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sudaif
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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. a-b 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 non-multiples, 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, a-b 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
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adiagr
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Post subject: Re: tricky Question! Posted: Thu Jun 17, 2010 12:50 pm |
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sudaif wrote: Help!
Is the sum of integers a and b divisible by 7? 1. a is not divisible by 7 2. a-b 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 non-multiples, 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, a-b 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) a-b 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) a-b = 7k from (2) thus b= (7q+r)-7k = 7(q-k)+r so b works out to be not divisible by 7. a+b = (7q+r) + (7z+r) ......say z = q-k 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.?
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sudaif
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Post subject: Re: tricky Question! Posted: Thu Jun 17, 2010 2:13 pm |
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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 (a-b) 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.
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mschwrtz
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Post subject: Re: tricky Question! Posted: Sun Jun 27, 2010 2:01 am |
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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.
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muktarashmi
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Post subject: Re: tricky Question! Posted: Thu Apr 07, 2011 3:32 pm |
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if a - b is divisible by 7 then, a/7 and b/7 must produce remainders that are equal - b/c only then will (a-b) 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 a-b is divisble by 7 then a/7 & b/7 must have same remainder?
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jnelson0612
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Post subject: Re: tricky Question! Posted: Wed Apr 13, 2011 3:13 pm |
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muktarashmi wrote: How is that if a-b 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 a-b 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
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jp.jprasanna
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Post subject: Re: tricky Question! Posted: Tue Mar 13, 2012 9:49 am |
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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 (a-b) 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...
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jnelson0612
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Post subject: Re: tricky Question! Posted: Mon Apr 02, 2012 9:06 am |
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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 (a-b) 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
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NMencia09
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Post subject: Re: tricky Question! Posted: Wed Apr 04, 2012 11:14 am |
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la2ny
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Post subject: Re: tricky Question! Posted: Mon Apr 09, 2012 4:47 pm |
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jnelson0612 wrote: muktarashmi wrote: How is that if a-b 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 a-b 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.
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tim
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Post subject: Re: tricky Question! Posted: Wed Apr 25, 2012 5:39 pm |
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Posts: 2242 Location: Southwest Airlines, seat 21C
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yes, OA is C..
when you take a-b you subtract the remainders. if a and b have equal remainders, then the remainder of a-b is 0, making it divisible by 7..
_________________ Tim Sanders Manhattan GMAT Instructor
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