GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Sep 2019, 00:15 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # Is the sum of integers a and b divisible by 7?

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Senior Manager  Joined: 18 Aug 2009
Posts: 307
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Is the sum of integers a and b divisible by 7?  [#permalink]

### Show Tags

4
14 00:00

Difficulty:   75% (hard)

Question Stats: 52% (01:55) correct 48% (01:29) wrong based on 352 sessions

### HideShow timer Statistics

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

_________________
Never give up,,,
Math Expert V
Joined: 02 Sep 2009
Posts: 58115
Re: Data Sufficiency, Number Properties  [#permalink]

### Show Tags

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.

(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.

_________________
##### General Discussion
Senior Manager  Joined: 18 Aug 2009
Posts: 307
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Re: Data Sufficiency, Number Properties  [#permalink]

### Show Tags

Thanks for fast explanation, this way, I think it better to understand, rather the explanation suggested my MGMAT.
_________________
Never give up,,,
Intern  Joined: 08 Mar 2011
Posts: 9
Re: divisibility & primes  [#permalink]

### Show Tags

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)
Intern  Joined: 08 Sep 2010
Posts: 42
Re: divisibility & primes  [#permalink]

### Show Tags

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.?
_________________
Don't forget to give me ****KUDOS**** Retired Moderator Joined: 20 Dec 2010
Posts: 1646
Re: divisibility & primes  [#permalink]

### Show Tags

1
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.
*************************************
_________________
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1836
Re: divisibility & primes  [#permalink]

### Show Tags

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).
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
Director  Joined: 01 Feb 2011
Posts: 588
Re: divisibility & primes  [#permalink]

### Show Tags

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 .

Intern  Joined: 08 Sep 2010
Posts: 42
Re: divisibility & primes  [#permalink]

### Show Tags

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.
_________________
Don't forget to give me ****KUDOS**** Intern  Joined: 01 Jun 2010
Posts: 22
Location: United States
Schools: Harvard Business School (HBS) - Class of 2014
GMAT 1: 730 Q47 V44 GPA: 3.53
MGMAT Number Properties Ch 10 #10  [#permalink]

### Show Tags

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?
_________________
HBS Class of 2014
GMAT Tutor G
Joined: 24 Jun 2008
Posts: 1836
Re: MGMAT Number Properties Ch 10 #10  [#permalink]

### Show Tags

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'.
_________________
GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com
CEO  B
Joined: 17 Nov 2007
Posts: 3237
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40 Re: MGMAT Number Properties Ch 10 #10  [#permalink]

### Show Tags

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
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | Limited GMAT/GRE Math tutoring in Chicago
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9646
Location: Pune, India
Re: MGMAT Number Properties Ch 10 #10  [#permalink]

### Show Tags

2
1
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.
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Manager  Affiliations: Project Management Professional (PMP)
Joined: 30 Jun 2011
Posts: 128
Location: New Delhi, India
Re: Data Sufficiency, Number Properties  [#permalink]

### Show Tags

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.

(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.

Thanks Bunuel... great explanation.. really helps
_________________
Best
Vaibhav

If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks
Manager  Joined: 01 Nov 2010
Posts: 211
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Re: MGMAT Number Properties Ch 10 #10  [#permalink]

### Show Tags

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.

Karishma, can you please check your answer since OA is C not E.
and i am also getting C as answer.
_________________
kudos me if you like my post.

Attitude determine everything.
all the best and God bless you.
Intern  Joined: 01 Mar 2012
Posts: 19
Concentration: Operations, Finance
GMAT 1: 740 Q49 V41 GPA: 3.3
WE: Engineering (Manufacturing)
Re: MGMAT Number Properties Ch 10 #10  [#permalink]

### Show Tags

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.
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9646
Location: Pune, India
Re: MGMAT Number Properties Ch 10 #10  [#permalink]

### Show Tags

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.

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).
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Non-Human User Joined: 09 Sep 2013
Posts: 12528
Re: Is the sum of integers a and b divisible by 7?  [#permalink]

### Show Tags

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________ Re: Is the sum of integers a and b divisible by 7?   [#permalink] 21 Jul 2019, 03:27
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  