It is: 34
1020 and 11594=2×17=34
The Correct Answer is 34The Correct Answer is 34The Correct Answer is 34The Correct Answer is 34
hjds,usct8;
Don't know
Find the HCF of 135 and 165 prime factor method
19 ÷ 6 = 3 r 1 6 ÷ 1 = 6 r 0 → hcf(19, 6) = 1.
Write the numbers next to each other. Now divide all three numbers by any common factor greater than 1 and repeat until no common factor greater than 1 exists. Multiply together the factors by which you divided to get the HCF: ___2__|__1624___552__1276 ___2__|____812___276___638 ______|____406___138___319 No more common factors (other than 1) → hcf(1624, 552, 1276) = 2 × 2 = 4
The HCF of 40 & 12 is... 4
hcf(18, 30) = 6.
It is 34 because 34 is the largest number that will divide evenly into 1020 and 11594 leaving no remainders
To find the highest common factor (HCF) of two numbers using the division method, follow these steps: For 1020 and 11594: Step 1: Divide the larger number by the smaller number. Divide 11594 by 1020: 11594 ÷ 1020 = 11 remainder 414 Step 2: Now, divide the divisor from the previous step (1020) by the remainder obtained (414). 1020 ÷ 414 = 2 remainder 192 Step 3: Repeat the division process with the previous divisor (414) and the new remainder (192). 414 ÷ 192 = 2 remainder 30 Step 4: Continue the process until you obtain a remainder of 0. 192 ÷ 30 = 6 remainder 12 30 ÷ 12 = 2 remainder 6 12 ÷ 6 = 2 remainder 0 Step 5: The last divisor with a remainder of 0 is the HCF of the two numbers. Therefore, the HCF of 1020 and 11594 is 6.
The GCF of 12 and 20 is 4.
For the division of quantities
3
hcf(87, 102) = 3. factorization method: 87 = 3 x 29 102 = 2 x 3 x 17 hcf = 3
Find the HCF of 135 and 165 prime factor method
1. HCF by factorisation Method: HCF is the product of common factors of all the numbers. Example: HCF of 24, 48, 60 First find the prime factors of these numbers 2 |24 2|48 2|60 24= 2*2*2*3 2|12 2|24 2|30 48= 2*2*2*2*3 2|6 2|12 3|15 60= 2*2*3*5 3|3 2|6 5|5 HCF= 2*2*3=12 |1 3|3 |1 |1 2. Division Method: Divide the larger number by the smaller number. Now divide the devisor by the remainder. Continue the process till the remainder is zero. Now this last devisor is the HCF of those two numbers. Repeat the process between this HCF and the other number. 48)60(1 48 12)48(4 48 0 12)60(5 60 0 Ans: 12
Yes.First find the HCF of two of the numbers, then find the HCF of that answer and the third number.In this way you could find the HCF of as many numbers as you want.
882/2 = 441/3 = 147/3 = 49/7 = 7/7 = 13150/2 = 1575/3 = 525/3 = 175/5 = 35/5 = 7/7 = 12 x 3 x 3 x 7 x 7 = 8822 x 3 x 3 x 5 x 5 x 7 = 31502 x 3 x 3 x 7 = 126, the GCF
19 ÷ 6 = 3 r 1 6 ÷ 1 = 6 r 0 → hcf(19, 6) = 1.
As a product of its prime factors: 2*5*7 = 70 Note that at least two or more numbers are needed to find their HCF