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sudaif

Post subject: tricky Question! Posted: Thu Jun 17, 2010 11:54 am 


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Help!
Is the sum of integers a and b divisible by 7? 1. a is not divisible by 7 2. ab is divisible by 7
a + b can be divisible by 7 in two scenarios Either a and b are both multiples of 7, OR both are nonmultiples, such that when you sum the remainders from dividing each one by 7, they equal to 7.
statement 1) a is not divisible by 7. this doesn't tell us anything about b. insufficient.
statement 2) a  b is divisible by 7. We can conceptually think about this....either both are multiples or both are not. we can pick numbers and we quickly realize that just because a  b is divisible by 7, does NOT mean that a + b is also divisible by 7. Example: a = 17, b = 3 or a = 7, b = 7. Thus insufficient.
statement 1 + statement 2: since a is not divisible by 7, then b must not be divisible by 7. Now if we pick numbers, say a = 17, b = 3, ab is divisible by 7 but a + b is not divisible by 7. Also, statement above implies that the remainder from a/7 and the remainder from b/7 must have been equal...so that with a  b the remainders summed to zero.
How do I quickly check if there are ANY sets of numbers which will allow the difference and the sum to be divisible by 7. OA is C





adiagr

Post subject: Re: tricky Question! Posted: Thu Jun 17, 2010 12:50 pm 


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Posts: 88

sudaif wrote: Help!
Is the sum of integers a and b divisible by 7? 1. a is not divisible by 7 2. ab is divisible by 7
a + b can be divisible by 7 in two scenarios Either a and b are both multiples of 7, OR both are nonmultiples, such that when you sum the remainders from dividing each one by 7, they equal to 7.
statement 1) a is not divisible by 7. this doesn't tell us anything about b. insufficient.
statement 2) a  b is divisible by 7. We can conceptually think about this....either both are multiples or both are not. we can pick numbers and we quickly realize that just because a  b is divisible by 7, does NOT mean that a + b is also divisible by 7. Example: a = 17, b = 3 or a = 7, b = 7. Thus insufficient.
statement 1 + statement 2: since a is not divisible by 7, then b must not be divisible by 7. Now if we pick numbers, say a = 17, b = 3, ab is divisible by 7 but a + b is not divisible by 7. Also, statement above implies that the remainder from a/7 and the remainder from b/7 must have been equal...so that with a  b the remainders summed to zero.
How do I quickly check if there are ANY sets of numbers which will allow the difference and the sum to be divisible by 7. OA is C Hi, It is tricky!! (1) by plugging in Nos.:
(a,b) = (8,1), (9,2), (18,4) (5,2) ab in all above cases is divisible by 7 but not a+b (Not conclusive I admit) (2) Algebra:a= 7q+r (r is less than 7 and not divisible by it) ...from (1) ab = 7k from (2) thus b= (7q+r)7k = 7(qk)+r so b works out to be not divisible by 7. a+b = (7q+r) + (7z+r) ......say z = qk will 2r be divisible by 7, by any chance? only if r is a multiple of 7, which it is not. so a+b will not be divisible by 7.
Let me admit In exam I would have gone for approach (1) and my intuition. Any better approach pls.?





sudaif

Post subject: Re: tricky Question! Posted: Thu Jun 17, 2010 2:13 pm 


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Posts: 125

yeah...actually i was thinking about it and realized that i was overlooking the effect of something that i had observed if a  b is divisible by 7 then, a/7 and b/7 must produce remainders that are equal  b/c only then will (ab) be divisible by 7. That is, the remainder from a/7 minus remainder from b/7 must equal zero. however... we know that there are no two SAME integers whose sum could equal 7. thus, a+b will NEVER be divisible by 7. therefore, statement 1 + statement 2 > sufficient. its a bit tricky to explain...although the concept is v straight fwd.





mschwrtz

Post subject: Re: tricky Question! Posted: Sun Jun 27, 2010 2:01 am 


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Posts: 504

That's right sudaif. Notice that any odd number would work just as 7 does, since no odd number is equal to two times some remainder.





muktarashmi

Post subject: Re: tricky Question! Posted: Thu Apr 07, 2011 3:32 pm 


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Posts: 15

if a  b is divisible by 7 then, a/7 and b/7 must produce remainders that are equal  b/c only then will (ab) be divisible by 7. That is, the remainder from a/7 minus remainder from b/7 must equal zero. ....................................I still dont understand this concept. How is that if ab is divisble by 7 then a/7 & b/7 must have same remainder?





jnelson0612

Post subject: Re: tricky Question! Posted: Wed Apr 13, 2011 3:13 pm 


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Posts: 2615

muktarashmi wrote: How is that if ab is divisble by 7 then a/7 & b/7 must have same remainder? Let's test some numbers to make this more clear. Let's pretend that a is 10. 10 is not divisible by 7; when I divide I get 1 remainder 3. If I want (10  b) to be divisible by 7, I have to say that b is 3. When 3 is divided by 7 I get 0 remainder 3. Test out other b values of 1, 2, 4, 5, 6, 7, 8. None of them have a remainder of 3 when divided by 7, and none of them will fit the parameters of ab is divisible by 7 if a is 10. Notice that whatever remainder I get for a needs to be duplicated in b so that when I subtract b from a the remainder will be removed and I will be left with a multiple of 7. Please write back if this is not more clear.
_________________ Jamie Nelson ManhattanGMAT Instructor





jp.jprasanna

Post subject: Re: tricky Question! Posted: Tue Mar 13, 2012 9:49 am 


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Posts: 203

sudaif wrote: yeah...actually i was thinking about it and realized that i was overlooking the effect of something that i had observed if a  b is divisible by 7 then, a/7 and b/7 must produce remainders that are equal  b/c only then will (ab) be divisible by 7. That is, the remainder from a/7 minus remainder from b/7 must equal zero.
I totally understand this part sudaif wrote: however... we know that there are no two SAME integers whose sum could equal 7. thus, a+b will NEVER be divisible by 7. therefore, statement 1 + statement 2 > sufficient. its a bit tricky to explain...although the concept is v straight fwd. I don't completely get the above... can some1 please elaborate...





jnelson0612

Post subject: Re: tricky Question! Posted: Mon Apr 02, 2012 9:06 am 


ManhattanGMAT Staff 

Posts: 2615

jp.jprasanna wrote: sudaif wrote: yeah...actually i was thinking about it and realized that i was overlooking the effect of something that i had observed if a  b is divisible by 7 then, a/7 and b/7 must produce remainders that are equal  b/c only then will (ab) be divisible by 7. That is, the remainder from a/7 minus remainder from b/7 must equal zero.
I totally understand this part sudaif wrote: however... we know that there are no two SAME integers whose sum could equal 7. thus, a+b will NEVER be divisible by 7. therefore, statement 1 + statement 2 > sufficient. its a bit tricky to explain...although the concept is v straight fwd. I don't completely get the above... can some1 please elaborate... Can you tell us more about what you don't understand? It sounds as if you do understand part of this statement. Thanks!
_________________ Jamie Nelson ManhattanGMAT Instructor





NMencia09

Post subject: Re: tricky Question! Posted: Wed Apr 04, 2012 11:14 am 


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la2ny

Post subject: Re: tricky Question! Posted: Mon Apr 09, 2012 4:47 pm 


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Posts: 14

jnelson0612 wrote: muktarashmi wrote: How is that if ab is divisble by 7 then a/7 & b/7 must have same remainder? Let's test some numbers to make this more clear. Let's pretend that a is 10. 10 is not divisible by 7; when I divide I get 1 remainder 3. If I want (10  b) to be divisible by 7, I have to say that b is 3. When 3 is divided by 7 I get 0 remainder 3. Test out other b values of 1, 2, 4, 5, 6, 7, 8. None of them have a remainder of 3 when divided by 7, and none of them will fit the parameters of ab is divisible by 7 if a is 10. Notice that whatever remainder I get for a needs to be duplicated in b so that when I subtract b from a the remainder will be removed and I will be left with a multiple of 7. Please write back if this is not more clear. Jamie, I understand the math, but what I'm not clear on is how we can determine divisibility of 7 by having equal remainders between the 2 terms (a  b). Maybe I'm just bogged down in this question too much, but I can't see this. The problem makes more sense to me if I just see it as plugging in numbers.





tim

Post subject: Re: tricky Question! Posted: Wed Apr 25, 2012 5:39 pm 


ManhattanGMAT Staff 

Posts: 4923 Location: Southwest Airlines, seat 21C

yes, OA is C..
when you take ab you subtract the remainders. if a and b have equal remainders, then the remainder of ab is 0, making it divisible by 7..
_________________ Tim Sanders Manhattan GMAT Instructor





krishnan.anju1987

Post subject: Re: tricky Question! Posted: Fri Aug 17, 2012 3:59 pm 


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Posts: 125

la2ny wrote: jnelson0612 wrote: muktarashmi wrote: How is that if ab is divisble by 7 then a/7 & b/7 must have same remainder? Let's test some numbers to make this more clear. Let's pretend that a is 10. 10 is not divisible by 7; when I divide I get 1 remainder 3. If I want (10  b) to be divisible by 7, I have to say that b is 3. When 3 is divided by 7 I get 0 remainder 3. Test out other b values of 1, 2, 4, 5, 6, 7, 8. None of them have a remainder of 3 when divided by 7, and none of them will fit the parameters of ab is divisible by 7 if a is 10. Notice that whatever remainder I get for a needs to be duplicated in b so that when I subtract b from a the remainder will be removed and I will be left with a multiple of 7. Please write back if this is not more clear. Jamie, I understand the math, but what I'm not clear on is how we can determine divisibility of 7 by having equal remainders between the 2 terms (a  b). Maybe I'm just bogged down in this question too much, but I can't see this. The problem makes more sense to me if I just see it as plugging in numbers. Hi, Here are my two cents on this. Hope it helps. now, we understand that 1 and 2 are insufficient alone. Now let's consider the statements together. 1) says that a is not divisible by 7. Hence, lets take a=24, 28 or something. Now ab is divisible by 7 which implies ab will give a number divisible by 7. for e.g if a=24, then b could be 3, 10 or so, for ab to be divisible by 7. Here if we note, all b does is remove the remainder obtained when a is divided by 7 from a. Since, a is not divisible by 7, a will be taking some value between x and x+7 where x is a random number. So a cannot be x and x+7. a could be x+1, x+2, x+3, x+4, x+5, x+6,. Based on b, b will have to remove this 1,2,3,4,5,6 from a(Note it can also remove something like 10 from 24 which is 7+3 and hence a way to remove the extra 3 from a) Now , lets consider a+b. since b removes the extra number from 1 to make it divisible by 7, it can never add the complement of that factor to a to make it divisible by 7. As an example, see below. It would be easier to understand the above explanation with this example. For e.g, a=10, b=3 and its complement( The number required to be added to a to make it divisible by 7) would be c(b)=73 B cant be 3 and 4 at the same time and thus can never make ab and a+b divisible by 7 at the same time.





tim

Post subject: Re: tricky Question! Posted: Tue Aug 21, 2012 11:39 am 


ManhattanGMAT Staff 

Posts: 4923 Location: Southwest Airlines, seat 21C

thanks!
_________________ Tim Sanders Manhattan GMAT Instructor






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