If x, y, and z are all positive integers, what is the

E for me too. Same reasoning as above. Can you pls check the OA?

Yeah I made a Typo in the question stem! sorry

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

7^1 = 7/4 ------>R3
7^2 = 9/4 ------>R1
7^3 = 3/4 ------>R3
7^4 = 1/4------->R1

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.

Sufficient.

C.

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.

For example, 7^n / 6:

7^1 = 7 / 6 -----> R1
7^2 = 49 / 6 ---->R1
7^3 = 343 / 6 --> R1

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).

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.

If x, y and z are all positive integers, what is the remainder when 7xyz is divided by 4?
(i) yz = 3
(ii) x is odd

I dont know the OA

Merging similar topics.

Notice that it should read 7^(xyz) instead of 7xyz.

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.

I am concerned about what yz represents here. Possibilities
1) product of y and z
2) representation of last two digits.

Representation of last two digit makes little sense.
Ok. Thanks Brunnel.

xyz means x*y*z. If it were otherwise it would be explicitly stated.

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.

That is what I thought so. Thanks for the clarification.

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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.