-2)3-5(-2)2+4p = 10

-8-20+4p = 0

-28+4p = 0

4p = 28.
Last two digit = 64 (6 and 4).
16. (b)
p = 7.
20. (a)
39×55×57×24×13872×9871
100
Simplifying two times by 4 and 5. So, to get
last two digit we have to multiply 20(4×5).
x-a = 0
 x = a.

x3-3x2a+2a2x+b = 0

a3-3a2a+2a2a+b = 0
-1
+1
+2
+1
+2
+1

a3-3a3+2a3+b = 0







b=0
39 ×11 × 57 ×
6 ×13872 ×9871
5
1  1  2  1  2  1
=
5
= 
4
 4
5
52
5051
21. (d)
 
11
505152

11
Cyclicity of 11 is 10
51  52
2

=2
10
10
Actual Remainder =
= 5-4 = 1
So, Actual last two digits 1×20 = 20 (2 and 0)
6
(50) 2 36

3
11
11
=3
25. (d)
35
3334
22. (b)
 
=
 34    35   2
7
6
n = 1, 2, 3, 4, .........., n is natural number.
8n-4n
6
n = 1, 2, 3, ...... n is a natural number)
put n = 2.
=
 332
expression = 82-42 =64-16 =48
7

So, function is divisible by 4.
4
=
7
23. (c)
8n-4n is divisible by 48.
26. (c)
(x+2) & (x-1) are the factor of x3+10x2+mx+n
Actual R 4.
so put x = -2 & x=1 respectively
(an+bn) is always divisible by (a+b) when n is
odd power.
 (-2)3+10 (-2)2+m(-2)+n = 0
Then, 17+29 = 46.
Factor of 46 (1, 2, 23, 46)
 -2m +n+32 = 0
 -2m + n = -32
...... (i)
put x= 1
So, (1741+2941) is perfectly divisible by 23.
 (1)2 +10(1)2+m.1+n = 0
24. (c)
mn-nm = m+n.
consider m =2 and n = 5
Then, 25-52 = 5+2
7
= 7
1+10+m+n = 0
m+n = -11
...... (ii)
Slov. (i) and (ii) we get
m = 7 and n= -18.
The option (c) is correct answer.
7