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How many integer values of x and y satisfy the expression 4x + 7y = 3
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29 Jun 2017, 10:39
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How many integer values of x and y satisfy the expression 4x + 7y = 3 where x<1000 and y < 1000?. a) 284 b) 286 c) 285 d) 290 e) 296
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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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13 Jul 2017, 18:39
shreyashree wrote: first we need to find out a solution which is valid.
4x + 7y = 3 x = 1, y = 1 is a valid solution.
The valid values of x will be in an Arithmetic Progression where the common difference is the coefficient of y. y = (34x)/7
=> Valid values of x are = ... 15, 8, 1, 6, 13, 20...
We are given x < 1000 => Valid values of x are = 995, 988... 15, 8, 1, 6, 13, 20.... 986, 993 (I substituted x with a few numbers around 990 & 990 and checked the resultant value of (3 4x) for divisibility by 7 to get the values of 993 and 995 respectively)
(perhaps there is a better method to do this faster  please suggest)
@x=993, y = 567 @x=995,y = 569 (x is the limiting factor)
Number of values that satisfy the given condition: = (Last term  First Term)/Common Difference + 1 = (993  (995))/7 + 1 = 284 + 1 = 285
So, there are 285 valid integer values of x and y that satisfy the given conditions. I believe that you started right, faltered a little bit in between, but still got the right answer. Kudos! Once we reach \(y=34x/7\) then we can conclude that values of x, which is multiple of 7, is the limiting factor. so we need to find multiples of 7 between 1000<x<1000
they are 994, 987,.......,0,.......987,994
calculating total no of multiples = (Last term  First Term)/Common Difference + 1 =[994(994)]/7 + 1 =284+1 =285
Hope this helps!




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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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29 Jun 2017, 10:58
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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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29 Jun 2017, 12:56
first we need to find out a solution which is valid.
4x + 7y = 3 x = 1, y = 1 is a valid solution.
The valid values of x will be in an Arithmetic Progression where the common difference is the coefficient of y. y = (34x)/7
=> Valid values of x are = ... 15, 8, 1, 6, 13, 20...
We are given x < 1000 => Valid values of x are = 995, 988... 15, 8, 1, 6, 13, 20.... 986, 993 (I substituted x with a few numbers around 990 & 990 and checked the resultant value of (3 4x) for divisibility by 7 to get the values of 993 and 995 respectively)
(perhaps there is a better method to do this faster  please suggest)
@x=993, y = 567 @x=995,y = 569 (x is the limiting factor)
Number of values that satisfy the given condition: = (Last term  First Term)/Common Difference + 1 = (993  (995))/7 + 1 = 284 + 1 = 285
So, there are 285 valid integer values of x and y that satisfy the given conditions.



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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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16 Jul 2017, 23:30
GMATAspirer09 wrote: shreyashree wrote: first we need to find out a solution which is valid.
4x + 7y = 3 x = 1, y = 1 is a valid solution.
The valid values of x will be in an Arithmetic Progression where the common difference is the coefficient of y. y = (34x)/7
=> Valid values of x are = ... 15, 8, 1, 6, 13, 20...
We are given x < 1000 => Valid values of x are = 995, 988... 15, 8, 1, 6, 13, 20.... 986, 993 (I substituted x with a few numbers around 990 & 990 and checked the resultant value of (3 4x) for divisibility by 7 to get the values of 993 and 995 respectively)
(perhaps there is a better method to do this faster  please suggest)
@x=993, y = 567 @x=995,y = 569 (x is the limiting factor)
Number of values that satisfy the given condition: = (Last term  First Term)/Common Difference + 1 = (993  (995))/7 + 1 = 284 + 1 = 285
So, there are 285 valid integer values of x and y that satisfy the given conditions. I believe that you started right, faltered a little bit in between, but still got the right answer. Kudos! Once we reach \(y=34x/7\) then we can conclude that values of x, which is multiple of 7, is the limiting factor. so we need to find multiples of 7 between 1000<x<1000
they are 994, 987,.......,0,.......987,994
calculating total no of multiples = (Last term  First Term)/Common Difference + 1 =[994(994)]/7 + 1 =284+1 =285
Hope this helps! Thank you! But wouldn't (34x) as a whole have to be a multiple of 7 and not just 4x(i.e x)?



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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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20 Jul 2017, 16:08
shreyashree wrote: first we need to find out a solution which is valid.
4x + 7y = 3 x = 1, y = 1 is a valid solution.
The valid values of x will be in an Arithmetic Progression where the common difference is the coefficient of y. y = (34x)/7
=> Valid values of x are = ... 15, 8, 1, 6, 13, 20...
We are given x < 1000 => Valid values of x are = 995, 988... 15, 8, 1, 6, 13, 20.... 986, 993 (I substituted x with a few numbers around 990 & 990 and checked the resultant value of (3 4x) for divisibility by 7 to get the values of 993 and 995 respectively)
(perhaps there is a better method to do this faster  please suggest)
@x=993, y = 567 @x=995,y = 569 (x is the limiting factor)
Number of values that satisfy the given condition: = (Last term  First Term)/Common Difference + 1 = (993  (995))/7 + 1 = 284 + 1 = 285
So, there are 285 valid integer values of x and y that satisfy the given conditions. You need to find the first integer solution that satisfies this equation. which is x = 1 and y =1, after that it can be observed that for the integral solution you can either take x or y as the limiting value to determine the solution. Since the equation is a sum of 4x and 7y it means if X will increase Y will decrease and vice versa to keep the value of sum to 3, without given limit of x < 1000 and y < 1000 it will have infinite solutions. so lets say you start with X, then for the next integral solution the value of x will change by +7 and value of y will change by 4 going towards the positive range of X Similarly the value of x will reduce by 7 and value of y will increase by +4 going towards negative range of X. So going towards +ve range of X, next values will be 1, 6, 13,20 and so on which you can express as multiple of 7 X= 7K1, where K = 0,1,2,.............142(You can check, which will give), max value will be 993 so total values of X = 1420+1 = 143 satisfying the condition x < 1000. Now similarly you can go towards the negative range of X and you will have 142 values X=7K1, where K=1,2,3.........................142, max value will be 995. In this case you can just use the value of K from the previous range and check. so total values of x in the negative range = 1(142)+1=142 total values of X = 143+142 = 285. Hope this approach is easy to follow.
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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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21 Jul 2017, 09:36
[quote="shreyashree"]first we need to find out a solution which is valid.
4x + 7y = 3 x = 1, y = 1 is a valid solution.
The valid values of x will be in an Arithmetic Progression where the common difference is the coefficient of y. y = (34x)/7
=> Valid values of x are = ... 15, 8, 1, 6, 13, 20... [quote] Can anyone please explain how the further values of x were determined (step by step), as I did not get it.



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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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22 Jul 2017, 08:38
Supermaverick wrote: How many integer values of x and y satisfy the expression 4x + 7y = 3 where x<1000 and y < 1000?.
a) 284 b) 286 c) 285 d) 290 e) 296 Hi Bunuel, I started with x=37y/4 and for every 4th integer value of y this condition gives an integer value of "x". From this I am getting #=500.(none of the option) Please reply, why is there a difference b/w the # of integer values that satisfy this equation.
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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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24 Jul 2017, 02:33
gmatexam439 wrote: Supermaverick wrote: How many integer values of x and y satisfy the expression 4x + 7y = 3 where x<1000 and y < 1000?.
a) 284 b) 286 c) 285 d) 290 e) 296 Hi Bunuel, I started with x=37y/4 and for every 4th integer value of y this condition gives an integer value of "x". From this I am getting #=500.(none of the option) Please reply, why is there a difference b/w the # of integer values that satisfy this equation. As per the earlier mentioned method in the string, the number of values for 'y' that satisfy this equation is 285. Using similar method we can find that the number of values of 'x' that satisfy the equation are 500, as you correctly mentioned. Hence in this case the number of values of 'y' that satisfy the equation will be the limiting factor, i.e. 285 values of 'y' can use only 285 values of 'x' to satisfy this equation. Moreover from 999 to +999 there will obviously be more multiples of 4 than 7. Hope this explanation helped.



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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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24 Jul 2017, 08:54
Dkingdom wrote: gmatexam439 wrote: Supermaverick wrote: How many integer values of x and y satisfy the expression 4x + 7y = 3 where x<1000 and y < 1000?.
a) 284 b) 286 c) 285 d) 290 e) 296 Hi Bunuel, I started with x=37y/4 and for every 4th integer value of y this condition gives an integer value of "x". From this I am getting #=500.(none of the option) Please reply, why is there a difference b/w the # of integer values that satisfy this equation. As per the earlier mentioned method in the string, the number of values for 'y' that satisfy this equation is 285. Using similar method we can find that the number of values of 'x' that satisfy the equation are 500, as you correctly mentioned. Hence in this case the number of values of 'y' that satisfy the equation will be the limiting factor, i.e. 285 values of 'y' can use only 285 values of 'x' to satisfy this equation. Moreover from 999 to +999 there will obviously be more multiples of 4 than 7. Hope this explanation helped. In short if we start wrongly, we need to find both. So, essentially, we should have looked for the limiting factor first. Thanks a lot mate.
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Long And A Fruitful Journey  V21 to V41; If I can, So Can You!! Preparing for RC my way My study resources:1. Useful Formulae, Concepts and TricksQuant2. eGMAT's ALL SC Compilation3. LSAT RC compilation4. Actual LSAT CR collection by Broal5. QOTD RC (Carcass)6. Challange OG RC7. GMAT Prep Challenge RC



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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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26 Jul 2017, 05:01
GMATAspirer09 wrote: shreyashree wrote: first we need to find out a solution which is valid.
4x + 7y = 3 x = 1, y = 1 is a valid solution.
The valid values of x will be in an Arithmetic Progression where the common difference is the coefficient of y. y = (34x)/7
=> Valid values of x are = ... 15, 8, 1, 6, 13, 20...
We are given x < 1000 => Valid values of x are = 995, 988... 15, 8, 1, 6, 13, 20.... 986, 993 (I substituted x with a few numbers around 990 & 990 and checked the resultant value of (3 4x) for divisibility by 7 to get the values of 993 and 995 respectively)
(perhaps there is a better method to do this faster  please suggest)
@x=993, y = 567 @x=995,y = 569 (x is the limiting factor)
Number of values that satisfy the given condition: = (Last term  First Term)/Common Difference + 1 = (993  (995))/7 + 1 = 284 + 1 = 285
So, there are 285 valid integer values of x and y that satisfy the given conditions. I believe that you started right, faltered a little bit in between, but still got the right answer. Kudos! Once we reach \(y=34x/7\) then we can conclude that values of x, which is multiple of 7, is the limiting factor. so we need to find multiples of 7 between 1000<x<1000
they are 994, 987,.......,0,.......987,994
calculating total no of multiples = (Last term  First Term)/Common Difference + 1 =[994(994)]/7 + 1 =284+1 =285
Hope this helps! Lucid explanation, thank you! Do please post more questions like these if possible! I just have one more query : what about 1000<y<1000  why do we not use these limits here?



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Re: How many integer values of x and y satisfy the expression 4x + 7y = 3
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