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Re: If x, y, and z are all positive integers, [#permalink]

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08 Nov 2012, 01:57

3

This post received KUDOS

Maxswe wrote:

If x, y, and z are all positive integers, what is the remainder when 7xyz is divided by 4?

(1) yz = 3

(2) x is odd.

Easy way to answer this question maybe?

I think the OA is wrong. Please check source of question again.

Question needs remainder when 7xyz is divided by 4.

Statement 1: yz=3 => we are looking for remainder of 21x/4. this would depend on value of x. if x=1, remainder =1. But if x=2 remainder =2. Not sufficient.

Statment 2: x is odd => remainder of some odd number*yz/4. This would again depend on what odd number it is and also on values of y and z. Not sufficient.

Combining statement 1 and 2. we need remainder of 21x/4 where x is odd. This is still not sufficient as if x =1 remainder is 1 while if x=3 remainder is 3

Re: If x, y, and z are all positive integers, [#permalink]

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08 Nov 2012, 10:28

Vips0000 wrote:

Maxswe wrote:

If x, y, and z are all positive integers, what is the remainder when 7xyz is divided by 4?

(1) yz = 3

(2) x is odd.

Easy way to answer this question maybe?

I think the OA is wrong. Please check source of question again.

Question needs remainder when 7xyz is divided by 4.

Statement 1: yz=3 => we are looking for remainder of 21x/4. this would depend on value of x. if x=1, remainder =1. But if x=2 remainder =2. Not sufficient.

Statment 2: x is odd => remainder of some odd number*yz/4. This would again depend on what odd number it is and also on values of y and z. Not sufficient.

Combining statement 1 and 2. we need remainder of 21x/4 where x is odd. This is still not sufficient as if x =1 remainder is 1 while if x=3 remainder is 3

Re: If x, y, and z are all positive integers, [#permalink]

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08 Nov 2012, 20:34

Looking for remainder left after 7^(xyz) / 4...examine the cycle of remainders left when 7 to some power is divided by 4. Use cycle of powers for last digit of 7: 7, 9, 3, 1

It appears that 7^n divided by 4 alternates between odd and even powers in producing either a remainder of 3 or 1, respectively.

(1) yz = 3

7^(3x) / 4 --> ?

x may be any integer, odd or even, that produces an odd or even power when multiplied by 3. Remainder may be 3 or 1.

Insufficient.

(2) x = odd

7^(yz *some odd integer x) -->?

Product of yz may be any integer, odd or even, that produces an odd or even power when multiplied by odd integer x. Remainder may be 3 or 1.

Insufficient.

(1) & (2)

7^(3*some odd integer n) --> 7^(xyz = some odd integer)

Since we now know that xyz = some odd integer, then we know that the remainder of 7^(xyz) will produce a remainder of 3 when divided by 4, based on the cycle of remainders outlined above.

Re: If x, y, and z are all positive integers, [#permalink]

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08 Nov 2012, 20:53

seths22 wrote:

Looking for remainder left after 7^(xyz) / 4...examine the cycle of remainders left when 7 to some power is divided by 4. Use cycle of powers for last digit of 7: 7, 9, 3, 1

It appears that 7^n divided by 4 alternates between odd and even powers in producing either a remainder of 3 or 1, respectively.

(1) yz = 3

7^(3x) / 4 --> ?

x may be any integer, odd or even, that produces an odd or even power when multiplied by 3. Remainder may be 3 or 1.

Insufficient.

(2) x = odd

7^(yz *some odd integer x) -->?

Product of yz may be any integer, odd or even, that produces an odd or even power when multiplied by odd integer x. Remainder may be 3 or 1.

Insufficient.

(1) & (2)

7^(3*some odd integer n) --> 7^(xyz = some odd integer)

Since we now know that xyz = some odd integer, then we know that the remainder of 7^(xyz) will produce a remainder of 3 when divided by 4, based on the cycle of remainders outlined above.

Re: If x, y, and z are all positive integers, [#permalink]

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08 Nov 2012, 22:29

catennacio wrote:

Question is 7xyz, not 7^(xyz).

He edited the original question.

And on further reflection, I think my reasoning above is flawed. Just using the cycle of powers for 7 (i.e. last digit) won't provide accurate answers for remainders in all cases.

But just considering 7, 3, 9, 1 (last digits of 7) all give different remainders.

A more sound logic may be: since 7^n always produces an odd integer, the remainder will always be 1 less or more than a multiple of the divisor 4 (4n +1 --> R1 or 4n - 1 --> R3).

Re: If x, y, and z are all positive integers, what is the [#permalink]

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09 Nov 2012, 04:13

4

This post received KUDOS

Expert's post

If x, y, and z are all positive integers, what is the remainder when 7^(xyz) is divided by 4?

7^1=7 divided by 4 yields the remainder of 3; 7^2=49 divided by 4 yields the remainder of 1; 7^1=243 divided by 4 yields the remainder of 3; ...

As we can see, 7^(odd) divided by 4 yields the remainder of 3, while 7^(even) divided by 4 yields the remainder of 1. So, all we need to know to answer the question is whether xyz is odd or even.

(1) yz = 3. Not sufficient. (2) x is odd. Not sufficient.

Re: If x, y, and z are all positive integers, [#permalink]

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09 Nov 2012, 16:22

Vips0000 wrote:

Maxswe wrote:

If x, y, and z are all positive integers, what is the remainder when 7xyz is divided by 4?

(1) yz = 3

(2) x is odd.

Easy way to answer this question maybe?

I think the OA is wrong. Please check source of question again.

Question needs remainder when 7xyz is divided by 4.

Statement 1: yz=3 => we are looking for remainder of 21x/4. this would depend on value of x. if x=1, remainder =1. But if x=2 remainder =2. Not sufficient.

Statment 2: x is odd => remainder of some odd number*yz/4. This would again depend on what odd number it is and also on values of y and z. Not sufficient.

Combining statement 1 and 2. we need remainder of 21x/4 where x is odd. This is still not sufficient as if x =1 remainder is 1 while if x=3 remainder is 3

Hence ans E it is!

yz=3 does not mean its 21x. yz=3 means its 7^3x. if zy is odd and x is odd, it means 7 will always be to the "odd" power, which is only a remainder of 3.

Re: If x, y, and z are all positive integers, [#permalink]

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10 Nov 2012, 23:54

Astrocat15 wrote:

Vips0000 wrote:

Maxswe wrote:

If x, y, and z are all positive integers, what is the remainder when 7xyz is divided by 4?

(1) yz = 3

(2) x is odd.

Easy way to answer this question maybe?

I think the OA is wrong. Please check source of question again.

Question needs remainder when 7xyz is divided by 4.

Statement 1: yz=3 => we are looking for remainder of 21x/4. this would depend on value of x. if x=1, remainder =1. But if x=2 remainder =2. Not sufficient.

Statment 2: x is odd => remainder of some odd number*yz/4. This would again depend on what odd number it is and also on values of y and z. Not sufficient.

Combining statement 1 and 2. we need remainder of 21x/4 where x is odd. This is still not sufficient as if x =1 remainder is 1 while if x=3 remainder is 3

Hence ans E it is!

yz=3 does not mean its 21x. yz=3 means its 7^3x. if zy is odd and x is odd, it means 7 will always be to the "odd" power, which is only a remainder of 3.

Aristocrat15- Note the question I responded to it is visible in my post(what is the remainder when 7xyz is divided by 4). The question was later edited. _________________

Re: If x, y, and z are all positive integers, what is the [#permalink]

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03 Dec 2013, 13:18

Here is the answer from Kaplan. I dont understand how he gets, If we increase x by 1, then the previous number is multiplied by 7.

This is a Value question. For sufficiency, we need to be able to determine one value for the remainder of 7xyz divided by 4. We need to determine the value of xyz. We are told that x, y, and z are all positive integers.

Evaluate the Statements:

Statement (1): We are given that yz = 3. We know that all of the variables are positive integers, so they must be 1 or greater. If we assume that x = 1, then we get:

The remainder when divided by 4 is 3.

If we increase x by 1, then the previous number is multiplied by 7. If the previous quotient has a remainder of 3 and we multiply that by 7, we get 21. Dividing this by 4, we get a remainder of 1. Increasing x by 1 again, we multiply the remainder of 1 by 7 and get 7. Dividing this by 4 will give us a remainder of 3. This pattern will continue indefinitely. We cannot determine one remainder from these conditions. Therefore, Statement (1) is Insufficient to answer the question. Eliminate choices (A) and (D).

Statement (2): We are told x is odd. This leaves us in the same situation as with Statement (1). When the number of 7s being multiplied is odd, we will get a remainder of 3. When the number of 7s is even, we will get a remainder of 1. Since we do not know the value of y or z, we do not know if the exponent will be odd or even. We cannot determine an answer from this information.

Statement (2) is Insufficient to answer the question. Eliminate choice (B).

Combined: We know that yz = 3 and x is odd. Since an odd number times an odd number is always odd, we know that the exponent will always be odd. An odd exponent will always give us a remainder of 3, as explained in the analysis of Statement (1). We can say that, with these conditions, the remainder will always be 3. Therefore, Statement (1) and Statement (2) combined are Sufficient to answer the question. Eliminate choice (E).

Re: If x, y, and z are all positive integers, what is the [#permalink]

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05 Dec 2013, 08:13

Expert's post

fullymooned wrote:

Here is the answer from Kaplan. I dont understand how he gets, If we increase x by 1, then the previous number is multiplied by 7.

This is a Value question. For sufficiency, we need to be able to determine one value for the remainder of 7xyz divided by 4. We need to determine the value of xyz. We are told that x, y, and z are all positive integers.

Evaluate the Statements:

Statement (1): We are given that yz = 3. We know that all of the variables are positive integers, so they must be 1 or greater. If we assume that x = 1, then we get:

The remainder when divided by 4 is 3.

If we increase x by 1, then the previous number is multiplied by 7. If the previous quotient has a remainder of 3 and we multiply that by 7, we get 21. Dividing this by 4, we get a remainder of 1. Increasing x by 1 again, we multiply the remainder of 1 by 7 and get 7. Dividing this by 4 will give us a remainder of 3. This pattern will continue indefinitely. We cannot determine one remainder from these conditions. Therefore, Statement (1) is Insufficient to answer the question. Eliminate choices (A) and (D).

Statement (2): We are told x is odd. This leaves us in the same situation as with Statement (1). When the number of 7s being multiplied is odd, we will get a remainder of 3. When the number of 7s is even, we will get a remainder of 1. Since we do not know the value of y or z, we do not know if the exponent will be odd or even. We cannot determine an answer from this information.

Statement (2) is Insufficient to answer the question. Eliminate choice (B).

Combined: We know that yz = 3 and x is odd. Since an odd number times an odd number is always odd, we know that the exponent will always be odd. An odd exponent will always give us a remainder of 3, as explained in the analysis of Statement (1). We can say that, with these conditions, the remainder will always be 3. Therefore, Statement (1) and Statement (2) combined are Sufficient to answer the question. Eliminate choice (E).

The correct answer is Choice (C).

For (1) we have \(7^{3x}\).

If they mean that if we increase x by 1 in \(7^{3x}\), then the previous number is multiplied by 7, then Kaplan is wrong.

If we increase x by 1 in \(7^{3x}\), then the previous number is multiplied by 7^3, not 7. _________________

Re: If x, y, and z are all positive integers, what is the [#permalink]

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09 May 2014, 03:59

Bunuel wrote:

BangOn wrote:

Maxswe wrote:

If x, y, and z are all positive integers, what is the remainder when 7^(xyz) is divided by 4?

(1) yz = 3. (2) x is odd.

7^(xyz) / 4

Even power raised to 7 will yield remainder 3 Odd power will yield 1

A) yz = 3 => which means xyz is odd. So remainder is 1. B) X is odd. Odd/Even both pssbl.

What is the mistake here?

For (1) xyz could be even if x is even.

Hope it's clear.

Bit confusing Bunuel because xyz here was interpreted by me as some three digit number with x in hundred's place , y in ten's place and z in one's place rather than X*Y*Z. Not sure if I am thinking too much into it. I do recollect seeing one question in this forum where xyz was interpreted as a three digit number.

gmatclubot

Re: If x, y, and z are all positive integers, what is the
[#permalink]
09 May 2014, 03:59

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