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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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26 May 2017, 04:21
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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? Answer Choices A. 12 B. 6 C. 4 D. 2 E. 0 Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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26 May 2017, 04:43
Reserving this space to post the official solution.
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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26 May 2017, 04:46
EgmatQuantExpert wrote: Q. x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? Answer Choices Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts Well, I don't 100% understand the question, but I'll take a stab at it. Since y<5, 1 is a possible value for y. Since 14=3, x would have to be 0 or 6 and that's not possible. So I'm going to try 1, which yields 14=5 so x could be 2. That will yield answer choice (D), and I don't see how you can reach zero since neither x nor y can be zero. I'm going with D.
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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26 May 2017, 04:58
eliaslatour wrote: Well, I don't 100% understand the question, but I'll take a stab at it.
Since y<5, 1 is a possible value for y. Since 14=3, x would have to be 0 or 6 and that's not possible.
So I'm going to try 1, which yields 14=5 so x could be 2. That will yield answer choice (D), and I don't see how you can reach zero since neither x nor y can be zero. I'm going with D. Nice try eliaslatour. Though I won't comment whether is correct or not, as I want the others to also give it a shot. I just wanted to point out one error in your analysis. When we write y < 5 and we remove the modulus from y, the range of y is not y < 5, but 5 < y < 5.Apart from that, kudos to your observation that since x and y are nonzero integers, the value of xy can never be 0. Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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26 May 2017, 05:25
EgmatQuantExpert wrote: Q. x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? Answer Choices Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy 5<x<5, 5<y<5 We need to solve for values of x & y to keep XY to minimum using x + 3 = y 4 by value substitution x =2 and y=1 fits the criteria keeping XY to minimum hence option D
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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28 May 2017, 04:45
I tried as mentioned below
Given :
5<x<5, 5<y<5 & x and y are non zero integers.
Therefore; x = {4,3,2,1,1,2,3,4} & y = {4,3,2,1,1,2,3,4}
x + 3 = y 4
Try plugging above values & above equation holds when 1. x=4 & y = 3 2. x=3 & y= 4 3. x=2 & y =1 4. x = 3 & y=2 5. x=4 & y = 3.
 xy will be max when x=2 & y = 1
Option D
This is how i arrived.
Thanx Narayana raju



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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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19 Jun 2017, 15:08
This is a good and tricky question. Given: lx+3l = ly4l  x & y are nonzero integers. lxl < 5 lyl < 5 Max of lxyl = ? As we have to find the max value, and the final max number will be negative as lxyl will be negative, we need to ensure that we get smallest value of negative possible which satisfies given equation. As per given information values of x and y can range as per below: 5 < x < 5 5 < y < 5 As we have to take least numbers, we will plug numbers from this range. Note we cannot take x & y as zero as per the given information, also it does not satisfy the given equation lx+3l = ly4l.Only values we can check for us 1 & 2 or 2 & 1 so maximum value we get is 2. Hence, Answer is D = 2
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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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29 Mar 2019, 11:40
x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy?
A. 12 B. 6 C. 4 D. 2 E. 0



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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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29 Mar 2019, 15:54
kiran120680 wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy?
A. 12 B. 6 C. 4 D. 2 E. 0 To maximize xy, we have to minimize xy Possible cases x + 3 = y 4(I) \(x+3=y4....x=y7\)........xy=y(y7)... y(y7) will be at least 6 even if we take y as 1. (II) \((x+3)=y4....x3=y4................y+x=1\).....let us take minimum possible value for this. Both are nonzero, so one of them 2 and other 1 will fit and xy=2(1)=2 so maximum value of xy=2. No need to check further as in the options the max value is 2.0 cannot be a value as both x and y are nonzero. D
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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29 Mar 2019, 23:29
kiran120680 wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy?
A. 12 B. 6 C. 4 D. 2 E. 0 Merging topics. Please search before posting.
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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30 Mar 2019, 00:00
kiran120680 wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy?
A. 12 B. 6 C. 4 D. 2 E. 0 simplify the expression x + 3 = y 4 x+3= y4 xy=7 or say x3=y4 (x+y)=1 x+y=1 so x=2 and y=1 possible and it satisfies x + 3 = y 4 lxyl ; 2 IMO D



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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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27 Apr 2019, 02:13
Archit3110 wrote: kiran120680 wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy?
A. 12 B. 6 C. 4 D. 2 E. 0 simplify the expression x + 3 = y 4 x+3= y4 xy=7or say x3=y4 (x+y)=1 x+y=1so x=2 and y=1possible and it satisfies x + 3 = y 4 lxyl ; 2 IMO D correct me if am wrong but the solution of eqution in red is x=3 and y=4 I tried a simlar approach and was struck for quite a bit . Press kudos if it helps!!



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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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27 Apr 2019, 02:19
chetan2u wrote: kiran120680 wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy?
A. 12 B. 6 C. 4 D. 2 E. 0 To maximize xy, we have to minimize xy Possible cases x + 3 = y 4(I) \(x+3=y4....x=y7\)........xy=y(y7)... y(y7) will be at least 6 even if we take y as 1. (II) \((x+3)=y4....x3=y4................y+x=1\).....let us take minimum possible value for this. Both are nonzero, so one of them 2 and other 1 will fit and xy=2(1)=2 so maximum value of xy=2. No need to check further as in the options the max value is 2.0 cannot be a value as both x and y are nonzero. D hello Egmat/Chetan sir, I tried solving the inequalities and landed up withe x=3 and y=4. Can you please suggest the right line of thinking when seeing such questions..I see your approach is clear...But seeing inequalities like this general tendency is to jump into solving them. Here after seeing the options it didnt help rather just plugging in values from options(assuming x<4 and y<5 wold have been far more beneficial)



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x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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27 Apr 2019, 04:29
I think this type of question is not too difficult if we solve things step by step. Below is my approach:
First things first, we have x < 5 and y < 5, which mean 5 < x < 5 and 5 < y < 5
Then, solve the equation:x + 3 = y 4
We apply this: A = B then A = B or A = B
=> x+3 = y4 or x+3 = 4y => x  y = 7 or x + y = 1
The possible values of x and y may be: 1. x  y = 7 > pairs of value (x,y) can be: (4;3) or (3;4) 2. x + y = 1 > (1;2) or (2;3) or (3;4)
Then xy can get these following values: 2, 6, 12 => 2 is the highest => D



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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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24 Oct 2019, 21:34
Hi.. I think the question is tricky and twisted. However, the answer 2 is definitely incorrect.
Why?
Because mod(xy) can have a maximum value of zero, right. Let's keep it as it will be useful.
Let's consider a pair of (x,y) as y=1, and x=0.
This pair definitely satisfies all the equations in the question.
So, a valid pair of (x,y) is (0,1), which will give a product of xy as zero.
Hence, mod(x,y)=0, which is the maximum attainable value.
Any thoughts?



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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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27 Oct 2019, 05:19
Harshit Srivastava wrote: Hi.. I think the question is tricky and twisted. However, the answer 2 is definitely incorrect.
Why?
Because mod(xy) can have a maximum value of zero, right. Let's keep it as it will be useful.
Let's consider a pair of (x,y) as y=1, and x=0.
This pair definitely satisfies all the equations in the question.
So, a valid pair of (x,y) is (0,1), which will give a product of xy as zero.
Hence, mod(x,y)=0, which is the maximum attainable value.
Any thoughts? hi it is given in the question that x and y are nonzero integers



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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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01 Nov 2019, 01:20
EgmatQuantExpert wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? Answer Choices A. 12 B. 6 C. 4 D. 2 E. 0 Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts Here is my approach to this question: Information gathering: We are given that the distance of x from 3 is equal to the distance of y from 4...(converting x + 3 = y 4 into a sentence)..(info 1) We are also given that distance of x from 0 is less than 5..(x < 5)..means 5<x<5 and distance of y from 0 is less than 5...(y < 5) 5<y<5 To find: Maximum possible value of xy Method: Plot points 5, 3, 0 , 4 and 5 on a number line...and start pairing numbers! pair 1: x = 4..which is 1 distance away from 3..hence y would also be 1 distance away from 4 as per info 1. For (4,3) xy = 12 = 12 Pair 2: x=2...which is 1 distance away from 3..hence y would also be 1 distance away from 4 as per info 1. For (2,3) xy = 6 = 6 Pair 3: x=1.....which is 2 distance away from 3..hence y would also be 2 distance away from 4 as per info 1. For (1,2) xy = 2 = 2 Since x and y are nonzero integers, there is no point testing x and y with 0 value. Among the 3 pairs, 2 is the greatest. Our answer is option D..2
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Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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01 Nov 2019, 02:13
This is 20 seconds problem imo We need to maximize xy; hence, check the given options from the bottom. We know that x and y are nonzero integers, eliminate E Now check for whether we can get xy=2, where 5<x<5 and 5<y<5 At x=1 and y=2, x + 3 = y 4 D EgmatQuantExpert wrote: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and y < 5, what is the maximum possible value of xy? Answer Choices A. 12 B. 6 C. 4 D. 2 E. 0 Thanks, Saquib Quant Expert eGMATRegister for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the realtime guidance of our Experts




Re: x + 3 = y 4, where x and y are nonzero integers. If x < 5 and
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