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When \(x\) is divided by 5, the remainder is 2. Is \(x\) divisible by 7? 1. \(x\) is a prime number 2. \(x + 3\) is a multiple of 10 Source: GMAT Club Tests  hardest GMAT questions I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did: I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any. 1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff. 2. I got the equation: x+3=10c, so x=10c3, where c is a constant. since we already know x=5a+2 from the question, we can set up the eqn as follows: 10c3=5a+2 10c=5a+5 2c=a+1 Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly?



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wcgmatclub wrote: When \(x\) is divided by 5, the remainder is 2. Is \(x\) divisible by 7?
1. \(x\) is a prime number 2. \(x + 3\) is a multiple of 10
I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did:
I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any.
1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff.
2. I got the equation: x+3=10c, so x=10c3, where c is a constant. since we already know x=5a+2 from the question, we can set up the eqn as follows: 10c3=5a+2 10c=5a+5 2c=a+1
Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly? The highlighted statements are not correct as a and c are not constants. in fact they are variables. 1. \(x\) is a prime number: x could be 7 or 17 or 37 or 47 and so on.................. 2. \(x + 3\) is a multiple of 10: x could be again 7 or or 17 or 37 or 47 and so on.................. Togather also same. so E.
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Re: M02, 12 [#permalink]
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05 Feb 2009, 21:18
my bad in mislabeling "a" and "c" as constants instead of variables, but are there any mistakes in terms of my logic and approach in solving this type of problems? I understand there're multiple solutions for both 1) and 2). But I'm trying to understand how to reach that conclusion outside of brutally plugging in different #'s until I find the answer. This doesn't seem very efficient use of time on the GMAT.
Thanks in advance.



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Re: M02, 12 [#permalink]
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06 Feb 2009, 19:54
wcgmatclub wrote: my bad in mislabeling "a" and "c" as constants instead of variables, but are there any mistakes in terms of my logic and approach in solving this type of problems? I understand there're multiple solutions for both 1) and 2). But I'm trying to understand how to reach that conclusion outside of brutally plugging in different #'s until I find the answer. This doesn't seem very efficient use of time on the GMAT.
Thanks in advance. Your approach for 1 is correct but 2 leads nowhere. Since you are looking for y, not a or c.
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Re: M02 #12 [#permalink]
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17 Mar 2010, 06:02
wcgc wrote: When \(x\) is divided by 5, the remainder is 2. Is \(x\) divisible by 7? 1. \(x\) is a prime number 2. \(x + 3\) is a multiple of 10 Source: GMAT Club Tests  hardest GMAT questions I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did: I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any. 1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff. 2. I got the equation: x+3=10c, so x=10c3, where c is a constant. since we already know x=5a+2 from the question, we can set up the eqn as follows: 10c3=5a+2 10c=5a+5 2c=a+1 Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly? From Statement 1: X is a prime number. from this we can say that x should be 7 or 17 or 47.... 7 is divisible by 7 but not 17  not sufficient From Statement 2: X+3 is multiple of 10 so again numbers should be 7, 17, 27... again 7 is divisible by 7 but not 17.. for these type of questions, its easier to solve by picking numbers instead of writing euqations..



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Re: M02 #12 [#permalink]
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17 Mar 2010, 10:06
wcgc wrote: When \(x\) is divided by 5, the remainder is 2. Is \(x\) divisible by 7? 1. \(x\) is a prime number 2. \(x + 3\) is a multiple of 10 Source: GMAT Club Tests  hardest GMAT questions I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any. Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly? As statement 1 if x is prime number than it cannot be divided by 7 except 7 itself so we cannot say fully that x is divisible by 7 or not. As per statement 2 is x+3 is multiple of 10 then x can be 7,17,27,37,47,57,67,77 ... so from this 7,77 are divisible by 7 and others are not so insufficient. Combininh also no use hence answer is E



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Re: M02 #12 [#permalink]
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19 Mar 2010, 07:34
Its E both are not sufficient
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Re: M02 #12 [#permalink]
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21 Mar 2011, 06:09
lionslion wrote: wcgc wrote: When \(x\) is divided by 5, the remainder is 2. Is \(x\) divisible by 7? 1. \(x\) is a prime number 2. \(x + 3\) is a multiple of 10 Source: GMAT Club Tests  hardest GMAT questions I got B, the official answer is E. I can see why the official answer is correct, but I'm not sure where my approach went wrong. Below is what I did: I simplified the given equation to x=5a+2, where a is a constant, we're looking for y in x=7b+y, where y is the value of the remainder, if there's any. 1. I simply plugged in different values of x and see if there're more than 1 prime # that fits the bill. Starting w/a=1, I got x=7, 12, 17. Since both 7 and 17 are prime # and 7 is divisible by 7 but 17 isn't. 1 is insuff. 2. I got the equation: x+3=10c, so x=10c3, where c is a constant. since we already know x=5a+2 from the question, we can set up the eqn as follows: 10c3=5a+2 10c=5a+5 2c=a+1 Now I got stuck. It appears to me from the eqn. that x can not be divisible by 7. So what is going on here? or am I simply not interpreting the eqn. correctly? From Statement 1: X is a prime number. from this we can say that x should be 7 or 17 or 47.... 7 is divisible by 7 but not 17  not sufficient From Statement 2: X+3 is multiple of 10 so again numbers should be 7, 17, 27... again 7 is divisible by 7 but not 17.. for these type of questions, its easier to solve by picking numbers instead of writing euqations.. I agree, it is much faster to just plug in numbers then to derive an equation in this case



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Re: M02 #12 [#permalink]
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21 Mar 2011, 23:51
lionslion wrote:
From Statement 1: X is a prime number. from this we can say that x should be 7 or 17 or 47.... 7 is divisible by 7 but not 17  not sufficient
From Statement 2: X+3 is multiple of 10 so again numbers should be 7, 17, 27... again 7 is divisible by 7 but not 17..
for these type of questions, its easier to solve by picking numbers instead of writing euqations..
Nicely explained. Going with equations is time consuming.
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Re: M02 #12 [#permalink]
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22 Mar 2011, 01:34
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x = 5k + 2, where k is an integer x = 7,12,19,22,27,32... So is x = 7n, where n is an integer ? From (1), x is prime number x = 7,17.. so (1) is not sufficient From (2) x+3 = 10m x = 7, 17, 27 , So (2) is not sufficient (1) and (2) are also not sufficient, so answer is E.
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Re: M02 #12 [#permalink]
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24 Mar 2012, 15:12
I got E, I simply tried a couple numbers and showed that in some cases it was divisible by 7 and others it wasn't.



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Re: M02 #12 [#permalink]
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25 Mar 2012, 06:12
E
When is divided by 5, the remainder is 2. Is divisible by 7?
Stmt says X when divided by 5 gives Remainder (R) = 2... that means X=(5*a+2) X can be 2, 7, 12, 17, 22... (putting a =0,1,2...) and asks is X divisible by 7?
1. X is a prime number
this says X is prime both 7 and 17 are prime, not sufficient
2. (x+ 3) is a multiple of 10 again taking 7 and 17, both (7+3) and (17+3) are multiple of 10. not sufficient.
Since same numbers are used for both 1 and 2, answer E



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Re: M02 #12 [#permalink]
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26 Mar 2012, 01:13
Its E..... I got it wrong because i was focusing on statement 2 in wrong way.....i was considering only multiples of 7



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Re: M02 #12 [#permalink]
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26 Mar 2012, 07:16
one observation i have made is , when both the options are saying same ( like wht we have now)... 1. If Option A is suffcient, then Option B would be sufficient too making answer D 2. If Option A is insuffcient, then Option B would be insufficient too making answer E.... I go with E for the above question n i agree with the above explantions....
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Re: M02 #12 [#permalink]
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26 Mar 2013, 05:15
When is divided by 5, the remainder is 2. Is divisible by 7?
1. is a prime number 2. x+3 is a multiple of 10
The numbers for two sets: SET A:.7,17,27,37 etc SET B:12,22,32,42 etc
for Case 1: we cannot draw a conclusion as SET A contains prime and non prime nos. for Case 2: the statement is false for SET B data.
Combining also we don't get a true correlation.
Hence IMO E.



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Re: M02 #12 [#permalink]
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26 Mar 2013, 19:05
qtn says x leaves reminder 2 when divided by 5 => 2,7,12,17,27,32,37,....
1. x is a prime number =>2,3,5,7,11,13,17,... 2. x+3 is a multiple of 10 => 7,17,27,37,...
Checking choice 1: Qtn & 1 => 2,7,11,... 2 is not divisible by 7 7 is divisible by 7 => choice 1 insuffecient
Checking choice 2: Qtn & 1 => 7,17,27,... 7 is divisible by 7 17 is not divisible by 7 => choice 2 insuffecient
Combine 1+2: 7,17,.. 7 is divisible by 7 17 is not divisible by 7 => choice 1+2 together is insuffecient
So E, hope it helps



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Re: M02 #12 [#permalink]
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27 Mar 2013, 06:01
5a + 2 = x (given)
1) X is a prime number, prime numbers 2,3,5,7,11
a = 0 in given equation makes X = 2. a = 1 in given equation makes X = 7.
two different answers, insufficient.
2) X+3 is a multiple of 10
If X = 7, then X+3 = 10, is a multiple of 10 If X =17, then X+3 = 20, is a multiple of 10
again two different answers, insufficient.
Combining 1 & 2 insufficient as, X can be 17 & 7
Solution E



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Re: M02 #12 [#permalink]
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27 Mar 2013, 11:39
for option 1 > I picked couple of examples like 7,17,37 which are prime numbers and leaves remainder as 2. In all three of them, first one 7 is divisible by 7 but rest 17 and 37 are not. Clearly, we can not confirm the question under investigation "is number divisible by 7 or not?" for option 2 > Again picked examples like 7, 17, 27, 37 and so on.. In all of them, first one is divisible by 7 not rest are not. So we can not again be pretty sure if the number is divisible by 7 or not. Therefore, I go with option E.
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