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Divisiblity and remainder help

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Divisiblity and remainder help

by alltimeacheiver » Thu Feb 10, 2011 11:34 pm
For a nonnegative integer n, if the remainder is 1 when 2n is divided by 3, then which of
the following must be true?
I. n is greater than zero.
II. 3n = (-3)n
III. √2n is an integer.
...A. I only
B. II only
C. I and II
D. I and III
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by Night reader » Fri Feb 11, 2011 12:12 am
2n/3=i {i is integer} +1/3

I. n is greater than zero --> n=1, 2*1/3=0 + 2/3 the remainder is not 1; eliminate
II. 3n = (-3)n --> n=0, -3*0=0 {non-negative, non-positive}, 2*(0)/3=0 + 0/3 the remainder is not 1; eliminate
III. √2n is an integer --> n=50, √2n=√100=10, 2*50/3=99+1/3 true, But n=0 and √2*0=0, 2*0/3=0 +0/3 eliminate

Answer --> none is true, possibly E
alltimeacheiver wrote:For a non-negative integer n, if the remainder is 1 when 2n is divided by 3, then which of the following must be true?
I. n is greater than zero
II. 3n = (-3)n
III. √2n is an integer

A. I only
B. II only
C. I and II
D. I and III
E. :)
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by alltimeacheiver » Fri Feb 11, 2011 12:25 am
thank you so much. I think I get nerves when c the questions. These things do came in my mind. Got picture to attempt it. thanks a ton
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by prachich1987 » Fri Feb 11, 2011 1:42 am
what would be the reminder if we divide -6 by 4?
will it be 2?
Thanks!
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by Night reader » Fri Feb 11, 2011 3:32 am
prachich1987 wrote:what would be the reminder if we divide -6 by 4?
will it be 2?
-6/4=i {i is integer} + r/6 --> i=-1, r=-2 <> -6=4*(-1) - 2 (NOT -1, because -2/4 <>-1/2 :) )
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by prachich1987 » Fri Feb 11, 2011 6:08 am
Night reader wrote:
prachich1987 wrote:what would be the reminder if we divide -6 by 4?
will it be 2?
-6/4=i {i is integer} + r/6 --> i=-1, r=-2 <> -6=4*(-1) - 2 (NOT -1, because -2/4 <>-1/2 :) )
Thanks night reader!
Thanks!
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by rros0770 » Fri Feb 11, 2011 9:48 am
alltimeacheiver- What's the OA ?
Is there a Choice (E) None? As there were only 4 choices posted below

I think A is the answer. Let me know if I missed something though:

We're told in the question stem that n is nonnegative (zero or positive) and 2n/3 leaves a remainder of 1, so:
n= 3(k) + 1

In order to end up with a remainder of 1, n would need to be greater then zero, which STATEMENT (I) tells us.
We know this to be true because we have been told that (n) is not negative and that there is a remainder, therefore (n) cannot be zero

If the stem asked us to solve for n, we would be unable to as Night Reader demonstrated, but it's just asking us which 'must be true.'
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by Night reader » Fri Feb 11, 2011 10:10 am
let me jump in>> the stem requires us to observe 2n/3=i {i is integer} +1/3
For a non-negative integer n, if the remainder is 1 when 2n is divided by 3, then which of the following must be true?
rros0770 wrote:alltimeacheiver- What's the OA ?
Is there a Choice (E) None? As there were only 4 choices posted below

I think A is the answer. Let me know if I missed something though:

We're told in the question stem that n is nonnegative (zero or positive) and 2n/3 leaves a remainder of 1, so:
n= 3(k) + 1

In order to end up with a remainder of 1, n would need to be greater then zero, which STATEMENT (I) tells us.
We know this to be true because we have been told that (n) is not negative and that there is a remainder, therefore (n) cannot be zero

If the stem asked us to solve for n, we would be unable to as Night Reader demonstrated, but it's just asking us which 'must be true.'
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by rros0770 » Fri Feb 11, 2011 10:32 am
Hey Night Reader-

I follow your logic, just missing why we've ruled out (A)
I agree with the calc you set up here: "2n/3=i {i is integer} +1/3"

From the stem:
(1)We know that (n) is not negative.
(2)We know that (n) is not zero, because if (n)= 0 then we get the below, and your calc wouldn't work:
2(0)/3 = 0 (and we are not left with a remainder of 1)

From the above two points, it can be derived that (n) must be a positive number.

Stem:
For a non-negative integer n, if the remainder is 1 when 2n is divided by 3, then which of the following must be true?

Choice (A) answers this. We are not being asked to solve for (n), if we were being asked to solve for (n), this would obviously not suffice as (n) could then be an infinite range of numbers
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by Ashish8 » Sun Feb 13, 2011 5:37 pm
I also got A.

I used n=5, since (5*2)/3 = remainder of 1.

This I as sufficient, because n has to be nonnegative integer
This also crosses out II.
As for III, I don't know if if its suppose to be sqrt(2*n) or n*sqrt(2), either way neither of these will be a integer when n = 5.
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by Geva@EconomistGMAT » Mon Feb 14, 2011 12:23 am
alltimeacheiver wrote:For a nonnegative integer n, if the remainder is 1 when 2n is divided by 3, then which of
the following must be true?
I. n is greater than zero.
II. 3n = (-3)n
III. √2n is an integer.
...A. I only
B. II only
C. I and II
D. I and III
for must be true questions, the goal is to try and find counter examples and eliminate. For each of the statements, ask "Must this be true?" Can I find an example where it is NOT true?" then set out to find a single counter example - a single n that satisfies the question stem, but does not satisfy the statement - thus proving that the statement is not a MUST.

I: Must n be greater than zero? we know that n is non-negative (stem), so that n is positive or ZERO. The only way stat. (1) is disproved is if n could EQUAL zero. could that be the case? If n=0, then the remainder when dividing 2n=0 by 3 is zero, which means that n-0 is not a valid plug in. Thus, whatever n is, it must be positive. Stat. I must be true, because we cannot find a counter example that satisfies the question stem.
Note that you can eliminate B here, since it does not include I.

II Must 3n equal (-3)n? The only way this works is if n=0, which we've already show is not a valid plug in. In this case, it means that II is NOT true: since n cannot be zero. Thus, we can eliminate any answer choice that claims that II is true: C is out.

III Must √2n be an integer? (I'm assuming that n is under the square root - otherwise, root 2 * int will neve be an integer and III is out immediately)
From the question stem, we know something about the number 2n: it gives a remainder of 1 when divided by 3. What could 2n be, then? it could be 4, 7, 10, 13, 16 - all of which give a reminder of I when divided by 3. IF 2n is 4 or 16, then it is a perfect square, and rot (2n) will be an integer. But if 2n is any of the other examples, then it's not a perfect square and root (2n) will NOT be an integer. Thus, III is NOT a must be true, and is eliminated.
Eliminate any answer choice which includes III: D is out.

A is the answer: only I is a must be true, since we cannot find a counter example that satisfies the question stem.
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by Night reader » Mon Feb 14, 2011 10:10 am
@Geva, according to your logic we should do seek the remainder condition. Then why we disregard that when n>0 we do not always have the remainder of 1? e.g. n=1, 2*1/3=0+2/3 OR n=4, 2*4/3=2 +2/3. In one case you say that the remainder counts, and in the other case you discount remainder and look after n must be positive. It's clear that when you multiply n which is non-negative by 2 (+ve) and then divide by 3 (+ve), you should receive the remainder which is +ve, then n is always non-negative and can be both 0 and >0

[spoiler]there's no answer among the choices given
[/spoiler]
Geva@MasterGMAT wrote:
alltimeacheiver wrote:For a nonnegative integer n, if the remainder is 1 when 2n is divided by 3, then which of
the following must be true?
I. n is greater than zero.
II. 3n = (-3)n
III. √2n is an integer.
...A. I only
B. II only
C. I and II
D. I and III
for must be true questions, the goal is to try and find counter examples and eliminate. For each of the statements, ask "Must this be true?" Can I find an example where it is NOT true?" then set out to find a single counter example - a single n that satisfies the question stem, but does not satisfy the statement - thus proving that the statement is not a MUST.

I: Must n be greater than zero? we know that n is non-negative (stem), so that n is positive or ZERO. The only way stat. (1) is disproved is if n could EQUAL zero. could that be the case? If n=0, then the remainder when dividing 2n=0 by 3 is zero, which means that n-0 is not a valid plug in. Thus, whatever n is, it must be positive. Stat. I must be true, because we cannot find a counter example that satisfies the question stem.
Note that you can eliminate B here, since it does not include I.

II Must 3n equal (-3)n? The only way this works is if n=0, which we've already show is not a valid plug in. In this case, it means that II is NOT true: since n cannot be zero. Thus, we can eliminate any answer choice that claims that II is true: C is out.

III Must √2n be an integer? (I'm assuming that n is under the square root - otherwise, root 2 * int will neve be an integer and III is out immediately)
From the question stem, we know something about the number 2n: it gives a remainder of 1 when divided by 3. What could 2n be, then? it could be 4, 7, 10, 13, 16 - all of which give a reminder of I when divided by 3. IF 2n is 4 or 16, then it is a perfect square, and rot (2n) will be an integer. But if 2n is any of the other examples, then it's not a perfect square and root (2n) will NOT be an integer. Thus, III is NOT a must be true, and is eliminated.
Eliminate any answer choice which includes III: D is out.

A is the answer: only I is a must be true, since we cannot find a counter example that satisfies the question stem.
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by Geva@EconomistGMAT » Mon Feb 14, 2011 12:32 pm
Night reader wrote:@Geva, according to your logic we should do seek the remainder condition. Then why we disregard that when n>0 we do not always have the remainder of 1? e.g. n=1, 2*1/3=0+2/3 OR n=4, 2*4/3=2 +2/3. In one case you say that the remainder counts, and in the other case you discount remainder and look after n must be positive. It's clear that when you multiply n which is non-negative by 2 (+ve) and then divide by 3 (+ve), you should receive the remainder which is +ve, then n is always non-negative and can be both 0 and >0

the remainder condition always holds. any n which does not satisfy the condition is irrelevant, as it's an invalid plug in. n cannot be 1 or 4, even though these are non-negative, because 1 and 4 do not satisfy the remainder condition.
It's easy to find values of n that satisfy the remainder condition AND I. n can be 2, or 5, or 8 - all of these are integers greater than zero that satisfy the remainder condition, and you will find an infinite number of values that do the same. That's not the point. These alone do not prove that n MUST be non-negative. You don't know that n must be greater than zero - the statements are not facts about n, but rather need to be proven. All you know about n is what the question stem tells you: n is a nonnegative integer which satisfies the remainder condition. your job is to ask whether this knowledge of n is enough to necesitate that n be greater than zero (I), or satisfy the equation in II, or the condition in III.

So, if n is a nonnegative integer that satisfies the remainder condition, does this mean that n is greater than zero? yes, it does - n cannot be negative (from the question stem), and n cannot be zero (because n=0 does not satisfy the remainder condition), so n must be greater than zero. I is a MUST.
Another way of looking at the same problem, which is equivalent to the must be true question above: In the following DS question, is the stat. (1) sufficient?

Is n >0?
(1) n is a non-negative integer, and the remainder is 1 when 2n is divided by 3.


II: Again: n is a nonnegative integer that satisfies the remainder condition, does this mean that n3n is equal to -3n? In other words, if n is a nonnegative integer that satisfies the remainder condition, does that mean than n is zero? No, it does not - qite the contrary. Weve seen that n can be 2, 5, 8, but not zero. So II is invalidated, and is not a MUST.

Etc.
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