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M05-20

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M05-20  [#permalink]

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New post 15 Sep 2014, 23:25
1
9
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A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

61% (01:21) correct 39% (01:38) wrong based on 181 sessions

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Re M05-20  [#permalink]

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New post 15 Sep 2014, 23:25
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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, 09:11
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, 17:03
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, 20:59
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, 05:51
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  [#permalink]

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New post 09 Dec 2018, 15:20
When positive integer p is divided by 7 the remainder is 2.

Algebraically this means
Step 1. p/7 = Q +2/7 (p divided by 7 equals some quotient + remainder 2 (always express over divisor))
Step 2. multiply by 7
7*(p/7 = Q +2/7)
P = 7Q +2 where Q can be 0
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Re: M05-20 &nbs [#permalink] 09 Dec 2018, 15:20
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