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Re: How many positive integer values of 'a' are possible such that [#permalink]
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\(\frac{a + 220 }{ a + 4} = \frac{a + 4 + 216 }{ a + 4} = 1 + \frac{216}{ a + 4}\)

=> \(216 = 2^3 * 3^3 = (3 + 1) (3 + 1) = 4 * 4 = 16\) factors

a + 4 can have all those factors in which a > 0. This simple a + 4 > 4.

So, a = 1, 2, 3 and 4 is not possible values.

Total values: 16 - 4 = 12

Answer C
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Re: How many positive integer values of 'a' are possible such that [#permalink]
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Asked: How many positive integer values of 'a' are possible such that \(\frac{a+220}{a+4}\) is an integer?

\(\frac{a+220}{a+4} = 1 + \frac{216}{a+4}\)
If \(1 + \frac{216}{a+4}\) is an integer, then \(\frac{a+220}{a+4}\) should be an integer too.

a+4 should be a factor of 216
\(216 = 2^3*3^3\)

Number of factors = Number of terms in the expression (1+2+2^2+2^3)(1+3+3^2+3^3) = 4*4 = 16
But a is a positive integer; a>0
a +4 > 4

Values 1, 2, 3, 4 are not allowed.
Therefore, number of values of a such that \(\frac{a+220}{a+4}\) is an integer = 16 - 4 = 12

IMO C
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Re: How many positive integer values of 'a' are possible such that [#permalink]
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