If n is a positive integer and r is the remainder....

[james_dennis]

Hi,

Couldn't find this problem after considerable searching...I got it right but it was an educated guess. Could someone explain why the answer is A?

DS: If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r?

...This means 3Q + r = 4 + 7n (where Q = quotient) and that r > 3 (ie 0,1, or 2)

Rephrased Question (RQ): r = 4 + 7n - 3Q = ?

Statement 1: n+1 is divisible by 3

...This means that n+1 has a 3 in its prime box and that n = even integer...I'm stuck after that. This seemed like it was sufficient information (if I knew what to do with it) so I guessed A

Statement 2: n > 20

...you can plug in a few values of n that satisfy this inequality and quickly see that r will change (ie if n = 30, r = 1; if n = 31, r = 2)....therefore INSUFFICIENT


Between A, C or E I guessed A using the reasoning above. Why is A correct? Thanks

Dennis

[adiagr]

Hi,

Couldn't find this problem after considerable searching...I got it right but it was an educated guess. Could someone explain why the answer is A?

DS: If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r?

...This means 3Q + r = 4 + 7n (where Q = quotient) and that r > 3 (ie 0,1, or 2)

Rephrased Question (RQ): r = 4 + 7n - 3Q = ?

Statement 1: n+1 is divisible by 3

...This means that n+1 has a 3 in its prime box and that n = even integer...I'm stuck after that. This seemed like it was sufficient information (if I knew what to do with it) so I guessed A

Statement 2: n > 20

...you can plug in a few values of n that satisfy this inequality and quickly see that r will change (ie if n = 30, r = 1; if n = 31, r = 2)....therefore INSUFFICIENT


Between A, C or E I guessed A using the reasoning above. Why is A correct? Thanks

Dennis

As divisibility of 4+7n by 3 has to be checked, separate out those terms which are already divisible by 3.


Arrange 4+7n as:

(6n+3) + (n+1)

Now (6n +3) is already divisible by 3.

We have to check divisibility of (n+1)

Statement 1: n+1 is divisible by 3

Thus (n+1) is divisible by 3 and as such remainder r will be zero.
Sufficient.

Statement 2: n > 20

n = 23, n+1 is divisible by 3 , remainder = 0

n = 24, n+1 not divisible by 3 , remainder = 1

Not Sufficient.


Answer is A.

[ugenderr]

adiagr, good explanation. thank you.

[RonPurewal]

that is a good solution, but i don't think it's realistic to expect most students to break 4 + 7n up into (3 + 6n) and (1 + n) without any obvious hints.

one way you can solve this problem without having to do that is to rephrase statement 1 as "n is one less than a multiple of 3", hence n = 3k - 1 where k is an integer.
then the expression in the question is 4 + 7(3k - 1), or 21k - 3. this is a multiple of 3, so the answer to the question will always be 0; sufficient.

--

a far, far easier way to proceed, though, is simply to plug in numbers, getting the same result each time, until you're convinced that the statement is sufficient.
it's not hard to generate numbers that satisfy statement 1:
2, 5, 8, 11, etc.
if you plug these into (4 + 7n) in succession and find the remainders upon division by 3, you'll find that all of those remainders are zero.