HCF (highest Common Factor)
Category : 6th Class
HCF (highest Common Factor)
Highest Common Factor is also called as Greatest Common Measure (GCM) or Greatest Common Divisor (GCD). H.C.F of two or more numbers is the greatest number which exactly divides each of the number. Therefore, HCF of two or more numbers is its highest common factors. Let us consider the HCF of 4, 8, 16 and 32. Factors of\[4=2\times 2,\text{ }8=2\times 2\times 2,16=2\times 2\times 2\times 2,32=2\times 2\times 2\times 2\times 2\] Therefore, \[2\times 2=4\]is the highest common factor which can exactly divide the numbers, 4, 8, 16, and 32.so, 4 is the HCF of 4, 8, 16 and 32. H.C.F is calculated by prime factorization and continued division methods.
HCF by Prime Factorization Method
The HCF of two or more numbers is obtained by the following steps:
Step 1: Find the prime factors of each one of the given number.
Step 2 : Find the common prime factors from prime factors of all the given numbers.
Step 4 : The product of the common prime factors is the HCF of given numbers. Let us consider the Greatest Common Measure of 36, 90, 72.
Step 1: The prime factors of \[36=2\times 2\times 3\times 3,90=2\times 5\times 3\times 3\text{ }\]and\[72=2\times 2\times 2\times 3\times 3\]
Step 2 : The common prime factors from all the prime factors \[=2\times 3\times 3\]
Step 3 : Therefore, HCF of 36, 90 and \[72=2\times 3\times 3=18\]
Find the HCF of 101, 573, and 1079 by prime factorization method?
(a) 4
(b) 1
(c) 2
(d) All of these
(e) None of these
Answer: (b)
Explanation
The prime factors of 101 = 1 and 101,573 = 1,3,191 but the common factor is 1. Therefore, the HCF of 101, 573 and 1079 = 1.
HCF by Continued Division Method
The HCF of two or more numbers is obtained by continued division. The greatest number is considered as dividend and smallest number is divisor. Follow the following steps to perform the HCF of the given numbers:
Step 1: Divide the greatest number by smallest.
Step 2: If remainder is zero, then divisor is the HCF of the given number.
Step 3: If remainder is not zero then, divide again by considering divisor as ne dividend and remainder as new divisor till remainder becomes zero.
Step 4 : The HCF of the numbers is last divisor which gives zero remainder.
The HCF of 45, 76 = ?
(a) 1
(b) 2
(c) 3
(d) 0
(e) None of these
Answer: (a)
| 45)76(1 45 31)45(1 14)31(2 28 3)14(4 12 2)3(1 2 1)2(2 2 0 |
(45 is smaller than 76, therefore, 75 is divided by 45)
(31 is the remainder therefore, considered as the new divisor)
(45 is the first divisor therefore, it is consederd as new dividend)
(Divisor 1, gives zero remainder therefore, required HCF is 1)
Find the HCF of 78 and 1786 by continued division method?
(a) 2
(b) 5
(c) 1
(d) All of these
(e) None of these
Answer: (a)
Explanation
| 78)1786(2 156 22)78(3 66 12)22(1 12 10)12(1 10 2)10(5 10 0 | HCF of 78 and 1786 = 2 |
HCF of more than Two Numbers
The HCF of more than two numbers is the HCF of resulting HCF of two numbers with third one number. Therefore, HCF of more than two numbers is obtained by finding the HCF of two numbers with third, fourth and fifth numbers.
Find the HCF of 56, 98 and 123?
(a) 2
(b) 1
(c) 3
(d) All of these
(e) None of these
Answer: (b)
Explanation
Step 1: The HCF of 98 and 56
| 56)98(1 56 |
| 42)56(1 |
| 42 |
| 14)42(3 |
| 42 |
| 0 |
The HCF of 98 and 56=14, therefore, the required HCF of 56, 98 and 123 the HCF of 14 and 123
HCF of 56, 98 and 123=1
HCF of Larger Numbers
The HCF of smaller number (one or two digit numbers ) is simply obtained by division but division of larger numbers take more time, therefore, the shortest method for finding the HCF larger numbers is performed by the following method.
Step: 1 Divide, all the given numbers by the common divisor which divided which divide all numbers exactly till last.
Step: 2 Divide the numbers which are obtained in step 1 by another divisor if divisible.
Step: 3 The required HCF is the product of common divisors.
Find the HCF of 3264 and 57384?
(a) 23
(b) 52
(c) 24
(b) all of these
(e) None of these
Answer: (c)
Explanation
| 2 | 3264,57384 |
| 2 | 1632,28692 |
| 2 | 816,14346 |
| 3 | 408,7173 |
| 136,2991 |
The product of common divisors\[~=2\times 2\times 2\times 3=24,\] hence the required
HCF is 24.
Properties of HCF and LCM
1. The product of HCF and LCM of two numbers a and b is always equal to their product, therefore,
(a) LCM of numbers, a and b \[\times \] HCF of a and \[b=\text{ }a\times b\]
(b) LCM \[\text{LCM=}\frac{\text{a }\!\!\times\!\!\text{ b}}{\text{HCF}}\]
(c) HCF of numbers a and \[\text{b=}\frac{\text{a }\!\!\times\!\!\text{ b}}{\text{LCM}}\]
2. HCF of two or more numbers is not greater than any of the numbers.
3. LCM of two or more numbers is not less than any of the numbers.
4. HCF of two co - prime numbers is 1.
5. LCM of co - primes is equal to their product.
6. The HCF of two or more numbers is always a factor of their LCM.
7. If x and y are whole numbers and q is the quotient of their division and r is the remainder then,
(a) The least subtracted number from x so that y divides the difference exactly = r
(b) The least added number to x so that y divides the sum exactly = y - r.
Applications on HCF
Two containers contain 500 litres and 1000 litres of water. The maximum capacity of pot which can measure the water of either tanker in exact numb of times is?
(a) 250 litres
(b) 500 litres
(c) 1000 litres
(d) All of these
(e) None of these
Answer: (b)
Explanation
The maximum capacity of pot can be used to measure the water in exact number of times = The HCF of 500 and 1000 = 500.
The measurement of a rectangular shaped iron piece is 3 m, 15 cm long, 2 m, cm wide and 1 m, 5 cm height. Find the longest tape which can measure the dimensions of the iron piece in exact number of times?
(a) 6cm
(b) 10cm
(c) 5cm
(d) All of these
(e) None of these
Answer: (c)
Explanation
The longest tape that can measure dimension of the iron piece in exact number of times = The HCF of the given dimension = 5.
The HCF of two numbers is 28 and their LCM is 336. If one of them is 112. Find the other number?
(a) 78
(b) 68
(c) 84
(d) All of these
(e) None of these
Answer: (c)
Explanation
Product of two numbers = Product of their LCM and HCF. Therefore,\[~112\times \] unknown number\[~=\text{ }336\times 28,\]therefore, unknown number \[\frac{~=\text{ }336\times 28}{112}=84.\]
The HCF of two numbers is 24, if their product is 4032 then their LCM is?
(a) 166
(b) 168
(c) 162
(d) All of these
(e) None of these
Answer: (b)
Explanation
The LCM of the numbers \[=\frac{\text{Product of numbers }\!\!~\!\!\text{ }}{\text{HCF of the numbers}}\text{=}\frac{\text{4032}}{\text{24}}\text{=168}\]
The greatest number which exactly divides the following numbers, 38, 95 and 171 is?
(a) 29
(b) 14
(c) 19
(d) All of these
(e) None of these
Answer: (c)
Explanation
The greatest number = HCF of the numbers = 19.
The HCF of an even and an odd number is?
(a) 0
(b) 2
(c) 1
(d) All of these
(e) None of these
Answer: (c)
Explanation
There is no common factor other than 1 for the HCF of an even and an odd number.
Find the least number which is exactly divisible by 4, 6 and 8?
(a) 24
(b) 94
(c) 20
(d) All of these
(e) None of these
Answer: (a)
Explanation
The least number which is exactly divisible by 4, 6 and 8 is their LCM = 24
Find the smallest number of five digits which is exactly divisible by 60, 90 and 80?
(a) 10080 .
(b) 10090
(c) 10100
(d) All of these
(e) None of these
Answer: (a)
Explanation
The LCM of 60, 80 and 90 is 720. The smallest five digit number which is exactly divisible by 60, 80 and 90 must be divisible their LCM. Division of10000 by 720 leaves 640 as remainder, therefore, smallest five digit number which is exactly divisible by 60, 80 and 90 = 10000 + (720 - 640) = 10080.
Find the smallest number which when divided by 6, 8,12,15 and 20 leaves the remainder 5?
(a) 1075
(b) 1085
(c) 1065
(d) All of these
(e) None of these
Answer: (b)
Explanation
The LCM of 6, 8, 12, 15 and 20 is 120. The smallest number is more than three digits, therefore, the required number = 1000 + (120 - 40) + 5 = 1085.
Four traffic light bulbs blink after every 5, 10, 15 and 20 seconds. If they start blinking simultaneously at 5 a.m. then after what period of time will they blink again simultaneously?
(a) 80 seconds
(b) 60 seconds
(c) 50 seconds
(d) All of these
(e) None of these
Answer: (b)
Explanation
All four traffic lights will blink simultaneously after 5 a.m is the LCM of the given period of time \[\begin{align} & \,\,\,\,\,\left. {\overline {\, \text{2} \,}}\! \right| \underline{\text{5,10,15,20}} \\ & =\left. {\overline {\, \text{5} \,}}\! \right| \underline{\text{5,5,15,10}}\,\,\,\,\text{=}\,\text{2 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 3 }\!\!\times\!\!\text{ 5 }\!\!\times\!\!\text{ = 60 seconds}\text{.} \\ & \,\,\,\,\overline{\text{1,1,3,2}\,\,\,\,\,\,\,\,} \\ \end{align}\]
In a factory three vessels produce sound simultaneously at 10. a.m. if the vessels produce sound after every 60, 120 and 180 minutes then at what period of time will they again produce sound simultaneously?
(a) 7 hours
(b) 8 hours
(c) 6 hours
(d) All of these
(e) None of these
Answer: (c)
Explanation
The vessels will produce sound again simultaneously = LCM of the given time period = 360 minutes = 6 hours.
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