If you are preparing for an exam and struggling with divisor problems, then you have come to the right place. Divisor problems involve finding the factors of a given number or determining if a number is divisible by another. These types of problems can be time-consuming and tricky, but with the right strategies, you can quickly solve them. In this resource, we have compiled a list of quick solution strategies that will help you tackle divisor problems with ease.
Multiple Choice Questions
Que 1: Find the sum of the divisors of 544 which are Perfect Squares.
(a) 32
(b) 64
(c) 21
(d) 42
Solution: 544= 25×17
Sum of Even divisors= (20+22+24)×170 =21
Option (c) is correct (Divisor Problems)
Que 2: Find the remainder when the product of even divisors of 1800 is divided by 11.
(a) 9
(b) 6
(c) 3
(d) 1
Solution: 1800= 23×32×52
=2×[22×32×52]
Total Number of factors of even divisors= (2+1)×(2+1)×(2+1)=27
So, the Product of even divisors= 2^{27}\times (2^{2}\times 3^{2}\times 5^{2})^{13}\times \sqrt{2^{2}\times 3^{2}\times 5^{2}}.
=(2^{2}\times 3\times 5)^{27}.
=60^{27}.
Here ‘mod’ represents the remainder. Use Remainder concepts to find it.
Now, 6027 mod 11=57 mod 11
=1252×5 mod 11
=42×5 mod 11
=80 mod 11
=3
Option (c) is correct (Divisor Problems)
Que 3: Which of the following has the most number of divisors?
(a) 99
(b) 101
(c) 176
(d) 182
Solution: 99= 32×111
101=1011
176=24×111
182=2×7×13
Number of divisors of 176= (4+1)×(1+1)=10 …….The highest
Option (c) is correct (Divisor Problems)
Que 4: Find the number of factors of 7200 that are not divisible by 24.
(a) 27
(b) 36
(c) 40
(d) 18
Solution: 7200=25×32×52
Total Factors= (5+1)×(2+1)×(2+1)=54
7200=24×(22×31×52)
Total factors which are divisible by 24= (2+1)×(1+1)×(2+1)= 18
Required number of factors= 54-18=36
Option (b) is correct
Que 5: Find the sum of the even divisors of 96 and odd divisors of 3600.
(a) 639
(b) 735
(c) 651
(d) 589
Solution: 96= 2×(24×31)
The sum of all even divisors of 96= 2×(20+21+22+23+24)×(30+31) =248
3600=24×32×52
The sum of all odd divisors of 3600= (30+31+32)×(50+51+52)= 403
Total=248+403= 651 (Divisor Problems)
Option (c) is correct
Que 6: Find the sum of the factors of 2450 which are divisible by 245.
(a) 3870
(b) 4410
(c) 5480
(d) 3960
Solution: 2450=245×(21×51)
Required sum= 245×(20+21)×(50+51)= 4410
Option (b) is correct
Que 7: What is the sixth divisor of a natural number that has exactly 6 divisors, where five of the divisors have a product of 648? (Divisor Problems)
(a) 9
(b) 6
(c) 8
(d) 1
Solution: Number of divisors= 6 (Even)
The product of 1st and 6th divisors of N= 1×N=N
The product of 2nd and 5th divisors= N
The product of 3rd and 4th divisors= N
Product of all disors= N3
648= 23×34
Here 32=9 is missing because the product of all divisors is a perfect cube.
So, the sixth divisor= 9
N= \sqrt[3]{648\times 9}=18.
Option (a) is correct
Que 8: Find the sum of the digits of the product of the common divisors of 26280 and 27300.
(a) 27
(b) 63
(c) 18
(d) 51
Solution: 26280=23×32×5×73
27300=22×3×52×7×13
Common part⇒ 22×3×5
All possible divisors⇒ (20, 21, 22)×(30, 31)×(50, 51)
⇒ (1, 2, 4)×(1, 3)×(1, 5)
⇒ (1×1, 2×1, 4×1, 1×3, 2×3, 4×3) ×(1, 5)
⇒(1, 2, 4, 3, 6, 12) ×(1, 5)
⇒(1×1, 2×1, 4×1, 3×1, 6×1, 12×1, 1×5, 2×5, 4×5, 3×5, 6×5, 12×5)
⇒ (1, 2, 4, 3, 6, 12, 5, 10, 20, 15, 30, 60)
In ascending order⇒ (1,2,3,4,5,6,10,12,15,20,30,60)
Common part⇒ 22×3×5
Total number of common divisors (n)= (2+1)×(1+1)×(1+1)=12
The sum of divisors= (20+21+22)×(30+31)×(50+51)= 7×4×6=168
Product of divisors⇒ (2^2\times 3\times 5)^{n/2}=60^6=46656000000.
Here 66=2162 so first find the square then add the digits.
The sum= 4+6+6+5+6=27
Option (a) is correct.
Que 9: Find the remainder when the sum of all the divisors of 6480 which are divisible by 18 is divided by 11.
(a) 3
(b) 4
(c) 5
(d) 6
Solution: 6480=24×34×5 (Divisor Problems)
6480=18×(23×32×5)
The sum of divisors that are divisible by 18= 18×(20+21+22+23)×(30+31+32)×(50+51)
=18×15×13×6
Remainder Calculation:
⇒(18×15×13×6) mod 11
⇒(7×4×2×6) mod 11
⇒(28×12) mod 11
⇒(6×1) mod 11
⇒6
Option (d) is correct.
Que 10: What is the sum of the numbers if the sum of their factors is 124?
(a) 100
(b) 123
(c) 89
(d) 251
Solution: 124=22×31
Let the number be N=pa×bb
Sum of the factors= (1+p+p2+p3+…+pa)×(1+q+q2+q3+….+qb)
124⇒ (22×31), (2×62), (1×124)
Only the case of (22×31) is possible
22×31=(1+3)×(1+2+22+23+24)⇒ N=31×24=48
22×31=(1+3)×(1+5+52)⇒ N=31×52=75
The Sum= 48+75= 123
Option (b) is correct.
Que 11: Which of the numbers A, B, C, and D, having 16, 28, 30, and 27 factors respectively, could potentially be a perfect cube?
(a) A and B
(b) B and C
(c) A, B, and C
(d) B and D
Solution: N=p3xq3yr3z
The number of factors= (3x+1)×(3y+1)×(3z+1)
16=24=4×4=(3+1)×(3+1)
28=22×7=(3+1)×(6+1)
30=2×3×5 or 6×5 or 3×10⇒ Not possible
27=33 or 3×9 ⇒ Not possible
Option (a) is correct.
Que 12: If a three-digit number ‘abc’ has 3 factors, how many factors does the six-digit number ‘abcabc’ have?
(a) 16
(b) 24
(c) 16 or 24
(d) 20 or 28
Solution: abc=p2
abcabc=abc×1001=p2×7×11×13
The number of factors=3×2×2×2=24 ……..Case(i)
If p=7 or 11 or 13
abcabc=73×11×13
The number of factors=4×2×2=16 ………Case (ii)
Option (c) is correct.
Que 13: How many numbers less than 100 cannot be expressed as multiples of a perfect square greater than 1?
(a) 61
(b) 56
(c) 52
(d) 65
Solution: Multiples of 4 {4,8,12,……, 96}⇒ 24 numbers
Multiples of 9 {9,18,27,……, 99}⇒ 11-{36,72}=9 numbers
Multiples of 16 {has already been counted as a multiple of 4}⇒ 0
Multiples of 25 {25,50,75}⇒ 3 numbers
Multiples of 49 {49,98}⇒ 2 numbers
Total multiples of perfect squares= 24+9+3+2=38
The remaining numbers= 99-38=61
Option (a) is correct.
Que 14: If a number N2 has 15 factors, how many factors can N have?
(a) 5 or 7
(b) 6 or 8
(c) 4 or 6
(d) 9 or 8
Solution: 15=3×5 or 15
3×5=(2+1)×(4+1)⇒p2q4
15=(14+1)⇒p14
So, N={pq2, p7}
Number of factors of N= 6 or 8
Option (b) is correct.
Que 15: What is the smallest number that has exactly 18 factors?
(a) 180
(b) 216
(c) 240
(d) None
Solution: 18=2×32
18={(1×18), (2×9), (3×6), (2×3×3)}
(1×18)⇒p17
(2×9)⇒p1q8
(3×6)⇒p2q5
(2×3×3)⇒p1q2r2
For smallest value, put p=5,q=2, and r=3
p1q2r2=51×22×32=180
Option (a) is correct.
LCM and HCF Concepts: Click Here