REMAINDER THEOREM

(SET – 02)
1. Find the remainder when 54124 is divided by 17. 7. What will be the remainder when 123456789 is
divided by 8?
54124 dks 17 ls foHkkftr djus ij izkIr 'ks"kiQy crkb,\
123456789 dks 8 ls foHkkftr djus ij izkIr 'ks"kiQy crkb,\
(a) 4 (b) 5
(a) 5 (b) 7
(c) 3 (d) 15
(c) 9 (d) None of these
2. Find the remainder when 21875 is divided by 17.
8. When 101+102+103+.........+1099+10100 is divided by6
21 875
dks 17 ls foHkkftr djus ij izkIr 'ks"kiQy crkb,\ the remainder will be?
(a) 12 (b) 13 101+102+103+.........+1099+10100 dks 6 ls Hkkx nsus ij izkIr
(c) 14 (d) 15 'ks"kiQy crkb,\
(a) 0 (b) 4
3. What will be the remainder when
1750×1748×1753×70×35 divided by 17. (c) 5 (d) None of these

1750×1748×1753×70×35 dks 17 ls Hkkx nsus izkIr 'ks"kiQy

9. Find the remainder when 555555......244 times is
divided by 37.
gSA
(a) 12 (b) 15 ;fn 555555......244 ckj] dks 37 ls Hkkx nsus ij izkIr 'ks"kiQ
(c) 13 (d) 14
crkb,\
(a) 1 (b) 3
4. What will be the remainder
(c) 5 (d) 0
(1750+1748+1753+70+35) is divided by 17?
10. Find the remainder when 888888......184 times is
(1750+1748+1753+70+35) dks 17 ls Hkkx nsus ij izkIr divided by 37.
'ks"kiQy crkb,\
;fn 888888......184 ckj] dks 37 ls Hkkx nsus ij izkIr 'ks"kiQ
(a) 10 (b) 0
crkb,\
(c) 5 (d) 6
(a) 1 (b) 8
68
5. What will be the remainder when 2 is divided by (c) 36 (d) 7
65?
11. The remainder when 260 is divided by 5.
268
dks 65 ls Hkkx nsus ij izkIr 'ks"kiQy D;k gksxk\
260 dks 5 ls foHkkftr djus ij 'ks"kiQy gksxk&
(a) 60 (b) 61
(a) 0 (b) 1
(c) 38 (d) 65
(c) 2 (d) None of these
99
6. When [51+(67) ] is divided by 68. Find the 12. The remainder when 784 is divided by 342
remainder?
784 dks 342 ls Hkkx nsus ij 'ks"kiQy D;k gksxk\
[51+(67)99] dks 68 ls Hkkx nsus ij izkIr 'ks"kiQy crkb,\
(a) 0 (b) 1
(a) 82 (b) 51
(c) 49 (d) 341
(c) 50 (d) None of these

1
13. What is the remainder obtain when 4128 is divided (c) 69 (d) 35
by 15
18. Consider a large number N = 123456789 ....
4128 dks 15 ls Hkkx nsus ij 'ks"kiQy D;k gksxk& 9899100. What is the remainder when first 100
(a) 14 (b) 2 digits of N is divided by 9

(c) 1 (d) 8 ,d cM+h la[;kN = 123456789 .... 9899100 ij fopkj

dhft, tc N ds izFke 100 vadksa dks 9 ls foHkkftr fd;k tkrk g
14. What is the remainder when 2250 is divided by 7.
rks 'ks"kiQy D;k gksxk\
2250 dks 7 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\ (a) 0 (b) 8
(a) 1 (b) 3 (c) 1 (d) 5
(c) 2 (d) None of these
19. What is the remainder when we divide 390+590 by
15. Let N = 421×1423×1425. What is the remainder 34
when N is divided by 12?
390+590 dks 34 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\
N = 421×1423×1425 dks 12 ls foHkkftr djus ij 'ks"kiQy D;k (a) 0 (b) 17
gksxk\
(c) 33 (d) 1
(a) 0 (b) 9
20. What is the remainder when (13 100 +17 100 ) is
(c) 3 (d) 6
divided by 25.
16. The remainder, when (1523+2323) is devided by 19
(13100+17100 dks 25 ls foHkkftr djus ij 'ks"kiQy D;k gksxk\
is :
(a) 0 (b) 2
tc (1523+2323) dks 19 ls foHkkftr fd;k tkrk gS rks 'ks"kiQy D;k
(c) 4 (d) 11
gksxk\
(a) 1 (b) 0 21. How many positive integers are there from to 1000
that leave a remainder of 3 on division by 7 and a
(c) 2 (d) 3
remainder of 2 on division by 4?
17. If x = (163+173+183+193), then x divided by 70 leave 0 vkSj 1000 ds chp ,slh fdruh /ukRed la[;k,a gksxh ftudks 7 rFk
a remainder of :
4 ls Hkkx nsus ij 'ks"kiQy Øe'k% 3 vkSj 2 izkIr gksrk gS&
;fn x = (163+173+183+193) rksx dks 70 ls foHkkftr djus ij (a) 32 (b) 36
'ks"kiQy D;k gksxk\ (c) 24 (d) 19
(a) 0 (b) 1

2
 SOLUTIONS

1  3  2  2  1
 12
17
1. (a)
 Hence Remainder = 12.
4. (b)

-1 -3 +1 +2 +1
(13)12
=     
17
1750 + 1748 + 1752 + 70 + 35
3 3 17
312 (34 )3  81 13
Actual Remainder =   
17 17 17 17 1  3  1  2  1 0
= =0
17 17
3
 4  64
=  Hence Remainder = 0.
17 17

Remainder = +4
268 268×(26)11
5. (b) =
65 65

875 875 875 +4 -1

 21  21  4
2. (b)   4×(64)11
17 17 17 = 65
16 4×(-1)11
=
65
5 5
Rem.
11
4

4  42   =
4  16 
17 17 17
1
=   4
4 65
=  = -4,
17
Remainder = 65-4 = 61
Actual Rem. = 17-4= 13
[51  (67)99 ] 17  (1)99
3. (a) 6. (c) 
68 68
-1 -3 +2 +2 +1

     17  1 18
  18
68 68
1750 × 1748 × 1753 × 70 × 35

17 Remainder = 68-18 = 50

3
 123456789 5555....244 times
7. (a) [since last 3 digit divisible by8] 9. (c)
8 37

5555.... 240 times is divisible by 37 became

8 789 98 any digit is made by repeating 6 times. The
72 number will be divisible by 37.
69
69 5555....240 times, 5555
5 Remainder 37

So, the remainder is 5.

5555
Remainder = 5
37
101  102  103  ....1099  10100
8. (b)
6
37 5555 15
10 37
R=4 185
6 185
05 Remainder
101  102 4  4
 2
6 6 7777....363 times
10. (b)
R = 2.
11

+4 +4 +4
7777....360 times 777
=
   11
101 + 102 + 103
777
6 = 7
11
Remainder = 0.
Remainder = 7
The remainder will be zero (0) after each three
number so the remainder is 0 upto the 99th 30 30
2 60  2 
2

term. So the remainder 10100 term will be

  1
11. (b)   =1
divided by 6 to get the remainder. 5 5 5

28

4 12. (b)
7  3


(343) 28

1
1
 342 342 342

10100 = 4 64
6 13. (b)
4  2


(16) 64 164
 1
15 15 15
Hence, the remainder = 4.
83 83
2  23  2  8
14. (c)  2
7 7

4
 1421 1423  1425 5  7  9 13100  17100
15. (c)  3 20. (b)
12 12 25

1523  2323 (4) 23  (4) 23 13100

16. (b)  0  (cyclicity of 25 is 20)
19 19 25

17. (a) (163+193)+ (173+183) 130

R 1
= (16+19) (162-16×19+193)+(17+18) (172-17×18+182) 25
= 35× odd +35× odd
= 35× even 17100 170
= R 1
= 70 k 25 25

70k 11
Now, R 0 Now,  2  Remainder .
70 25

18. (d) 1 to 50, there are 91 digits 21. (b) N = 7a+3 = 4b+2
We need 9 gidits more. 
51, 52, 53, 54, 5  total 9 digits. a=1 b=2
So least number = 10
54
So, The sum of 1 to 54 = [1  54]  5 Now in the form of 28k+10 gives remainder 3
2 and 2.

27  55  5
28×35+10  990 (k= 1, 2, 3, 4....35)
=  5 Rem. Such type total number = 1+35 = (36)
9


45 45
(10)

19. (a)
90
3 5

3
90 2
  5 2

34 34
= an+bn If 'n' is odd number, It is always divisible
by (a+b)

945 +2545 34
So, = R 0
34 34