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28 Aug 2016, 22:50
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NaeemHasan
First, understand - why do we add 1 to find the number of factors. We say that given N = 2^p * 3^q * 5^r ...
Total number of factors of N = (p + 1) * (q + 1) * (r + 1) * ...
Why the +1s? Because say p is 4, then you can have 2^1 / 2^2 / 2^3 / 2^4 in the factor but what about 2^0 (i.e. no 2 in the factor)? So in all you have 5 ways to take 2s - you can take no 2, one 2, two 2s, three 2s and all four 2s.
Now, if you want all factors which must have a 2, the first case (no 2s) is not acceptable to you. So you don't add 1 to p.
So in the generic case,
If you want factors that are divisible by 6 (i.e. they have both a 2 and 3), you will not add 1s to their powers.
You will say that number of factors = p * q * (r + 1)...
04 Nov 2016, 22:40
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enigma123
Numbers that divide 264600 are the factors of 264600
So we have to find out the numbers that are factors of 264600 but are not multiples of 6 (2*3)
a Number won't be a multiple of 6 if it last either 2 or 3 or both as it's prime factors
264600 = 2646*100 = 1323*2*2^2*5^2 = 9*147*(2^3)*(5^2) = (3^2)*49*3*(2^3)*(5^2) = (3^3)*(7^2)*(2^3)*(5^2)
Factors of (7^2)*(2^3)*(5^2) without a multiple of 3 = (2+1)*(3+1)*(2+1)=36
Factors of (7^2)*(2^3)*(5^2) without a multiple of 2 = (3+1)*(2+1)*(2+1)=36
Factors of (7^2)*(2^3)*(5^2) without primes 2 and 3 both = (2+1)*(2+1)=9
So total required numbers = 36+36-9 = 63
ANswer: Option D
06 Jun 2018, 15:19
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The prime factorization of 264600 leads to \(2^3*3^3*5^2*7^2\)
Now to find numbers that are NOT multiples of 6 and divide evenly into 264600 - we need to find numbers that do not contain 2 & 3 simultaneously.
That is - Numbers with 2 (But NOT 3) as factors + Numbers with 3 (BUT NOT 2) as factors + Numbers with neither 2 NOR 3 as factors
Thus,
(A) Numbers with 2 (BUT NOT 3) as factors = \(3*(2+1)*(2+1) = 27\)
(B) Similarly, numbers with 3 (BUT NOT 2) as factors - \(3*(2+1)*(2+1) = 27\)
(C) AND Numbers with neither 2 NOR 3 as factors - \((2+1)*(2+1) =9\)
Thus the answer is \((A)+(B)+(C) = 27+27+9=63\)
Now to find numbers that are NOT multiples of 6 and divide evenly into 264600 - we need to find numbers that do not contain 2 & 3 simultaneously.
That is - Numbers with 2 (But NOT 3) as factors + Numbers with 3 (BUT NOT 2) as factors + Numbers with neither 2 NOR 3 as factors
Thus,
(A) Numbers with 2 (BUT NOT 3) as factors = \(3*(2+1)*(2+1) = 27\)
(B) Similarly, numbers with 3 (BUT NOT 2) as factors - \(3*(2+1)*(2+1) = 27\)
(C) AND Numbers with neither 2 NOR 3 as factors - \((2+1)*(2+1) =9\)
Thus the answer is \((A)+(B)+(C) = 27+27+9=63\)
