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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:

  55% (hard)

Question Stats:

60% (02:03) correct 40% (02:25) wrong based on 126 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
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Re: M05-20  [#permalink]

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New post 09 Dec 2018, 16:20
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  [#permalink]

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New post 30 Jun 2019, 14:18
I understand the step in (1) where the numbers come out to be 30 and 72 because they fit the criteria, but what does the remainders 72 and 30 when they're divided by 8 have to do with it? Nowhere in the questions does it mention that the remainder of p/8 has to be 2.

Though, I do think it makes sense that (1) does not suffice because 30 and 72 could both be answers, so isn't sufficient.

Am I on the right track?

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Re: M05-20   [#permalink] 30 Jun 2019, 14:18
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