Find the greatest number that can be exactly divisible by 147, 161 ,210
Answers
Answer:
Step-by-step explanation:
H.C.F of 210,315,147,161 will be the largest number which divides these numbers.
Prime factorization of:
210=2×3×5×7
315=3^2×5×7
147=3×7^2
161=23×7
product of common factors of least power=7
The greatest number that can be exactly divisible by 147, 161 ,210 is 7
Given:
147, 161, 210
To find:
The greatest number that can be exactly divisible by 147, 161 ,210
Solution:
To find the greatest number that can be exactly divisible by 147, 161, 210
or HCF (147, 161, 210) write given numbers as product of prime numbers
147 = 3 × 7 × 7
161 = 7 × 23
210 = 2 × 3 ×5 × 7
In above factors HCF (147, 161 ,210) = factor which is common in all factors
⇒ HCF (147, 161 ,210) = 7
The greatest number that can be exactly divisible by 147, 161 ,210 is 7
#SPJ2