
Author 
Message 
sudaif

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


Course Students 

Posts: 125

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 


Students 

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 


Course Students 

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 


ManhattanGMAT Staff 

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 


Course Students 

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 


ManhattanGMAT Staff 

Posts: 2597

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 


Students 

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: 2597

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 


Course Students 

Posts: 36





la2ny

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


Course Students 

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: 4847 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 


Forum Guests 

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: 4847 Location: Southwest Airlines, seat 21C

thanks!
_________________ Tim Sanders Manhattan GMAT Instructor






Users browsing this forum: No registered users and 0 guests 



You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot post attachments in this forum



