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For positive integers 10 < K < Z < Q, what is the value of the remaind
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02 Dec 2018, 01:55
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For positive integers 10 < K < Z < Q, what is the value of the remainder when Q is divided by Z? Statement #1: Q = \(K^2\) Statement #2: Z = K + 1
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Re: For positive integers 10 < K < Z < Q, what is the value of the remaind
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02 Dec 2018, 10:06
From statement 1: Q = \(K^2\) From the question stem, Min value of K is 11 as K is an integer. But no info about Z. Insufficient. From statement 2: Z = K+1 Min value of K is 11. Then Z is 12. but Q can be any integer, hence insufficient. Combining both, \(K = Z1\) \(Q = K^2\) \(Q = Z^2+12Z\). When Q is divided by Z, The remainder is always as \(Z^2\) and 2Z is always divisible by Z. C is the asnwer.
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Re: For positive integers 10 < K < Z < Q, what is the value of the remaind
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02 Dec 2018, 16:43
Afc0892 wrote: When Q is divided by Z, The remainder is always as \(Z^2\) and 2Z is always divisible by Z.
C is the asnwer. To quickly clarify, I think you have it right except for this line. When we take Q=\(Z^2\)+1−2ZQ and divide it by Z we get Z2 R 1. The remainder is due to the +1 term divided by Z as we know the other are multiples of Z and Z is not one.



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For positive integers 10 < K < Z < Q, what is the value of the remaind
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Updated on: 03 Dec 2018, 05:29
aragonn wrote: Official Explanation: For this one, we’ll at least start out with picking numbers and see how far we get with this strategy. Statement #1: Q = \(K^2\) For simplicity, say that K = 20, so Q = 400. If Z = 40, then it goes evenly into 400, so the remainder is zero. On the other hand, if Z = 250, then it goes once into 400 with a remainder of 150. Two different choices of numbers give two different values of the remainder. This choice, alone and by itself, is not sufficient. Statement #2: Z = K + 1 For simplicity, let K = 10. This makes Z = 11, but that doesn’t matter. Under this statement, there’s no restriction on the value of Q. If Q = 60, then 10 divides evenly into 60, and the remainder is zero. If Q = 69, then 10 goes into it six times with a remainder of 9. Two different choices of numbers give two different values of the remainder. This choice, alone and by itself, is not sufficient. Combined statements:Let’s pick K = 11. Then Z = 12 and Q = 121. We know that 12 goes evenly into 120, so when we divide 121 by 12, we get a remainder of 1. It gets hard to pick numbers and do the squaring & division without a calculator. All other choices will result in the same remainder of 1. Remember that we can use picking numbers to disqualify an answer, to show that it leads to two different conclusions, but if the same numerical answer results every time, we need to verify that with algebra or logic. Remember the Difference of Two Squares pattern. We know that Q − 1 = \(K^2\)− 1 = (K + 1)(K − 1) Thus, Z = K + 1 always divides evenly into Q − 1. If we add one, the next integer up from Q − 1 is Q—the integer Q is exactly one more than the integer Q − 1. When we divide Q by Z = K + 1, Z always goes evenly into Q − 1, and the extra 1 is left over, so we always will have a remainder of 1. The two statements together allow us to find a unique numerical answer to the prompt questions. Combined, the statements are sufficient. Answer = (C)Thanks So much. I got it totally wrong. It is C
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For positive integers 10 < K < Z < Q, what is the value of the remaind
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03 Dec 2018, 05:22
Official Explanation: For this one, we’ll at least start out with picking numbers and see how far we get with this strategy. Statement #1: Q = \(K^2\) For simplicity, say that K = 20, so Q = 400. If Z = 40, then it goes evenly into 400, so the remainder is zero. On the other hand, if Z = 250, then it goes once into 400 with a remainder of 150. Two different choices of numbers give two different values of the remainder. This choice, alone and by itself, is not sufficient. Statement #2: Z = K + 1 For simplicity, let K = 10. This makes Z = 11, but that doesn’t matter. Under this statement, there’s no restriction on the value of Q. If Q = 60, then 10 divides evenly into 60, and the remainder is zero. If Q = 69, then 10 goes into it six times with a remainder of 9. Two different choices of numbers give two different values of the remainder. This choice, alone and by itself, is not sufficient. Combined statements:Let’s pick K = 11. Then Z = 12 and Q = 121. We know that 12 goes evenly into 120, so when we divide 121 by 12, we get a remainder of 1. It gets hard to pick numbers and do the squaring & division without a calculator. All other choices will result in the same remainder of 1. Remember that we can use picking numbers to disqualify an answer, to show that it leads to two different conclusions, but if the same numerical answer results every time, we need to verify that with algebra or logic. Remember the Difference of Two Squares pattern. We know that Q − 1 = \(K^2\)− 1 = (K + 1)(K − 1) Thus, Z = K + 1 always divides evenly into Q − 1. If we add one, the next integer up from Q − 1 is Q—the integer Q is exactly one more than the integer Q − 1. When we divide Q by Z = K + 1, Z always goes evenly into Q − 1, and the extra 1 is left over, so we always will have a remainder of 1. The two statements together allow us to find a unique numerical answer to the prompt questions. Combined, the statements are sufficient. Answer = (C)
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Re: For positive integers 10 < K < Z < Q, what is the value of the remaind
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03 Dec 2018, 16:03
aragonn wrote: For positive integers 10 < K < Z < Q, what is the value of the remainder when Q is divided by Z?
Statement #1: Q = \(K^2\)
Statement #2: Z = K + 1 Neither statement alone is going to help too much! Statement 1 doesn't tell us the value of Z. Statement 2 doesn't tell us the value of Q. You can't be 100% confident that each statement is insufficient just on that basis, but it's a good clue. Here's how I convinced myself: Statement 1: Suppose that Z is only 1 smaller than Q. In that case, the remainder of Q/Z should be 1. But if Z is 2 smaller than Q, the remainder should be 2. It's definitely possible for Z to be either of those numbers, since Q = K^2, which is much bigger than K  so there's a lot of 'wiggle room' for Z to have different values in between K and K^2. Statement 2: If Q is only 1 bigger than Z, the remainder is 1. If Q is 2 bigger than Z, the remainder is 2. Testing cases with numbers that are close together is useful for remainders! It's much easier to calculate the remainder of 99999999/99999998 than the remainder of 99999999/375. Not sufficient. Putting the two statements together, I'd like to test cases, since the math is getting a bit intense. Let's start with the easiest number we're allowed to use: K = 11. In that case, Z = 12 and Q = 11^2, which is 121. 121/12 has a remainder of 1. If K = 12, Z = 13 and Q = 12^2 = 144. 144/13 has a remainder of 1 as well. At this point, you could either choose C and move on, or you could try to come up with a mathematical justification to choose C. I thought about it like this: K goes into Q exactly K times, with a remainder of 0. Since Z is slightly bigger than K, but only by a tiny bit, we should be able to fit it into Q 1 fewer time. So, K+1 should go into Q K1 times. That gives you (K+1)(K1) = K^21, which has 1 left over when dividing into K^2.
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Re: For positive integers 10 < K < Z < Q, what is the value of the remaind
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03 Dec 2018, 16:03






