How many distinct integers are factors of 90?
A) 6
B) 8
C) 9
D) 10
E) 12
OA: E
Solution:
Approach 1: List down the factors. 1, 2, 3, 6, 5, 9, 10, 15, 30, 45, 90 = 12
Approach 2: I just learnt about this method. So I thought I'd share it with you'll.
Prime factorization of 90 = 2^1 * 3^2 * 5^1
Now, we add 1 to each exponent and multiply them.
(1+1) * (2+1) * (1+1) = 12
It works with any number. I tried it. Let's take the example of 50. It's factors are 1, 2, 5, 10, 25, 50 = 6
or 2^1 * 5^2 = (1+1) * (2+1) = 6
I'm quite weak with math, so I thought I'd share it with others like me. Hope it helped
Distinct Factors
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- thephoenix
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I get what u've done rahul...but is this some sort of formula ?
could you please explain the main concept behind this?
P.s.: thanks for sharing!
could you please explain the main concept behind this?
P.s.: thanks for sharing!
rahul.s wrote:How many distinct integers are factors of 90?
A) 6
B) 8
C) 9
D) 10
E) 12
OA: E
Solution:
Approach 1: List down the factors. 1, 2, 3, 6, 5, 9, 10, 15, 30, 45, 90 = 12
Approach 2: I just learnt about this method. So I thought I'd share it with you'll.
Prime factorization of 90 = 2^1 * 3^2 * 5^1
Now, we add 1 to each exponent and multiply them.
(1+1) * (2+1) * (1+1) = 12
It works with any number. I tried it. Let's take the example of 50. It's factors are 1, 2, 5, 10, 25, 50 = 6
or 2^1 * 5^2 = (1+1) * (2+1) = 6
I'm quite weak with math, so I thought I'd share it with others like me. Hope it helped
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- Mom4MBA
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In general, for any number N, which can be written as N = (m^a) x (n^b) x (p^c)...
where m,n, p ... are the prime factors of N and a,b,c .... are positive integers
the number of factors is equal to (a+1)(b+1)(c+1).....
where m,n, p ... are the prime factors of N and a,b,c .... are positive integers
the number of factors is equal to (a+1)(b+1)(c+1).....
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Is there any restriction to keep in mind wiith this formula??
or whenever any question is asked abt the # of factors all we do is apply the formula and get the answer???
for any # ???
be it 10 digits long???
or whenever any question is asked abt the # of factors all we do is apply the formula and get the answer???
for any # ???
be it 10 digits long???
Mom4MBA wrote:In general, for any number N, which can be written as N=m^a x n^b x p^c..
where m,n p ... are the prime factors of N and a,b,c .... are positive integers
the number of factors is equal to (a+1)(b+1)(c+1).....
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Yes, this will always work, provided you write your prime factorization in the normal way -- that is, using powers whenever a prime is repeated. So if you have a number like 36, you would first need to write the prime factorization as (2^2)(3^2), and then, adding one to each power and multiplying, we find that 36 has 3*3 = 9 positive divisors. If you write the prime factorization of 36 in a different way, say as (2)(3)(2)(3), then the method will not work.bhumika.k.shah wrote:Is there any restriction to keep in mind wiith this formula??
or whenever any question is asked abt the # of factors all we do is apply the formula and get the answer???
for any # ???
be it 10 digits long???
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