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31 Jul 2025, 07:06
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soniasw16
I’m not really using any special formula here, just logic.
Since 72 = 2^3 * 3^2, any factor divisible by 2 must include at least one 2. That means the exponent of 2 can be 1, 2, or 3 (3 choices), and the exponent of 3 can be 0, 1, or 2 (3 choices). So I get 3*3 = 9 even factors.
Same result as total factors (12) minus the 3 odd ones = 9.
10 Jan 2026, 23:47
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We can simply remember this method for problems where we are asked to find the number of factors of n that are divisible by x. Instead of using the usual method for finding the number of factors of n, we apply the same method to n/x. The number of factors of n/x directly gives the required answer.
07 Feb 2026, 02:14
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If we do prime factorization we have that it is
2^3 * 3^2 hence we know it follows the formula
2^x * 3^y
if we look, x must either be 1,2 or 3, if x is 0 then the number can never be even
however y can be either 0,1 or 2, hence x has possible values it can be while y has 3 possible values it can be
hence we can combine them in 3 * 3 = 9 ways
2^3 * 3^2 hence we know it follows the formula
2^x * 3^y
if we look, x must either be 1,2 or 3, if x is 0 then the number can never be even
however y can be either 0,1 or 2, hence x has possible values it can be while y has 3 possible values it can be
hence we can combine them in 3 * 3 = 9 ways
Bunuel
