tatz wrote:
PKN wrote:
Bunuel wrote:
How many factors greater than 1 do 120, 210, and 270 have in common?
(A) One
(B) Three
(C) Six
(D) Seven
(E) Thirty
We have LCM(120,210,270)=30
Writing 30 in it's prime factorization form :- \(30=2^1*3^1*5^1\)
No of different factors of 30:- (1+1)*(1+1)*(1+1)=8
These 8 factors is including 1, but we are told to determine the factors greater than 1.
So the required number of factors: 7
Ans. (D)
I think you wanted to tell HCF / GCD of 120, 210, 270 is 30. Rather than LCM.
Also, can you please clarify how did you think of taking HCF and then finding its total number of factors rather than prime factorising all numbers and finding the common factors.
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I think you wanted to tell HCF / GCD of 120, 210, 270 is 30. Rather than LCM.
You are correct, calculated GCD.(NOT LCM)
Quote:
Also, can you please clarify how did you think of taking HCF and then finding its total number of factors rather than prime factorising all numbers and finding the common factors.
Thought process:-
1. We are told to determine the maximum no of common factors of three numbers. (All factors must be greater than 1)
2. If a number is divisible by its highest factor, then that number is also divisible by all of the factors of the highest factor. (
3. What is the magnitude of highest factor of 'n' nos of numbers? GCD is the answer. Then calculate GCD. This is the highest number which divides all the numbers under discussion.
4.Now apply 2 above. Calculate the no of factors of the GCD. The no of factors of a number is calculated using prime factorization method as explained.
5. Since the no of factors is inclusive of 1, so the required no of factors will be 1 less the total no of factors.
Thanking you.
_________________
Regards,
PKN
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