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

Question

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

Bunuel · Accepted Answer

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.

ENAFEX · Answer

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

subhashghosh · Answer

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.

amanat · Answer

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

- answer C Hope this helps

SrinathVangala · Answer

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.

gracie · Answer

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
C

ScottTargetTestPrep · Answer

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

RickyTseng · Answer

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

then (8n+3) - (3n+2) = 5n+1 also divisible by integer p
now we have 3n+2, 5n+1, 8n+3

then do the calculation again: (5n+1) - (3n+2) = 2n-1
so we have 2n-1, 3n+2, 5n+1, 8n+3 all divisible by p

then do the calculation again: (3n+2) - (2n-1) = n+3
so we get n+3, 2n-1, 3n+2, 5n+1, 8n+3 all divisible by p

since n+3 divisible by p, we can imply (n+3) + (n+3) = 2n+6 divisible by p
so (2n+6) - (2n-1) = 7 and we get the answer.

adewale223 · Answer

Since I know if I subtract the two terms the result will be divible by p

So 3n + 2 -8n -3

-5n-1

Second stage

-5n -1 -8n -3

-13n -4

Third stage

-13n - 4 - 8n - 3

-21n - 7

Now that is divisible by 7

Answer choice C

Abhishek009 · Answer

If  n = 4  ,  3n+2 = 14  &  8n+3 = 35

HCF (14 , 35) = 7 , Answer must be (C)

chakreee93 · Answer

Integers 3n+2 and 8n+3 are divisible by an integer p. Therefore 8*(3n+2) and 3(8n+3) are divisible by 3. Which are 24n+16 and 24n+9 .The difference is 7 which should also be multiple of p. since the option has only 7 which divides .Answer is 7 (option c)

bumpbot · Answer

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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

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