The GCD (Greatest Common Divisor), also known as the HCF (Highest Common Factor), of two or more numbers is the largest number that divides all of them exactly, without leaving a remainder.
Key Points:
- It is the greatest number that is a factor of each of the numbers.
- For two numbers a and b, the GCD is the largest number d such that both a ÷ d and b ÷ d are integers.
Read More: Interesting Facts about GCD
Solved Questions on Greatest Common Divisor (Hard)
Question 1: The GCD of two numbers is 30 and their LCM is 600. If one of the numbers is 150, what is the other number?
Solution:
Formula Used: GCD × LCM = Number 1×Number 2
Substituting the known values: 30 × 600 = 150 × A
Calculate 30 × 600: 18000 = 150 × A
Solving for A: A = 18000/150 = 120
Answer: The other number is 120.
Question 2: A gardener has two lengths of rope: one measuring 90 meters and the other 150 meters. What is the largest length of rope that can be cut from both lengths without any leftovers?
Solution:
To find the largest length of rope that can be cut from both lengths of rope (90 meters and 150 meters) without any leftover, we need to determine the greatest common divisor (GCD) of the two lengths.
90 = 2 x 3 x 3 x 5
150 = 2 x 3 x 5 x 5
HCF (90, 150) = 2 x 3 x 5 = 30
The HCF of 90 and 150 is 30, so the gardener can cut the rope into lengths of 30 meters each, and no rope will be left over.
Question 3: The lengths of three different pieces of fabric are in the ratio of 3:5:7. If the greatest common divisor of these lengths is 4 meters, what are the actual lengths of the pieces of fabric? How does this GCD help in determining the minimum cutting length for uniform pieces?
Solution:
The lengths of the three pieces of fabric in the ratio 3:5:7, with a GCD of 4 meters, can be calculated as follows:
- First piece: 3 × 4 = 12
- Second piece: 5 × 4 = 20
- Third piece: 7 × 4 = 28
Actual lengths:
- 12 meters, 20 meters, 28 meters
The GCD of 4 meters determines the minimum cutting length for uniform pieces, allowing for cuts of 4 meters without waste.
Question 4: The GCD of two numbers is 12 and their LCM is 240. And the sum of the numbers is 108, what are the numbers?
Solution:
Formula Used: GCD × LCM = Number 1 × Number 2
Product of the numbers: 2880
Sum of the Numbers: 108
let one of the numbers be x, then the other number will be (108 - x)
now we get,
x( 108 - x) = 2880
108x - x2 = 2880
rearranging the terms
x2 - 108x + 2880 = 0
by factorizing the above equation
x2 - 60x - 48x + 2880 = 0
x( x - 60) - 48( x - 60 ) =0
(x -48) (x - 60) = 0
By solving the above equation we get x = 48, 60
Answer: The numbers are 48, 60.
Question 5: Find the GCD of the polynomials P(x) = 2x3 + 6x2 + 4x and Q(x) = 4x2 + 8x.
Solution:
Factor P(x):
First, we can factor out the common term 2x: P(x) = 2x(x2 + 3x + 2)
Next, we can factor the quadratic x2 + 3x + 2 = (x + 1)(x + 2)
So,
P(x) = 2x(x + 1)(x + 2).
Factor Q(x):
Again, we can factor out the common term 4x:
Q(x) = 4x(x + 2)
Identify the common factors
The common factors are:
- The term 2x from P(x) and 4x from Q(x).
- The term (x+2) appears in both factorizations.
Therefore, the GCD is:
GCD(P(x), Q(x)) = 2x(x + 2).
Question 6: Find the GCD of the polynomials P(x) = x4 − 6x3 + 11x2 − 6x and Q(x) = x2 − 3x.
Solution:
Factor P(x):
First, we can factor out the common term x:P(x) = x(x3 − 6x2 + 11x − 6)
Next, we can factor the cubic polynomial x3 − 6x2 + 11x − 6
Using synthetic division or factoring, we find:x3 − 6x2 + 11x − 6 = (x − 1)(x − 2)(x−3)
So,
P(x) = x(x - 1)(x - 2)(x - 3)
Factor Q(x):
We can factor out the common term x:Q(x) = x(x − 3)
Identify the common factors:
The common factors are:
- The term x from both P(x) and Q(x).
- The term (x−3) appears in both factorizations.
Therefore, the GCD is:
GCD(P(x), Q(x)) = x(x - 3)
Question 7: Prove that the GCD of two consecutive integers is always 1.
Solution:
Proof: The GCD of Two Consecutive Integers is Always 1
Let n be an integer. The two consecutive integers are n and n+1.
- Assume d is a common divisor of n and n+1. This means:
- d/n
- d/(n + 1)
- Since d divides both, it must also divide their difference: (n + 1) - n = 1 Thus, d divides 1.
- The only positive integer that divides 1 is 1 itself.
Conclusion: Therefore, the GCD of n and n+1 is:
GCD(n, n + 1) = 1
Question 8: Find the GCD of 1001 and 143 using the Euclidean algorithm.
Solution:
- Divide 98 by 56:
- 98 ÷ 56 = 1(quotient)
- Remainder = 98−(56 ×1 ) = 42
- Next Step: Use the remainder to find the GCD of 56 and 42:
- GCD(56, 42)
- Divide 56 by 42:
- 56 ÷ 42 = 1
- Remainder = 56 − (42 × 1) = 14
- Next Step: Use the remainder to find the GCD of 42 and 14:
- GCD(42, 14)
- Divide 42 by 14:
- 42 ÷ 14 = 3
- Remainder = 42 − (14 × 3) = 0
Since the last remainder is 0, the GCD is the other number:
GCD(56, 98) = 14
Also Read : Short Tricks to solve HCF
GCD Practice Problems (Hard)
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GCD Practice Questions (Easy Level)
GCD Practice Questions (Medium Level)