Sep 23 08:00 AM PDT  09:00 AM PDT Join a free 1hour webinar and learn how to create the ultimate study plan, and be accepted to the upcoming Round 2 deadlines. Save your spot today! Monday, September 23rd at 8 AM PST Sep 26 07:00 AM PDT  08:00 AM PDT a chat session focused on MBA recruitment of international students at US schools. Sep 28 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn two proprietary ways to PreThink assumptions and ace GMAT CR in 10 days. Sep 29 07:00 AM PDT  09:00 AM PDT Attend this Free Algebra Webinar and learn how to master Inequalities and Absolute Value problems on GMAT.
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
22 Oct 2009, 20:04
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) ab is divisible by 7
Official Answer and Stats are available only to registered users. Register/ Login.
_________________




Math Expert
Joined: 02 Sep 2009
Posts: 58115

Re: Data Sufficiency, Number Properties
[#permalink]
Show Tags
22 Oct 2009, 20:30
mirzohidjon wrote: Is the sum of integers a and b divisible by 7? 1) a is not divisible by 7 2) ab 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) \(ab=7q\) > \(a+b=?\) > Nor sufficient (1)+(2) \(a=7p+r\) and \(ab=7q\) > \(b=a7q\). \(a+b=(7p+r)+(a7q)=7p+r+7p+r7q=7(2pq)+2r\) > \(7(2pq)\) is divisible by 7. Since \(r\) is remainder from a being divided by 7, \(r\) and thus \(2r\) is not divisible by 7. So, \(a+b=7(2pq)+2r\) is not divisible by 7. Sufficient. Answer: C.
_________________




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
22 Oct 2009, 21:48
Thanks for fast explanation, this way, I think it better to understand, rather the explanation suggested my MGMAT.
_________________



Intern
Joined: 08 Mar 2011
Posts: 9

Re: divisibility & primes
[#permalink]
Show Tags
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 ab 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
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) ab = 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
24 Jun 2011, 14:38
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) ab = 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. ab 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 ab is divisible by 7. Means, b is also NOT divisible by 7. a=9 b=2 ab=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 ab=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
Joined: 24 Jun 2008
Posts: 1836

Re: divisibility & primes
[#permalink]
Show Tags
24 Jun 2011, 16:58
If a+b and ab 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+ab = 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 ab 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
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
ab is divisible by 7
a b a+b divisible by 7 14 7 yes 15 1 no
together
we have ab 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.



Intern
Joined: 08 Sep 2010
Posts: 42

Re: divisibility & primes
[#permalink]
Show Tags
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.
_________________
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
GPA: 3.53

MGMAT Number Properties Ch 10 #10
[#permalink]
Show Tags
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) ab 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 ab is divisible by 7. This tells us a and b have the same remainder when divided by 7: if ab is divisible by 7, then the remainder of ab 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?
_________________



GMAT Tutor
Joined: 24 Jun 2008
Posts: 1836

Re: MGMAT Number Properties Ch 10 #10
[#permalink]
Show Tags
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) ab 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 ab is divisible by 7. This tells us a and b have the same remainder when divided by 7: if ab is divisible by 7, then the remainder of ab 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) ab is divisible by 6then 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
Joined: 17 Nov 2007
Posts: 3237
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: MGMAT Number Properties Ch 10 #10
[#permalink]
Show Tags
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
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9646
Location: Pune, India

Re: MGMAT Number Properties Ch 10 #10
[#permalink]
Show Tags
06 Apr 2012, 02:26
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) ab 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 ab is divisible by 7. This tells us a and b have the same remainder when divided by 7: if ab is divisible by 7, then the remainder of ab 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) ab 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 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
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) ab 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) \(ab=7q\) > \(a+b=?\) > Nor sufficient (1)+(2) \(a=7p+r\) and \(ab=7q\) > \(b=a7q\). \(a+b=(7p+r)+(a7q)=7p+r+7p+r7q=7(2pq)+2r\) > \(7(2pq)\) is divisible by 7. Since \(r\) is remainder from a being divided by 7, \(r\) and thus \(2r\) is not divisible by 7. So, \(a+b=7(2pq)+2r\) is not divisible by 7. Sufficient. Answer: C. 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: 08272012
GPA: 3.8
WE: Marketing (Manufacturing)

Re: MGMAT Number Properties Ch 10 #10
[#permalink]
Show Tags
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) ab 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 ab is divisible by 7. This tells us a and b have the same remainder when divided by 7: if ab is divisible by 7, then the remainder of ab 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) ab 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.
_________________
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
GPA: 3.3
WE: Engineering (Manufacturing)

Re: MGMAT Number Properties Ch 10 #10
[#permalink]
Show Tags
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) ab 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 ab is divisible by 7. This tells us a and b have the same remainder when divided by 7: if ab is divisible by 7, then the remainder of ab 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
Joined: 16 Oct 2010
Posts: 9646
Location: Pune, India

Re: MGMAT Number Properties Ch 10 #10
[#permalink]
Show Tags
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) ab 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).
_________________
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 >



NonHuman User
Joined: 09 Sep 2013
Posts: 12528

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






