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If a and b are positive integers such that a/b = 2.86, which

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If a and b are positive integers such that a/b = 2.86, which [#permalink] New post 08 Sep 2010, 20:22
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If a and b are positive integers such that a/b = 2.86, which of the following must be a divisor of a?

A. 10
B. 13
C. 18
D. 26
E. 50
[Reveal] Spoiler: OA

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Re: Prime Factor [#permalink] New post 08 Sep 2010, 20:31
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vigneshpandi wrote:
If a and b are positive integers such that a/b = 2.86, which of the following must be a divisor of a?

1. 10
2. 13
3. 18
4. 26
5. 50


\(\frac{a}{b}=2.86=\frac{286}{100}=\frac{143}{50}\) --> \(b=\frac{50a}{143}=\frac{50a}{11*13}\), for \(b\) to be an integer \(a\) must have all the factors of 143 (50 has none of them). Hence \(a\) must be divisible by both 11 and 13.

Answer: B.
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Re: Prime Factor [#permalink] New post 22 Sep 2013, 09:13
Bunuel wrote:
vigneshpandi wrote:
If a and b are positive integers such that a/b = 2.86, which of the following must be a divisor of a?

1. 10
2. 13
3. 18
4. 26
5. 50


\(\frac{a}{b}=2.86=\frac{286}{100}=\frac{143}{50}\) --> \(b=\frac{50a}{143}=\frac{50a}{11*13}\), for \(b\) to be an integer \(a\) must have all the factors of 143 (50 has none of them). Hence \(a\) must be divisible by both 11 and 13.

Answer: B.


Hi Bunuel,

I'm trying to follow MGMAT's method:
we know that a/b = 2.86
2.86 => 2 and 86/100 or 43/50
and we know that r/b = 43/50
hence 50r = 43b
from that we conclude that b must be a multiple of 50 and 43 a multiple of r.
What am I doing wrong?
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Re: If a and b are positive integers such that a/b = 2.86, which [#permalink] New post 22 Sep 2013, 10:14
You're doing everything correct. I did the same method, then got stuck near the end as you did. Here's how you would finish the problem.

Since b = 50x, and R = 43x, and a/b = 2 + R/b

a = 2b + R.

So a is equal to 143 (x = 1), 283 (x = 2), etc... In each of these, a is a multiple of 11 and 13.


Looking back, Bunuel's method is method is much easier, as it ignores calculations involving the remainder.

Last edited by grant1377 on 23 Sep 2013, 01:29, edited 1 time in total.
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Re: If a and b are positive integers such that a/b = 2.86, which [#permalink] New post 23 Sep 2013, 00:57
grant1377 wrote:
You're doing everything correct. I did the same method, then got suck near the end as you did. Here's how you would finish the problem.

Since b = 50x, and R = 43x, and a/b = 2 + R/b

a = 2b + R.

So a is equal to 143 (x = 1), 283 (x = 2), etc... In each of these, a is a multiple of 11 and 13.


Looking back, Bunuel's method is method is much easier, as it ignores calculations involving the remainder.


Hi mate, thanks for the reply. I got a question.
if x = 1, then 50x would be 50 and 43x would be 43, hence 93.
Can you explain in more detail please?
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Re: If a and b are positive integers such that a/b = 2.86, which [#permalink] New post 23 Sep 2013, 00:58
Also, can I substitute MGMAT's method to Bunuel's? i.e. does this method apply in general?
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Re: If a and b are positive integers such that a/b = 2.86, which [#permalink] New post 23 Sep 2013, 01:18
Expert's post
Skag55 wrote:
grant1377 wrote:
You're doing everything correct. I did the same method, then got suck near the end as you did. Here's how you would finish the problem.

Since b = 50x, and R = 43x, and a/b = 2 + R/b

a = 2b + R.

So a is equal to 143 (x = 1), 283 (x = 2), etc... In each of these, a is a multiple of 11 and 13.


Looking back, Bunuel's method is method is much easier, as it ignores calculations involving the remainder.


Hi mate, thanks for the reply. I got a question.
if x = 1, then 50x would be 50 and 43x would be 43, hence 93.
Can you explain in more detail please?


It's 2B, not B --> A = 2B + R = 2*50x + 43x = 143x.
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Re: If a and b are positive integers such that a/b = 2.86, which [#permalink] New post 13 Apr 2015, 09:34
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Re: If a and b are positive integers such that a/b = 2.86, which [#permalink] New post 16 Apr 2015, 19:09
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Hi All,

The prompt gives us a couple of facts to work with:
1) A and B are positive INTEGERS
2) A/B = 2.86

We can use these facts to figure out POSSIBLE values of A and B. The prompt asks us for what MUST be a divisor of A. Since we're dealing with a fraction, A and B could be an infinite number of different integers, so we have to make both as SMALL as possible; in doing so, we'll be able to find the divisors that ALWAYS divide in (and eliminate the divisors that only SOMETIMES divide in).

The simplest place to start is with...
A = 286
B = 100
286/100 = 2.86

These values are NOT the smallest possible values though (since they're both even, we can divide both by 2)...

A = 143
B = 50
143/50 = 2.86

There is no other way to reduce this fraction, so A must be a multiple of 143 and B must be an equivalent multiple of 50. At this point though, the value of B is irrelevant to the question. We're asked for what MUST divide into A....

Since A is a multiple of 143, we have to 'factor-down' 143. This gives us (11)(13). So BOTH of those integers MUST be factors of A. You'll find the match in the answer choices.

Final Answer:
[Reveal] Spoiler:
B


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Re: If a and b are positive integers such that a/b = 2.86, which   [#permalink] 16 Apr 2015, 19:09
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