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M30-16

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M30-16  [#permalink]

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New post 16 Sep 2014, 00:45
2
14
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

40% (01:52) correct 60% (01:16) wrong based on 181 sessions

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Re M30-16  [#permalink]

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New post 16 Sep 2014, 00:45
4
3
Official Solution:


If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?

Notice that two statements together are obviously sufficient to answer the question. When you see such question you should be extremely cautious when choosing C for an answer. Chances are that the question is a "C trap" question ("C trap" is a problem which is VERY OBVIOUSLY sufficient if both statements are taken together).

(1) \(a + 3b = 61\).

Let the greatest common divisor of \(a\) and \(b\) be \(d\), then \(a=md\) and \(b=nd\), for some positive integers \(m\) and \(n\). So, we'll have \((md)+3(nd)=d(m+3n)=61\). Now, since 61 is a prime number (61=1*61) then \(d=1\) and \(m+3n=61\) (vice versa is not possible because \(m\) and \(n\) are positive integers and therefore \(m+3n\) cannot equal to 1). Hence we have that the \(GCD(x, y)=d=1\). Sufficient.

(2) \(5a - b = 1\). Rearrange: \(5a=b+1\).

\(5a\) and \(b\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1 (for example 20 and 21 are consecutive integers, thus only common factor they share is 1). So, \(5a\) and \(b\) don't share any common factor but 1, thus \(a\) and \(b\) also don't share any common factor but 1. Hence, the \(GCD(x, y)\) is 1. Sufficient.


Answer: D
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Re M30-16  [#permalink]

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New post 09 Nov 2015, 22:12
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I think this is a high-quality question and I agree with explanation.
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Re: M30-16  [#permalink]

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New post 10 Nov 2015, 04:54
I got this problem right, but it took me 3 mins.
I plugged in different values of a and b: a=1, 4, 7, 10, .. then b= 20, 19, 18, 17, .. but it was lengthy and tiresome.

The explanation by Bunuel is just awesome. Such a time saver, but it needs an eye of a mathematician I think.

Thanks.
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Re: M30-16  [#permalink]

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New post 17 Dec 2016, 05:28
The easiest way is the concept of ODD EVEN..
st 1 -61 is ODD and also prime so numbers should be ODD and EVEN and hence GCD is 1
similarly for st 2 both should be ODD and EVEN
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Re: M30-16  [#permalink]

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New post 05 Jan 2017, 03:00
yaash

Sorry if I misunderstood, what do you exactly mean by should be odd and even ?

Your solution seems really fast and efficient but I did not fully got it

Many thanks for your help,
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Re: M30-16  [#permalink]

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New post 05 Jan 2017, 03:06
nickimonckom wrote:
yaash

Sorry if I misunderstood, what do you exactly mean by should be odd and even ?

Your solution seems really fast and efficient but I did not fully got it

Many thanks for your help,


That solution is not correct. yaash implies that the GCD of an odd number and of an even number is always 1, which is obviously wrong. For example, the GCD of 3 and 6 is 3. This is a hard question and has no silver bullet solution. Please refer to the official solution above.
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Re: M30-16  [#permalink]

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New post 10 Feb 2017, 01:20
From this problem I derive that the GCD of an odd number and an even number is always 1. Is this correct?
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Re: M30-16  [#permalink]

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New post 10 Feb 2017, 01:23
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M30-16  [#permalink]

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New post 10 Feb 2017, 23:58
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My way to solve this question:

If \(a\) and \(b\) are positive integers, what is the greatest common divisor of \(a\) and \(b\)?

(1) \(a + 3b = 61\)

a =61 -3b ..........Plug in numbers

b=1 ......a=58.......GCD= 1
b=2.......a=55.......GCD= 1
b=3.......a=52.......GCD= 1
b=4.......a=49.......GCD= 1
b=5.......a=46.......GCD= 1

It is safe to conclude that GCD=1

Sufficient

(2) \(5a - b = 1\)

b =5a -1

a=1 ........b=4.......GCD= 1
a=2.........b=9.......GCD= 1
a=3.........b=14.....GCD= 1
a=4.........b=19.....GCD= 1

It is safe to conclude that GCD=1

Sufficient

Answer: D
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Re: M30-16  [#permalink]

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New post 16 Jan 2019, 02:32
I answered this question incorrectly. I understood my error after reading Bunuel solution.
Please allow me to provide an alternative explanation of the second part. Which is easier for me to understand.

...Alternative explanation to Bunuel's solution of part 2...

(2) 5a−b=1. Rearrange: 5a=b+1.

Still using a=md and b=nd. We have 5(md)-(nd)=d(5d-n)=1. Since d and (5d-n) need to be positive Integers, the only possible solution to the equation is that both are equal to 1. Therefore d=1 and GCD(x,y)=d=1

Bunuel. Many Thanks for your solution
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Re: M30-16  [#permalink]

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New post 18 Jan 2019, 07:51
If d is a divisor of both a and b, then it is also a divisor of any linear combination ma+nb of these numbers a and b. Therefore it is a divisor of 5a + (-1)b, and hence of 1 (as 5a−b=1). The only positive divisor of 1 is 1.
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Re: M30-16   [#permalink] 18 Jan 2019, 07:51
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