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Is the sum of integers a and b divisible by 7?

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Is the sum of integers a and b divisible by 7?  [#permalink]

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New post 22 Oct 2009, 20:04
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A
B
C
D
E

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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

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Re: Data Sufficiency, Number Properties  [#permalink]

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New post 22 Oct 2009, 20:30
10
4
mirzohidjon wrote:
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

It has explanation in MGMAT test, but it is quite vague, and difficult to understand for me.
Please, elaborate...


(1) \(a=7p+r\) --> know nothing about b. Not sufficient.

(2) \(a-b=7q\) --> \(a+b=?\) --> Nor sufficient

(1)+(2) \(a=7p+r\) and \(a-b=7q\) --> \(b=a-7q\).

\(a+b=(7p+r)+(a-7q)=7p+r+7p+r-7q=7(2p-q)+2r\) --> \(7(2p-q)\) is divisible by 7. Since \(r\) is remainder from a being divided by 7, \(r\) and thus \(2r\) is not divisible by 7. So, \(a+b=7(2p-q)+2r\) is not divisible by 7. Sufficient.

Answer: C.
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Re: Data Sufficiency, Number Properties  [#permalink]

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New post 22 Oct 2009, 21:48
Thanks for fast explanation, this way, I think it better to understand, rather the explanation suggested my MGMAT.
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Re: divisibility & primes  [#permalink]

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New post 24 Jun 2011, 13:15
I think it is C

Clearly both of those statements are insufficient if taken individually,

If we are considering both of 'em together! then,

a = 12, b = 5, then a+b is not divisible by 7

a= 6, b = -1, then a + b is not divisible by 7

consistent answer: no; therefore, I guess it is C

Note that in both of the above cases a-b is divisible by 7 (Stmt II) and a is not divisible by 7 (statment I)
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Re: divisibility & primes  [#permalink]

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New post 24 Jun 2011, 14:14
fluke wrote:
So, the answer is C as we have proven that "a+b" is NOT divisible by 7 using both statements. What's the confusion or there isn't any?


Hi,

Please tell me if im thinking right:

generally the rule is: multiple of a number + or - multiple of a number = multiple of a number.

(1) says a = not a multiple of 7.
(2) a-b = multiple of 7

but since we don't know the exact values of the a or b, we cant say whether a+b is divisible by 7. Hence (c) gives us a definitive NO answer.?
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Re: divisibility & primes  [#permalink]

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New post 24 Jun 2011, 14:38
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386390 wrote:
fluke wrote:
So, the answer is C as we have proven that "a+b" is NOT divisible by 7 using both statements. What's the confusion or there isn't any?


Hi,

Please tell me if im thinking right:

generally the rule is: multiple of a number + or - multiple of a number = multiple of a number.

(1) says a = not a multiple of 7.
(2) a-b = multiple of 7

but since we don't know the exact values of the a or b, we cant say whether a+b is divisible by 7. Hence (c) gives us a definitive NO answer.?


The rule you stated is correct. But, we need to take that rule one step further.

Using 1 and 2, we can definitely say that "a+b" is NOT divisible by 7, irrespective of what value "a" or "b" bears.

Just because we can DEFINITELY answer the question as NO, the statements together become sufficient.

Question is: Is a+b divisible by 7?
Answer: No, "a+b" is not divisible by 7.

If we can answer the question asked in a stem as a definite YES or a definite NO, only then the statement(s) will be sufficient.

If the answer is: Maybe a+b is divisible by 7, then the statements become INSUFFICIENT.

In this case: a+b is definitely NOT divisible by 7. No matter what value you associate a or b with.

1. a is not divisible by 7.
a=5
b=2
a+b is divisible by 7.
****************
a=5
b=3
a+b is NOT divisible by 7.

Thus, NOT SUFFICIENT.

2. a-b is divisible by 7.
a=14
b=7
a+b=21 is divisible by 7.
a=9
b=2
a+b=11 is NOT divisible by 7.

Thus, NOT SUFFICIENT.

Together:
a is NOT divisible by 7.
BUT a-b is divisible by 7.
Means, b is also NOT divisible by 7.

a=9
b=2
a-b=7; divisible by 7.
But, a+b=11; NOT divisible by 7.

You can as many examples as you want that satisfy both statements and you will find that a+b is never divisible by 7.

a=23
b=2
a-b=21; divisible by 7.
a+b=25; NOT divisible by 7.

Thus, answer is "C"
*****************************************
You can try this with other examples but you will never find a value for a or b, such that, a+b is divisible by 7. Just make sure that you don't violate any condition stated in the stem, statement 1 or statement 2.
*************************************
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Re: divisibility & primes  [#permalink]

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New post 24 Jun 2011, 16:58
If a+b and a-b are *both* multiples of 7, then their sum must be a multiple of 7 (if you add two multiples of 7, you always get a multiple of 7), so if both are multiples of 7, then so is a+b+a-b = 2a, and a would thus also be a multiple of 7. We know from Statement 1 that's not true, so it's impossible for a+b and a-b to both be multiples of 7. Thus with both Statements we know a+b is *not* a multiple of 7 and the answer must be 'no', so the answer is C.

I'd add that I see far too many prep company questions where you have sufficient information to give a 'no' answer to the question. Such questions are exceedingly rare on the actual GMAT - there's only one such question in the two official guides combined (OG12 and Quant Review).
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Re: divisibility & primes  [#permalink]

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New post 24 Jun 2011, 17:38
1. Not sufficient
a is not divisible by 7

a b a+b divisible by 7
13 1 yes
13 2 no

2. Not sufficient

a-b is divisible by 7

a b a+b divisible by 7
14 7 yes
15 1 no

together

we have a-b is divisible by 7 and a is not divisible by 7
=> b is also not divisible by 7


a b a+b is divisible by 7
15 1 no
22 15 no

Sufficient to answer .

Answer is C.

answer is C.
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Re: divisibility & primes  [#permalink]

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New post 25 Jun 2011, 07:41
Thanks for confirming Fluke and Ian.

@ IanStewart

I think that's what put me off. I was by default looking for sufficient YES answer. Whereas this one deals with a definitive NO. But thanks for highlighting that. At least now this will sit in the back of my head.
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MGMAT Number Properties Ch 10 #10  [#permalink]

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New post 16 Dec 2011, 00:39
Math masters:

I have a quick question on MGMAT Number Properties Chapter 10 Question 10.

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
2) a-b is divisible by 7

So...I can get the right answer, but in the explanation I saw somethign that got me thinking about the answer if the question had used the number "6" instead of "7"

"1 and 2 combined tell us that a is not divisible by 7, but a-b is divisible by 7. This tells us a and b have the same remainder when divided by 7: if a-b is divisible by 7, then the remainder of a-b is zero. Therefore, the remainders of a and b must be equal"

The answer to the original question is no. But if the if the target # is 6, is the answer appears to be yes. It seems that the answer is yes for any even # and no for any odd #. Confirm?
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Re: MGMAT Number Properties Ch 10 #10  [#permalink]

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New post 16 Dec 2011, 02:48
alphastrike wrote:
Math masters:

I have a quick question on MGMAT Number Properties Chapter 10 Question 10.

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
2) a-b is divisible by 7

So...I can get the right answer, but in the explanation I saw somethign that got me thinking about the answer if the question had used the number "6" instead of "7"

"1 and 2 combined tell us that a is not divisible by 7, but a-b is divisible by 7. This tells us a and b have the same remainder when divided by 7: if a-b is divisible by 7, then the remainder of a-b is zero. Therefore, the remainders of a and b must be equal"

The answer to the original question is no. But if the if the target # is 6, is the answer appears to be yes. It seems that the answer is yes for any even # and no for any odd #. Confirm?


If you replace '7' everywhere in the question with '6', or with any other even number greater than 2, the answer would not be 'yes' using both Statements - it would sometimes be 'yes' and sometimes 'no', so the answer to the question would be E. So if we had this question:

Is the sum of positive integers a and b divisible by 6?
1) a is not divisible by 6
2) a-b is divisible by 6


then using both Statements it might be that a = 7 and b = 1, in which case the answer to the question is 'no', or it might be that a = 9 and b = 3, in which case the answer to the question is 'yes'.
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Re: MGMAT Number Properties Ch 10 #10  [#permalink]

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New post 03 Apr 2012, 07:16
It's C:

1) a+b is divisible by 7 for a=1, b=6 and is not divisible by 7 for a=1, b=5. Insufficient
2) a+b is divisible by 7 for a=7, b=7 and is not divisible by 7 for a=6, b=6. Insufficient
1&2) from 2 we know that a and b has the same remainder. from 1 we know that the remainder can be {1...6}
if a = 7n+r and b=7m+r then

a + b = 7n+r + 7m + r = 7(n+m) + 2r

2r = {1..6}*2 = {2,4,6,8,10,12} - none of the possible remainders is divisible by 7, so a+b is not divisible by 7. Sufficient
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Re: MGMAT Number Properties Ch 10 #10  [#permalink]

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New post 06 Apr 2012, 02:26
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alphastrike wrote:
Math masters:

I have a quick question on MGMAT Number Properties Chapter 10 Question 10.

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
2) a-b is divisible by 7

So...I can get the right answer, but in the explanation I saw somethign that got me thinking about the answer if the question had used the number "6" instead of "7"

"1 and 2 combined tell us that a is not divisible by 7, but a-b is divisible by 7. This tells us a and b have the same remainder when divided by 7: if a-b is divisible by 7, then the remainder of a-b is zero. Therefore, the remainders of a and b must be equal"

The answer to the original question is no. But if the if the target # is 6, is the answer appears to be yes. It seems that the answer is yes for any even # and no for any odd #. Confirm?


Here is the logical explanation to the different answers (C and E) in the two cases (7 and 6).
First, let me explain why the answer is (C) in the original question.

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
So a is of one of the following formats: 7n+1, 7n+2, 7n+3, 7n+4, 7n+5 and 7n+6

2) a-b is divisible by 7
The format of b is the same as the format of a i.e. if a = 7n + 3, then b = 7m + 3 since a and b vary by a multiple of 7
e.g. if a = 43 (= 7*6+1), then b could be 22 (= 7*3 + 1). The difference between them has to be a multiple of 7 so the remainder obtained by dividing a by 7 will be the same as the remainder obtained by dividing b by 7.

Using both statements, when you sum a and b, you get,
a+b = 7n+1 + 7m+1 = 7(n+m) + 2
or
a+b = 7n+2 + 7m+2 = 7(n+m) + 4
or
a+b = 7n+3 + 7m+3 = 7(n+m) + 6
etc
I hope you see that the remainder will be some even number (either 2 or 4 or 6 or 8 or 10 or 12). The remainder will never be 7 so the sum will never be a multiple of 7.

What happens if we replace 7 by 6?
Same logic.
a+b = 6n+1 + 6m+1 = 6(n+m) + 2
or
a+b = 6n+2 + 6m+2 = 6(n+m) + 4
or
a+b = 6n+3 + 6m+3 = 6(n+m) + 6
Let me stop here. Did you see something interesting? The remainder in the last case is 6 which means a+b is divisible by 6. In some cases, a+b will not be divisible by 6 and in some cases, it will be.
Hence your answer will be (E).
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Re: Data Sufficiency, Number Properties  [#permalink]

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New post 23 May 2012, 08:50
Bunuel wrote:
mirzohidjon wrote:
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

It has explanation in MGMAT test, but it is quite vague, and difficult to understand for me.
Please, elaborate...


(1) \(a=7p+r\) --> know nothing about b. Not sufficient.
(2) \(a-b=7q\) --> \(a+b=?\) --> Nor sufficient

(1)+(2) \(a=7p+r\) and \(a-b=7q\) --> \(b=a-7q\).

\(a+b=(7p+r)+(a-7q)=7p+r+7p+r-7q=7(2p-q)+2r\) --> \(7(2p-q)\) is divisible by 7. Since \(r\) is remainder from a being divided by 7, \(r\) and thus \(2r\) is not divisible by 7. So, \(a+b=7(2p-q)+2r\) is not divisible by 7. Sufficient.

Answer: C.

Thanks Bunuel... great explanation.. really helps
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Re: MGMAT Number Properties Ch 10 #10  [#permalink]

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New post 24 May 2012, 05:16
VeritasPrepKarishma wrote:
alphastrike wrote:
Math masters:

I have a quick question on MGMAT Number Properties Chapter 10 Question 10.

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
2) a-b is divisible by 7

So...I can get the right answer, but in the explanation I saw somethign that got me thinking about the answer if the question had used the number "6" instead of "7"

"1 and 2 combined tell us that a is not divisible by 7, but a-b is divisible by 7. This tells us a and b have the same remainder when divided by 7: if a-b is divisible by 7, then the remainder of a-b is zero. Therefore, the remainders of a and b must be equal"

The answer to the original question is no. But if the if the target # is 6, is the answer appears to be yes. It seems that the answer is yes for any even # and no for any odd #. Confirm?


Here is the logical explanation to the different answers (C and E) in the two cases (7 and 6).
First, let me explain why the answer is (C) in the original question.

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
So a is of one of the following formats: 7n+1, 7n+2, 7n+3, 7n+4, 7n+5 and 7n+6

2) a-b is divisible by 7
The format of b is the same as the format of a i.e. if a = 7n + 3, then b = 7m + 3 since a and b vary by a multiple of 7
e.g. if a = 43 (= 7*6+1), then b could be 22 (= 7*3 + 1). The difference between them has to be a multiple of 7 so the remainder obtained by dividing a by 7 will be the same as the remainder obtained by dividing b by 7.

Using both statements, when you sum a and b, you get,
a+b = 7n+1 + 7m+1 = 7(n+m) + 2
or
a+b = 7n+2 + 7m+2 = 7(n+m) + 4
or
a+b = 7n+3 + 7m+3 = 7(n+m) + 6
etc
I hope you see that the remainder will be some even number (either 2 or 4 or 6 or 8 or 10 or 12). The remainder will never be 7 so the sum will never be a multiple of 7.

What happens if we replace 7 by 6?
Same logic.
a+b = 6n+1 + 6m+1 = 6(n+m) + 2
or
a+b = 6n+2 + 6m+2 = 6(n+m) + 4
or
a+b = 6n+3 + 6m+3 = 6(n+m) + 6
Let me stop here. Did you see something interesting? The remainder in the last case is 6 which means a+b is divisible by 6. In some cases, a+b will not be divisible by 6 and in some cases, it will be.
Hence your answer will be (E).



Karishma, can you please check your answer since OA is C not E.
and i am also getting C as answer.
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Re: MGMAT Number Properties Ch 10 #10  [#permalink]

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New post 24 May 2012, 23:53
alphastrike wrote:
Math masters:

I have a quick question on MGMAT Number Properties Chapter 10 Question 10.

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
2) a-b is divisible by 7

So...I can get the right answer, but in the explanation I saw somethign that got me thinking about the answer if the question had used the number "6" instead of "7"

"1 and 2 combined tell us that a is not divisible by 7, but a-b is divisible by 7. This tells us a and b have the same remainder when divided by 7: if a-b is divisible by 7, then the remainder of a-b is zero. Therefore, the remainders of a and b must be equal"

The answer to the original question is no. But if the if the target # is 6, is the answer appears to be yes. It seems that the answer is yes for any even # and no for any odd #. Confirm?


Alphastrike you've pointed out something very interesting.
The reason is- AS THE SUM OF THE TWO REMAINDERS WHICH ARE EQUAL CAN'T BE AN ODD NUMBER, HENCE FOR ALL ODD NUMBERS THE ANSWER IS NO.
IT FOLLOWS LOGICALLY THAT FOR EVEN NUMBER i.e 6, IT MAY BE YES OR MAY BE NO.
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Re: MGMAT Number Properties Ch 10 #10  [#permalink]

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New post 25 May 2012, 00:02
321kumarsushant wrote:
Here is the logical explanation to the different answers (C and E) in the two cases (7 and 6).
[highlight]First, let me explain why the answer is (C) in the original question.[/highlight]

Are the sum of integers a and b divisible by 7?

1) a is not divisible by 7
So a is of one of the following formats: 7n+1, 7n+2, 7n+3, 7n+4, 7n+5 and 7n+6

2) a-b is divisible by 7
The format of b is the same as the format of a i.e. if a = 7n + 3, then b = 7m + 3 since a and b vary by a multiple of 7
e.g. if a = 43 (= 7*6+1), then b could be 22 (= 7*3 + 1). The difference between them has to be a multiple of 7 so the remainder obtained by dividing a by 7 will be the same as the remainder obtained by dividing b by 7.

Using both statements, when you sum a and b, you get,
a+b = 7n+1 + 7m+1 = 7(n+m) + 2
or
a+b = 7n+2 + 7m+2 = 7(n+m) + 4
or
a+b = 7n+3 + 7m+3 = 7(n+m) + 6
etc
I hope you see that the remainder will be some even number (either 2 or 4 or 6 or 8 or 10 or 12). The remainder will never be 7 so the sum will never be a multiple of 7.

[highlight]What happens if we replace 7 by 6?[/highlight]
Same logic.
a+b = 6n+1 + 6m+1 = 6(n+m) + 2
or
a+b = 6n+2 + 6m+2 = 6(n+m) + 4
or
a+b = 6n+3 + 6m+3 = 6(n+m) + 6
Let me stop here. Did you see something interesting? The remainder in the last case is 6 which means a+b is divisible by 6. In some cases, a+b will not be divisible by 6 and in some cases, it will be.
[highlight]Hence your answer will be (E).[/highlight]


Karishma, can you please check your answer since OA is C not E.
and i am also getting C as answer.


Please note that the answer to the original question is (C) as highlighted above. Later, I changed the question (6 in place of 7) and that's when you get the answer (E).
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Re: Is the sum of integers a and b divisible by 7?  [#permalink]

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Re: Is the sum of integers a and b divisible by 7?   [#permalink] 21 Jul 2019, 03:27
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