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Re: Integers 3n+2 and 8n+3 are divisible by an integer p. If p
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14 Mar 2012, 00:39
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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.
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
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
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02 Feb 2018, 12:02
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.