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How many positive integer values of 'a' are possible such that

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Hovkial
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How many positive integer values of 'a' are possible such that [#permalink] New post   20 Aug 2019, 10:00
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How many positive integer values of 'a' are possible such that \(\frac{a+220}{a+4}\) is an integer?

(A) 8

(B) 9

(C) 12

(D) 17

(E) 18
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Re: How many positive integer values of 'a' are possible such that [#permalink] New post   20 Aug 2019, 12:20
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\(\frac{a+220}{a+4}\)=\(1+ \frac{216}{a+4}\)

If \(\frac{a+220}{a+4}\) is an integer, then \(\frac{216}{a+4}\) must be an integer.
Hence a+4 must be a factor of 216.

\(216=2^3*3^3\)
total number of factors of 216= (3+1)(3+1)=16

a+4 can have 16 values.
But a>0, hence a+4 must be greater than 4. So, a+4 can't be 1, 2, 3 or 4 (all these are factors of 216, which are less than or equal to 4)

Total number of positive integer values of 'a' such that \(\frac{a+220}{a+4}\) is an integer= 16-4=12





Hovkial wrote:
How many positive integer values of 'a' are possible such that \(\frac{a+220}{a+4}\) is an integer?

(A) 8

(B) 9

(C) 12

(D) 17

(E) 18
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Re: How many positive integer values of 'a' are possible such that [#permalink] New post   20 Aug 2019, 13:12
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Hovkial wrote:
How many positive integer values of 'a' are possible such that \(\frac{a+220}{a+4}\) is an integer?

(A) 8

(B) 9

(C) 12

(D) 17

(E) 18


(a+220)/(a+4) = (a+4+216)/(a+4) = 1+216/(a+4)

216 = 2^3*3^3 . So overall there would be (3+1)*(3+1) = 16 factors. Out of these 16 factors 1,2,3,4 are not greater than 4. So, a+4 cannot take value of 1,2,3,4 as a is positive.

a+4 can take any other 12 values. Those 12 values will give an integer value to 216/(a+4). So, there are in all 12 values of a for the fraction to be an integer.
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Re: How many positive integer values of 'a' are possible such that [#permalink] New post   20 Jun 2021, 01:17
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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] New post   31 Jul 2023, 17:13
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]
31 Jul 2023, 17:13
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