HCF of 16 and 27 is the largest possible number which divides 16 and 27 without leaving any remainder. The methods to compute the HCF of 16, 27 are explained here.
To find the Highest Common Factor (HCF) of 16 and 32, we can follow these steps:Step 1: Factor the numbers First, we need to factor both numbers into their prime factors.- For 16: - 16 can be divided
Use Euclid's algorithm to find the HCF of 196 and 38220 View Solution Exams IIT JEE NEET UP Board Bihar Board CBSE Free Textbook Solutions KC Sinha Solutions for Maths Cengage Solutions for Maths DC Pandey Solutions for Physics HC Verma Solutions for Physics ...
HCF of 255 and 867 is the largest possible number which divides 255 and 867 without leaving any remainder. The methods to compute the HCF of 255, 867 are explained here.
百度试题 结果1 题目1. (a) Find the HCF of 42, 66 and 78.(b)Find the LCM of 9, 16 and 18. 相关知识点: 试题来源: 解析 1. (a) 6(b) 144 反馈 收藏
Highest Common Factor (HCF):The largest or greatest factor common to any two or more given natural numbers is termed asHCF of given numbers. Also known as GCD (Greatest Common Divisor). For example, HCF of 4, 6 and 8 is 2. 4 = 2 × 2 ...
HCF of co-prime numbers 4 and 15 was found as follows by factorisation:4 = 2 × 2 and 15 =3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0.Is the answer correct? If not, what is the correct HCF?
The HCF of 56 and 57 is 1. Learn to calculate the Highest Common Factor of 56 and 57 using prime factorisation, long division method and listing common factors with simple steps, at BYJU’S.
18 has two prime factors: 21 and 32 (since 18 = 2 * 3 * 3).28 have two prime factors: 22 and 71 (since 28 = 2 * 2 * 7). Now, from both numbers, take the least power of each prime factor.The smallest power of 2 is 21.The smallest power of 3 is 31.There is no ...
Find the HCF and LCM of the following pairs of numbers.36 and 4566 and 132 12, 18 and 20 相关知识点: 试题来源: 解析 \left( 36,45 \right)=9\left( 66,132 \right)=66\left( 12,18,20 \right)=2\left[ 36,45 \right]=180\left[ 66,132 \right]=132\left[ 12,18,20 \right]=...