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| *Application of Remainder theory |
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Rey Fernandez
MGMAT STAFF
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darden11, please contact your friend to find the source of this question. Once you give that, we'll be able to reply. Also please follow the protocol of one question per post and of making the subject of your post the first few words of the question. Many thanks.
Rey |
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relentlesspursuito700plus
Guest
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These may NOT be correct, but here's my attempt. If I am wrong, please point out where my logic fails. I think this should be a good exercise for me and for someone correcting my potential mistakes (but hopefully I'm right).
I am actually confused by the first question. I am confused about how to interpret the answer choices for a question like this where the statements seems to contradict each other.... you'll see what I mean. First: The question asks, “What is the remainder when the positive integer x is divided by 7?” I would rephrase this to: “Is x divisible by 7?” BUT I would remind myself that I actually need to be able answer what exactly that remainder is. 1) X-1 is divisible by 7 2) x+1 is divisible by 7 Statement 1: If x-1 is divisible by 7, x by itself cannot be a multiple of seven (so not 7). This statement is telling me that x is always one greater than a multiple of 7. So x could be 8, 15, 22, 29, etc. Take these integers and divide it by 7 and you will find that the remainder is, of course, always 1. Statement 2: Do the same thing for statement 2 and you get 6 as the remainder (20, 27, etc). But this contradicts statement 1. Is that possible? But they each demonstrate that x isn't divisible by 7. So is they answer D because I was able to say "no" by using each statement separately? Or is it E becuase the answers contradict? [NOTE: This is where I am unsure of my logic] I'm not sure so I move on and combine what I know and see what happens. Here we have consecutive integers (x-1), x, (x+1), where the first term and third term are divisible by 7. If X+1 and X-1 are all multiples of 7, then you can multiply everything in this set by 7 to get: (7x-7), (7x), (7x+7). Obviously all the terms are divisible by 7, but that's because they are all multiplied by 7. This shows that x can be any of the integeres I listed above, 8, 15, or 20, 27, which all lead me to the same answer. The answer is still "no" x isn't divisible by 7. So I chose D. Second: What is the remainder when x ^ 4 – y ^ 4 is divided by 3 (1) When x-y is divided by 3, remainder is 0 (2) When x+y is divided by 3, remainder is 2 We don’t now what x ^ 4 – y ^ 4 is equal to but we can still factor this out. It takes several factoring but you can strip it down to (x^2+y^2)(x^2-y^2). Then you can take X^2-y^2 and strip it down again to (x+y)(x-y). Since statement 1 tells us one of the factors is divisible by three (a prime number), the entire expression must also be divisible by three according to prime factorization rule. Statement two isn’t very helpful since there are many integers that will yield a remainder of 2 when divided by 3, so x+y could be 5 or even 14. We don’t know. Therefore A. If x is a positive integer is x divisible by 7 1) X+29 is divisible by 10 2) x + 10 is divisible by 29 Statement 1 tell us that x + 29 (a prime number) is divisible by 10. Since 29 isn’t a multiple of 10, x + 29 must be a multiple of 10. This means x could be 1, 11, 21, or any other integer that has a unit’s digit of 1. If x was 21, yes it is divisible by 7, but if it’s 11, no it’s not divisible by 7. Insufficient. Statement 2 tells us the reverse. X + 10 must be a multiple of 29. This means that x is equal to an integer that is 10 less than 29, meaning it is 19N. If N is 7 then x would be a multiple of 7 if N is not a multiple of 7, then x isn’t a multiple of seven either. So this is insufficient by itself. But, both combined, if X+10 = 19(9) +10(9)= 29(9). We have a value of x that has a unit’s digit of 1 . So C. Both together are sufficient. |
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P
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700plus
I don't see how you get C as Sufficient for the last problem (If x is a positive integer is x divisible by 7). I would hate to disagree but at this point I think the answer is E. Could you please explain the last step where you put Statements 1 and 2 together to form answer C. Thanks |
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Guest
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The first question is garbage because the two statements contradict each other. There is no positive integer that you can both add 1 and subtract 1 from and is divisible by 7. Zero is the only number that goes evenly into seven when you add or subtract one from but you said that its a positive integer.
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| *Application of Remainder theory |
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