In aptitude-based competitive and entrance exams, remainder problems are commonly asked. These problems involve finding the remainder of a division operation and require a strong understanding of basic arithmetic principles. These problems can be challenging, but with practice and a solid grasp of mathematical concepts, they can be tackled successfully. In this discussion, we will explore some important remainder problems and strategies for solving them.
Note: Please read the simplification concepts before starting this.
Simplification Concepts: Click Here
Multiple Choice Questions
Que 1: N=123456…. is a 51-digit number then find the remainder when N is divided by 99?
(a) 0
(b) 1
(c) 98
(d) None
Solution: From 1 to 9 total of 9 digits, then pairs of two digits. Let k numbers of two digits be required to complete 51 digits. (Remainder Problems)
9+2k=51
k=21
So the last number= 9+k=30
N=12345……..282930
Apply the divisibility rule of 99:
01+23+45+67+89+(10+11+12+…….+30)
Here the yellow part is in A.P. (Arithmetic Progression)
⇒[(01+89)+(23+67)+45]+{(10+30)/2}×21
⇒[90+90+45]+420
Now, (90+90+45+420) mod 99=(-9-9+45+24)=51
Option(d)
Que 2: If N=66666…… is a 700-digit number, then find the remainder when N is divided by 137.
(a) 0
(b) 1
(c) 90
(d) 136
Solution: 10001=73×137
AAAA×AAAA=AAAAAAAA
This means that every 8-digit number in which all the digits are the same is divisible by 137.
Now eliminate all the 8-digit pairs from the left side and make a new number of the remaining digits. (Remainder Problems)
700 mod 8=4
New number=6666
6666 mod 137=90
Option(c)
Que 3: Find the remainder when 1! ×2! ×3! ×4! ×5! ×…… ×12! is divided by 13.
(a) 1
(b) 2
(c) 8
(d) 12
Solution: 1! mod 13=1 (Remainder Problems)
2! mod 13=2
3! mod 13=6
4! mod 13=-2
5! mod 13=5×(-2)=-10 or -10+13=3
6! mod 13=6×3 mod 13=5
7! mod 13=7×5 mod 13=-4
8! mod 13=8×(-4) mod 13=-32+39=7
9! mod 13=9×7 mod 13=-2
10! mod 13=10×(-2) mod 13=-20+26=6
11! mod 13=11×6 mod 13=1
12! mod 13=12×1 mod 13=-1
⇒ 1×2×6×(-2)×3×5×(-4)×7×(-2)×6×1×(-1) mod 13
⇒ 2×6×2×3×5×4×7×2×6 mod 13
⇒ (-2)×2×2×(-1) mod 13
⇒ 8
Option:(c)
Que 4: Find the remainder when 111222333444…..999 is divided by 13?
(a) 6
(b) 8
(c) 9
(d) 11
Solution: (111-222+333-444+555-666+777-888+999) mod 13
⇒ (111+111+111+111+111) mod 13
⇒ 555 mod 13
⇒ 9 (Remainder Problems)
Option (c) is correct.
Que 5: N=45678947+76543278, find the remainder when N is divided by 73.
(a) 12
(b) 9
(c) 7
(d) 4
Solution: Here, Except for the sum of the digits in the units place, the sum of all the remaining digits is the same i.e. 11.
The sum of unit digits= 7+8=15
Concept: An eight-digit number in which all the digits are the same is always divisible by 73.
Remainder= 15-11=4 (Remainder Problems)
Option:(d)
Que 6: Find the remainder when N=1121231234….78 is divided by 101?
(a) 31
(b) 44
(c) 69
(d) 100
Solution: 1000001 is divisible by 101
(-112123+123412)+(-345123+456123)+(-456712+345678)
= 11289+111000-111034
=11289-34
=11255 (Remainder Problems)
11255 mod 101=44
Option:(b)
Que 7: Find the remainder when 1×2×3×4×5×…..×30 is divided by 61?
a) 4
(b) 11
(c) 24
(d) 53
Solution: 59! mod 61=1
⇒ 30!×(31×32×33×………×59) mod 61
⇒ 30!×(-30!) mod 61
let the remainder be k (Remainder Problems)
⇒ -k2 mod 61
Now, only option (b) satisfies the condition.
⇒ -112 mod 61
⇒ -121 mod 61
⇒ 1
Option:(b)
Que 8: 4444⁵⁵⁵⁵+XXXX⁴⁴⁴⁴, when divided by 9, leaves the remainder of 5 then finds the sum of all possible values of x?
(a) 12
(b) 9
(c) 6
(d) None
Solution: Totient of 9=32×(1-1/3)=6
E⇒Exponent
B⇒Base
Find E mod 6 and B mod 9
4444⁵⁵⁵⁵+XXXX⁴⁴⁴⁴= 4444⁵⁵⁵⁵+X⁴⁴⁴⁴.1111⁴⁴⁴⁴
⇒(75+X⁴.4⁴) mod 9
⇒[(-2)5+X⁴.(-2)2] mod 9
⇒[4+X⁴.4] mod 9
Now X could be 1,2,3,4,5,6,7,8,9 (Remainder Problems)
But only for 2 and 7, the remainder will be 5.
[4+2⁴.4] mod 9= 5
[4+7⁴.4] mod 9= 5
So, the sum= 2+7=9
Option (b) is correct.
Que 9: N is the smallest multiple of 11 which when divided successively by 6, 7, and 8 leaves remainders 4, 2, and 1 respectively. Find the remainder when N is divided by 13.
(a) 8
(b) 9
(c) 10
(d) 11
Solution: Concept of the Successive Remainder
8+1=9
9×7+2=65
65×6+4=394
N=LCM(6,7,8)×k+394
=(23×3×7)×k+394
=(8×3×7)×k+394
Now, [(8×3×7)×k+394] mod 11=0
(3k+9) mod 11=0
k=8 (Remainder Problems)
So, N=(8×3×7)×8+394
[(8×3×7)×8+394] mod 13= 5+4=9
Option (b) is correct.
Que 10: Find the remainder when 1+12+123+1234+12345+…+10th term is divided by 13.
(a) 0
(b) 5
(c) 8
(d) 12
Solution: (1+12) mod 13=0
123 mod 13= -7
1234 mod 13= 234-1=233 mod 13= -1
12345 mod 13= 345-12= 333 mod 13= 8
123456 mod 13= 456-123= 333 mod 13= 8
1234567 mod 13= 567-234+1= (333+1) mod 13= 9
12345678 mod 13= 678-345+12= (333+12) mod 13= 7
123456789 mod 13= 789-456+123= (333+123) mod 13= 8-7=1
12345678910 mod 13= 910-678+345-12= 565 mod 13= 6
(0-7-1+8+8+9+7+1+6) mod 13
Remainder= 5
Option (b) is correct.
Que 11: Find the remainder when 12345…. a 150-digit number is divided by 33.
(a) 12
(b) 19
(c) 29
(d) 32
Solution: 9+2k=150
k= 70.5
Last digits ⇒9+70=79 and then 8
N=12345…78798
First, use the divisibility rule of 99
[(01+23+45+67+89)+(10+11+12+…+79)]×10+8
⇒ [{(90×2+45)+(89×35)}×10+8] mod 99
⇒ [{(-9×2+45)+(-10×35)}×10+8] mod 99
⇒ [(27-53)×10+8] mod 99
⇒ [-260+8] mod 99
⇒ [-62+8] mod 99
⇒ [-54] mod 99
⇒ -54+99= 45
45 mod 33=12
Option (a) is correct.
Que 12: Find the remainder when 15101 is divided by 324.
(a) 256
(b) 243
(c) 317
(d) 79
Solution: 324=34×22
15101=34×54×1597
15101 mod 324= 54×1597 mod 4
Note: After calculating the new remainder, we must multiply it by 81.
⇒ 54×1597 mod 4
⇒ 14×(-1)97
⇒ -1
⇒ -1+4=3
Now, the final remainder= 3×81=243
Option (b) is correct.
Que 13: Find the sum of the digits of the remainder obtained when 13101+17101 is divided by 225.
(a) 6
(b) 13
(c) 11
(d) 9
Solution: 13101+17101= (15-2)101+(15+2)101
Here we use Binomial expansion: (Remainder Problems)
(15-2)101=101C0.(15)101+101C1.(15)100.(-2)1+…….+101C100.(15)1.(-2)100+101C101.(-2)101
(15+2)101=101C0.(15)101+101C1.(15)100.(2)1+…….+101C100.(15)1.(2)100+101C101.(2)101
Add both the expansions after dividing by 225:
⇒ 2×101C100.(15)1.(2)100
⇒ 2×101×15×2100
⇒ 101×15×2101
(101×15×2101) mod 225= [(101×2101) mod 15]×15
⇒[(101×2101) mod 15]×15
⇒[(-4×2101) mod 15]×15
⇒[(-2103) mod 15]×15
⇒[(-8) mod 15]×15
⇒7×15
⇒105
The sum of the digits= 1+0+5=6
Option (a) is correct.
Que 14: Find the remainder when 19200+23200 is divided by 49.
(a) 1
(b) 29
(c) 37
(d) 48
Solution: Totient of 49= 72×(1-1/7)= 42
200 mod 42=32 (Remainder Problems)
(21-2)32+(21+2)32=2×32C32.(2)32
⇒233 mod 49
⇒[(210)3×8] mod 49
⇒[443×8] mod 49
⇒[-53×8] mod 49
⇒[-1000] mod 49
=29
Option (b) is correct.
Que 15: Find the remainder when 30! is divided by 2035.
(a) 440
(b) 935
(c) 715
(d) 605
Solution: 2035= 5×11×37
30!= 2035×t+R
Here R is the remainder and t is Quotient. (Remainder Problems)
Concept: The part of 30! which is exactly divisible by 2035 is also exactly divisible by 5, 11, and 37.
According to Wilson Theorem: 35! mod 37=1
(35×34×33×32×31×30!) mod 37=1
(-2×-3×-4×-5×-6×k) mod 37=1
(-24×30×k) mod 37=1
(13×-7×k) mod 37=1
(-91×k) mod 37=1
(20×k) mod 37=1
This is true for k=13
30! mod 37=13 ……….(i)
Now divide the R (remainder) by 37
So, R= 37p+13
R mod 11=0
(37p+13) mod 11=0
(4p+2) mod 11=0
p=5, 16,…..
R mod 5= 0
(37p+13) mod 5=0
(2p+3) mod 5=0
p= 6, 11, 16….
Since for p=16, both conditions are satisfied
R= 37×16+13= 605
Option (d) is correct.
Que 16: What is the remainder when 790 is divided by 15?
(a) 7
(b) 4
(c) 1
(d) 11
Solution: Totient of 15 = 3×5×(1-1/3)×(1-1/5)=8
So, 78k mod 15=1
Here k is a positive integer
(788×72) mod 15=(1×49) mod 15=4
Option (b) is correct.
Que 17: Find the remainder when 1542537859431276681 is divided by 27.
(a) 1
(b) 6
(c) 17
(d) 20
Solution: 1+542+537+859+431+276+681= 3327
3327 mod 27= 6
Note: 999 is divisible by 27, so here we can apply the divisibility rule of 999.
Option (b) is correct.
Que 18: 6725+5413+7108-3257+173×35-247×53+205-1234=?
(a) 7924
(b) 7864
(c) 7904
(d) 7834
Solution: First, use the divisibility by 9 rule on options. Options (a) and (d) have the same remainders when divided by 9.
Now, Apply the divisibility rule of 11 on options.
7924 mod 11=(-7+9-2+4)=4
7864 mod 11=(-7+8-6+4)=-1=10
7904 mod 11=(-7+9-0+4)=6
7834 mod 11=(-7+8-3+4)=2
(6725+5413+7108-3257+173×35-247×53+205-1234) mod 11
⇒[4+1+2-1+(-3)×2-5×(-2)+7-2] mod 11
⇒[6-6+10+5] mod 11
⇒4
Option (a) is correct.
Que 19: Find the remainder when 32^{32^{32}} is divided by 9.
(a) 1
(b) 4
(c) 7
(d) 8
Solution: 32 mod 9=5
Totient (t) of 9= 32×(1-1/3)=6
3232 mod 6=4
So, 54 mod 9=4
Option (b) is correct.
Que 20: What is the remainder when a number N, whose sum of digits is 23, and the remainder when divided by 11 is 7, is divided by 33?
(a) 7
(b) 29
(c) 13
(d) 16
Solution: N mod 9= 23 mod 9=5
N=11k+7
11k+7 mod 9=5
2k-2 mod 9=5
k=8
N=11k+7=95
95 mod 33=-4 or 29
Option (b) is correct.