Question

6
23071
24
270 - 6 = 264 must be completely divisible by
the number.
Similarly, 426 – 6= 420 must be completely divisible
by the number.
HCF of 264 and 420 is the required largest number
which divides 270 and 426, leaving 6 as the remainder.
The HCF of 264 and 420 which divides 270 and 426
leaving the remainder 6 is 12.
. The required largest number is 12.
156 264 (1
108 ) 155 (1
-156
43
Properties of HCF
• The HCF of two or more numbers is the greatest of their common factors.
. If one number is a factor of another, then the smaller number is the HCF.
• The HCF of two co-prime numbers is always 1.
Exercise 2.6
1. Find the HCF of the following (the numbers are given in the prime factorisation form)
a. 2 x 2 x 3 x 3 x 17 and 2 x 3 x 17
b. 2 x 2 x 3 x 3 x 11 and 2 x 2 x 7 x 11 x 11
L
ine the fact​

Answer 1

Answer:

a. 2×3×17=6×17=102

b. 2×2×11=4×11=44

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