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M05-20

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M05-20  [#permalink]

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New post 16 Sep 2014, 00:25
1
9
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

59% (01:21) correct 41% (01:39) wrong based on 175 sessions

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Re M05-20  [#permalink]

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New post 16 Sep 2014, 00:25
1
1
Official Solution:


When positive integer \(p\) is divided by 7 the remainder is 2: \(p=7q+2\), so p can be: 2, 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, ...

(1) \(p\) is divisible by 2 and 3. This statement tells that \(p\) is a multiple of 6, so \(p\) could be 30 (answer NO) or 72 (answer YES). Not sufficient.

(2) \(p \lt 100\). Clearly insufficient.

(1)+(2) \(p\) can still be 30 (answer NO) or 72 (answer YES). Not sufficient.


Answer: E
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Re: M05-20  [#permalink]

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New post 28 Sep 2016, 10:11
Is there an algebraic solution to this, rather than having to write the first 10 possible values of p?
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Re: M05-20  [#permalink]

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New post 11 Jan 2017, 18:03
rajarams wrote:
Is there an algebraic solution to this, rather than having to write the first 10 possible values of p?


Experts - Please suggest if this can be solved by some other method

Thanks
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Re: M05-20  [#permalink]

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New post 11 Jan 2017, 21:59
ankur2710 wrote:
rajarams wrote:
Is there an algebraic solution to this, rather than having to write the first 10 possible values of p?


Experts - Please suggest if this can be solved by some other method

Thanks


Check here: when-positive-integer-p-is-divided-by-7-the-remainder-is-75299.html
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Re: M05-20  [#permalink]

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New post 05 Oct 2018, 06:51
Bunuel

"This statement tells that pp is a multiple of 6, so pp could be 30 (answer NO) or 72 (answer YES). Not sufficient " So here 30 or 72 are derived doing manual check such as 6,12,.....30...36....72 or any other way you directly arrived to these 2 number ?
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Re: M05-20 &nbs [#permalink] 05 Oct 2018, 06:51
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