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Integers 3n+2 and 8n+3 are divisible by an integer p. If p

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Integers 3n+2 and 8n+3 are divisible by an integer p. If p  [#permalink]

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New post 13 Mar 2012, 18:10
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Question Stats:

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Integers 3n+2 and 8n+3 are divisible by an integer p. If p is not equal to 1, then p equals to?

A. 2
B. 8
C. 7
D. 11
E. 6
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Re: Integers 3n+2 and 8n+3 are divisible by an integer p. If p  [#permalink]

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New post 13 Mar 2012, 23:39
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SergeNew wrote:
Integers 3n+2 and 8n+3 are divisible by an integer p. If p is not equal to 1, then p equals to?

A. 2
B. 8
C. 7
D. 11
E. 6


Trial and error approach would probably be the fastest for this problem.

One can also do: 8n+3=even+odd=odd, which means that p must be odd, so the answer is either 7 or 11. Now, since both numbers are divisible by p then their sum must also be divisible by p: (3n+2)+(8n+3)=11n+5. 11n+5 is not a multiple of 11 (it's 5 more than multiple of 11), hence p can only be 7.

Answer: C.

Hope it's clear.
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Re: Integers 3n+2 and 8n+3  [#permalink]

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New post 13 Mar 2012, 18:42
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Not sure if there is a shortest way to do this.

I just substituted values from 1 to 4 for n in both the equations 3n+2 and 8n+3.

When n=1, 3n+2= 5, 8n+3= 11 - No common divisor
When n=2, 3n+2= 8, 8n+3= 19 - No common divisor
When n=3, 3n+2= 11, 8n+3= 27 - No common divisor
When n=4, 3n+2=14 , 8n+3= 35 - divisible by 7

Answer C
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Re: Integers 3n+2 and 8n+3  [#permalink]

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New post 13 Mar 2012, 20:04
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I would also do the same, however, please note that 8n+3 is odd (because 8n is even and odd + even = odd). So the task is reduced to picking out from 7 and 11 among the answer choices now.
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Re: Integers 3n+2 and 8n+3 are divisible by an integer p. If p  [#permalink]

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New post 17 May 2013, 07:23
I used a more algebraic approach

From the question we can gather that

3n + 2 = pq and 8n + 3 = px

If we re-arrange in terms of n we get

n= (pq-2) / 3 and n= (px-3) / 8

equating the two equations we get

8(pq-2) = 3(px-3)

which simplifies to

p= 7 / (8q-3x)

therefore p is a multiple of 7 :-D - answer C

Hope this helps
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Re: Integers 3n+2 and 8n+3 are divisible by an integer p. If p  [#permalink]

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New post 17 May 2013, 09:37
SergeNew wrote:
Integers 3n+2 and 8n+3 are divisible by an integer p. If p is not equal to 1, then p equals to?

A. 2
B. 8
C. 7
D. 11
E. 6



3n + 2 is not divisible by 3 so 6 is ruled out.

8n + 3 is odd and not divisible by both 3 and 8 .So only 7 and 11 are left out.

I used trial and error after this. :P
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Integers 3n+2 and 8n+3 are divisible by an integer p. If p  [#permalink]

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New post 31 Jan 2018, 18:29
SergeNew wrote:
Integers 3n+2 and 8n+3 are divisible by an integer p. If p is not equal to 1, then p equals to?

A. 2
B. 8
C. 7
D. 11
E. 6


let x and y be quotients
3n+2=px→24n+16=8px
8n+3=py→24n+9=3py
subtracting,
7=8px-3py→7/p=8x-3y
if 8x-3y is an integer, then
p=7
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Re: Integers 3n+2 and 8n+3 are divisible by an integer p. If p  [#permalink]

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New post 02 Feb 2018, 11:02
SergeNew wrote:
Integers 3n+2 and 8n+3 are divisible by an integer p. If p is not equal to 1, then p equals to?

A. 2
B. 8
C. 7
D. 11
E. 6


Let’s list the pairs (3n + 2, 8n + 3) for some values of n, starting from n = 0:

(2, 3), (5, 11), (8, 19), (11, 27), (14, 35), …

We see that the first four pairs are relatively prime (i.e. they have no common factors besides 1), but the fifth pair (14, 35) has the common factor of 7. Thus, p is 7.

Answer: C
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Re: Integers 3n+2 and 8n+3 are divisible by an integer p. If p &nbs [#permalink] 02 Feb 2018, 11:02
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