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SWAPNILP
Hi i would like to know why ans is B here...since third digit is missing how can one determine what wud b the reminder? please explain
Quote:
\(x\) and \(y\) are positive integers. What is the remainder when \(x\) is divided by \(2^2\)?

(1) \(y = 7\)

(2) \(x = 3^{78y}\)

Hi..
78y is not a 3-digit number but 78*y and you can say this because it is given that y is a positive integer.

so let us see the question..

(1) \(y = 7\)
Nothing about y..
insuff

(2) \(x = 3^{78y}\)
Now, you should check few multiples of 3 and you will find a pattern ..
3^1 divided by 4 leaves 3 as remainder
3^2 =9 leaves 1 as remainder
3^3=27 leaves 3 as remainder and so on.. so pattern is 3,1,3,1...
\(x = 3^{78y}\), and this has an even power 78y, so answer will be that remainder is 1...
sufff

B

Ofcourse other way is binomial expansion.. \(x = 3^{78y}=(4-1)^{78y}\)...
In expansion. all terms except \((-1)^{78y}\) will be multiple of 4, so remainder = \((-1)^{78y}=1\)
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SWAPNILP
Hi i would like to know why ans is B here...since third digit is missing how can one determine what wud b the reminder? please explain
Quote:
\(x\) and \(y\) are positive integers. What is the remainder when \(x\) is divided by \(2^2\)?

(1) \(y = 7\)

(2) \(x = 3^{78y}\)

Hi..
78y is not a 3-digit number but 78*y and you can say this because it is given that y is a positive integer.

so let us see the question..

(1) \(y = 7\)
Nothing about y..
insuff

(2) \(x = 3^{78y}\)
Now, you should check few multiples of 3 and you will find a pattern ..
3^1 divided by 4 leaves 3 as remainder
3^2 =9 leaves 1 as remainder
3^3=27 leaves 3 as remainder and so on.. so pattern is 3,1,3,1...
\(x = 3^{78y}\), and this has an even power 78y, so answer will be that remainder is 1...
sufff

B

Ofcourse other way is binomial expansion.. \(x = 3^{78y}=(4-1)^{78y}\)...
In expansion. all terms except \((-1)^{78y}\) will be multiple of 4, so remainder = \((-1)^{78y}=1\)

thanks chetan2u
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Hi SWAPNILP
I don't want to provide the official explanation soo soon. Wanted to give a chance for others to think as well. :)

But here is a very big clue for you-
In Gmat a lot of problems depend on pattern recognision. So check the patterns when powers of 3 are divided by 4 mentioned in the question stem.

Edit- oops! Looks like chetan2u gave an awesome response (maybe a little too soon :D).


Hi blitzkriegxX,
I wanted to clarify on 78y as a 3-digit number or a product 78*y.
I think in that process, the solution too came out. :)
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chetan2u


Hi blitzkriegxX,
I wanted to clarify on 78y as a 3-digit number or a product 78*y.
I think in that process, the solution too came out. :)


Hey chetan2u
No worries. :)
And you're right. I did mean the product 78*y.
Is this how the product part is expressed in official gmat questions?
Or should I edit my question to 78*y ?
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chetan2u


Hi blitzkriegxX,
I wanted to clarify on 78y as a 3-digit number or a product 78*y.
I think in that process, the solution too came out. :)


Hey chetan2u
No worries. :)
And you're right. I did mean the product 78*y.
Is this how the product part is expressed in official gmat questions?
Or should I edit my question to 78*y ?

Hi..
you are perfectly fine with the way you have written, 78y would mean product unless specified otherwise, and y as positive integer also makes it much more clearer, because if 78y were a 3-digit number, y would be a digit, and not integer.
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What is the question source?

chetan2u
Quote:
I wanted to clarify on 78y as a 3-digit number or a product 78*y.
I think in that process, the solution too came out.

By this you mean the solution is B as well? 3^{(195·4)+1}, 3^{(195·4)+2}
I confused the exponents and thought it was a third digit and so selected C
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blitzkriegxX
What is the question source?

chetan2u
Quote:
I wanted to clarify on 78y as a 3-digit number or a product 78*y.
I think in that process, the solution too came out.

By this you mean the solution is B as well? 3^{(195·4)+1}, 3^{(195·4)+2}
I confused the exponents and thought it was a third digit and so selected C

Yes, I mean answer is B..
Tye initial statement says that X and y are integers. So if y is an integer, 78y would mean 78*y.
Had it been given that y is a digit, it could have been a 3-digit number. But, here too it is generally given that 78y is a 3-digut number or 78y is an integer etc
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Apologies, I wasn't clear. I meant in the scenario that it is a 3 digit number of 78[y]. Hence the \(3^{(195·4)+1}, 3^{(195·4)+2}\) when y =1 and y=2 and so on, is the answer still B?
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chetan2u
Apologies, I wasn't clear. I meant in the scenario that it is a 3 digit number of 78[y]. Hence the \(3^{(195·4)+1}, 3^{(195·4)+2}\) when y =1 and y=2 and so on, is the answer still B?

No, then the answer will be C.
Reason is that we are dividing X by 2^*2 or 4..
Now X is 3 to the power something.

When you divide 3 with odd power that is 3^1 or 3^3 or 3^781, the remainder will always be 3..
But with even power that is 3^2 or 3^782, the remainder will be 1..
So we have to know whether y is even or odd..

Statement I tells us that it is odd, hence remainder is 1

Thus combined the two statements are sufficient

C
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But how one can know whether 78y (a three digit no.) or 78*y? It's second time that i got confused in these problems.
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But how one can know whether 78y (a three digit no.) or 78*y? It's second time that i got confused in these problems.

If it were a three-digit number it would have been explicitly mentioned. Otherwise, 78y ALWAYS means 78*y.
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(2) \(x = 3^{78y}\)
Now, you should check few multiples of 3 and you will find a pattern ..
3^1 divided by 4 leaves 3 as remainder
3^2 =9 leaves 1 as remainder
3^3=27 leaves 3 as remainder and so on.. so pattern is 3,1,3,1...
\(x = 3^{78y}\), and this has an even power 78y, so answer will be that remainder is 1...
sufff

how do we know the remainder is 1...?
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(2) \(x = 3^{78y}\)
Now, you should check few multiples of 3 and you will find a pattern ..
3^1 divided by 4 leaves 3 as remainder
3^2 =9 leaves 1 as remainder
3^3=27 leaves 3 as remainder and so on.. so pattern is 3,1,3,1...
\(x = 3^{78y}\), and this has an even power 78y, so answer will be that remainder is 1...
sufff

how do we know the remainder is 1...?


3 to the odd power gives 3 as the remainder, while 3 ^even gives 1 as the remainder. ( We have already checked the pattern)
Now 3^{78y} has a power 78y, which is even, so the remainder is 1.
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\(x\) and \(y\) are positive integers. What is the remainder when \(x\) is divided by \(2^2\)?

(1) \(y = 7\)

(2) \(x = 3^{78y}\)

Clearly, the two statements combined provide sufficient information.
Eliminate E.
If C is the correct answer, 100% of test-takers will answer the problem correctly.
Implication:
The correct answer CANNOT BE C.
Eliminate C.
Statement 1 is on its own is clearly insufficient, since it provides no information about x.
Eliminate A and D.


Even without understanding WHY Statement 2 is sufficient on its own, a savvy test-taker can determine that the correct answer must be B.
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