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M31-47

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M31-47  [#permalink]

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New post 20 Jun 2015, 10:58
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If \(a\) and \(b\) are single-digit positive numbers and \(\frac{a}{b}\) is NOT a recurring decimal, what is the value of \(a\)?



(1) \(-\frac{1}{3} > -\frac{a}{b} > -\frac{4}{5}\)

(2) \(b\) is equal to the sum of its positive divisors excluding \(b\) itself.

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Re M31-47  [#permalink]

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New post 20 Jun 2015, 10:58
Official Solution:


If \(a\) and \(b\) are single-digit positive numbers and \(\frac{a}{b}\) is NOT a recurring decimal, what is the value of \(a\)?

\(a\) and \(b\) are single-digit positive numbers means that \(a\) and \(b\) can be 1, 2, 3, 4, 5, 6, 7, 8, or 9.

(1) \(-\frac{1}{3} > -\frac{a}{b} > -\frac{4}{5}\)

Simplify by multiplying by -1: \(\frac{1}{3} < \frac{a}{b} < \frac{4}{5}\)

Convert to decimals \(0.(3) < \frac{a}{b} < 0.8\).

Since \(\frac{a}{b}\) is NOT a recurring decimal, then it can be 0.4 (\(a = 2\), \(b = 5\)), 0.5 (\(a = 1\), \(b = 2\)), ... Not sufficient.

(2) \(b\) is equal to the sum of its positive divisors excluding \(b\) itself.

From single digit numbers only 6 satisfies this condition: \(6 = 1 + 2 + 3\). Since \(\frac{a}{b}\) is NOT a recurring decimal, then \(\frac{a}{6}\) can be \(\frac{3}{6} = 0.5\), \(\frac{6}{6} = 1\), or \(\frac{9}{6} = 1.5\). Not sufficient.

(1)+(2) From (2) \(\frac{a}{b}\) can be \(\frac{3}{6} = 0.5\), \(\frac{6}{6} = 1\), or \(\frac{9}{6} = 1.5\). Only one of which is between 0.(3) and 0.8, namely 0.5. Sufficient.


Answer: C
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Re: M31-47  [#permalink]

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New post 01 Sep 2016, 00:27
Hi the answer explanation states:

From single digit numbers only 6 satisfies this condition: 6=1+2+3.

However, 3 is also valid: 3 = 1+2.

What am I missing here? Thanks in advance :)
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Re: M31-47  [#permalink]

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New post 01 Sep 2016, 02:35
z0rpia wrote:
Hi the answer explanation states:

From single digit numbers only 6 satisfies this condition: 6=1+2+3.

However, 3 is also valid: 3 = 1+2.

What am I missing here? Thanks in advance :)


(2) says: b is equal to the sum of its positive divisors excluding b itself.

2 is not a divisor of 3. So, 3 does NOT equal to the sum of its positive divisors excluding b itself. 3 has only one positive divisor excluding 3 itself namely 1.
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Re: M31-47  [#permalink]

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New post 02 Sep 2016, 01:50
Bunuel wrote:

Since \(\frac{a}{b}\) is NOT a recurring decimal, then it can be 0.4 (\(a = 2\), \(b = 5\)), 0.5 (\(a = 1\), \(b = 2\)), ... Not sufficient.




Does 3/5 & 7/10 satisfy fact 1? I think so as they are 0.6 & 0.7 respectively. Did not you consider them fro certain reason?
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Re: M31-47  [#permalink]

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New post 02 Sep 2016, 02:55
Mo2men wrote:
Bunuel wrote:

Since \(\frac{a}{b}\) is NOT a recurring decimal, then it can be 0.4 (\(a = 2\), \(b = 5\)), 0.5 (\(a = 1\), \(b = 2\)), [highlight]... Not sufficient.[/highlight]




Does 3/5 & 7/10 satisfy fact 1? I think so as they are 0.6 & 0.7 respectively. Did not you consider them fro certain reason?


a/b can be 3/5. You can see ... part in the solution above, which indicates that there are some other values of a/b possible.

a/b cannot be 7/10, because 10 is not a single digit number.
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M31-47  [#permalink]

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New post 30 Oct 2018, 03:49
A quick tip -

To make sure of statement 2 faster. You can use terminating decimals theory. Because recurring decimals are basically non-terminating. So check if the denominator of any number is having multiples of 2 or 5 after u reduce fraction to lowest form.

Example-
We can simply eliminate 1/6 , 2/6 , 4/6 , 5/6, 7/6 and 8/6 because they have no denominator divisible by 2 and 5.
4/6 when reduced gives 2/3.
8/6 gives 4/3. and so on.


Hey Bunuel
To check statement 2 and make sure that we do have a recurring decimal. We can use terminating decimals theory right?
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M31-47 &nbs [#permalink] 30 Oct 2018, 03:49
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