HCF of 4 and 7=1 Step 3: Find the HCF of the DenominatorsNext, we find the HCF of the denominators 5 and 15.- The factors of 5 are: 1, 5- The factors of 15 are: 1, 3, 5, 15 The common factor is 5. Therefore,HCF of 5 and 15=5 Step 4: Calculate the HCF of the ...
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?
HCF of 10 and 15 is the largest possible number which divides 10 and 15 without leaving any remainder. The methods to compute the HCF of 10, 15 are explained here.
HCF of 15 and 18 is the largest possible number which divides 15 and 18 without leaving any remainder. The methods to compute the HCF of 15, 18 are explained here.
find out hcf of 5 and 7 answer: hence, hcf of 5 and 7 = 1 example 5: 2 given numbers are in the ratio 5:11. one of the hcf is number 7, what are the numbers? answer: assume the numbers to be 5m & 11m. here “m” is the highest common factor (hcf). the numbers ...
Learn Properties of HCF and LCM and the relation between LCM and HCF of natural numbers with examples. Formula to find HCF and LCM of fractions at BYJU'S.
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For example, 15 is the LCM of 3 and 5. To solve LCM of numbers there are two methods. They are 1. Factorization method 2. Division method 1. Factorization Method In this method, just like HCF, you have to write the numbers in the standard form. Then the product of prime numbers ...
Prime Factors of a Number Divisibility Tests HCF by Prime Factorization Before finding HCF by prime factorization we need to know the concept of the same. Let’s take a number say, 45. Now the factors of 45 are 1,3,5,9,15 and 45 itself. Now, apart from 3 and 5 the other numbers...
15 =3× 5 18 = 233 This means that3is the Highest Common Factor of 15 and 18 orHCF (15, 18) = 3. Example 2: Using a Table Let us use 15 and 18 again to see the difference of this method from theFactor Tree method. For the Table Method, we need to make a table with two...