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


Posts: 125
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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 Post subject: Re: tricky Question!
 Post Posted: Thu Jun 17, 2010 12:50 pm 
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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. 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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 Post subject: Re: tricky Question!
 Post 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 (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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 Post subject: Re: tricky Question!
 Post Posted: Sun Jun 27, 2010 2:01 am 
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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.


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


Posts: 2615
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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 Post subject: Re: tricky Question!
 Post Posted: Tue Mar 13, 2012 9:49 am 
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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 (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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 Post subject: Re: tricky Question!
 Post Posted: Mon Apr 02, 2012 9:06 am 
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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 (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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 Post subject: Re: tricky Question!
 Post Posted: Wed Apr 04, 2012 11:14 am 
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Course Students


Posts: 36
OA is C right?


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 Post subject: Re: tricky Question!
 Post Posted: Mon Apr 09, 2012 4:47 pm 
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Posts: 14
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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 Post subject: Re: tricky Question!
 Post Posted: Wed Apr 25, 2012 5:39 pm 
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ManhattanGMAT Staff


Posts: 4923
Location: Southwest Airlines, seat 21C
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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 Post subject: Re: tricky Question!
 Post Posted: Fri Aug 17, 2012 3:59 pm 
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Forum Guests


Posts: 125
la2ny wrote:
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.


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 a-b is divisible by 7 which implies a-b will give a number divisible by 7. for e.g if a=24, then b could be 3, 10 or so, for a-b 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)=7-3 B cant be 3 and 4 at the same time and thus can never make a-b and a+b divisible by 7 at the same time.


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


Posts: 4923
Location: Southwest Airlines, seat 21C
thanks!

_________________
Tim Sanders
Manhattan GMAT Instructor


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