Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 09:00 AM PDT  10:00 AM PDT Watch & learn the Do's and Don’ts for your upcoming interview Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE. Oct 27 08:00 PM EDT  09:00 PM EDT Strategies and techniques for approaching featured GMAT topics. One hour of live, online instruction
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 04 Aug 2013
Posts: 94
Location: India
GPA: 3
WE: Manufacturing and Production (Pharmaceuticals and Biotech)

a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
26 Nov 2014, 01:01
Question Stats:
52% (02:06) correct 48% (02:09) wrong based on 172 sessions
HideShow timer Statistics
a, b, and c are three distinct positive integers. What is the product abc? (1) a + b + c = 7 (2) ab + bc + ca = 14
Official Answer and Stats are available only to registered users. Register/ Login.



Senior Manager
Joined: 13 Jun 2013
Posts: 266

Re: a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
26 Nov 2014, 02:44
anceer wrote: a, b, and c are three distinct positive integers. What is the product abc?
1. a + b + c = 7 2. ab + bc + ca = 14 st.1 as a,b,c>0 thus all three of them must be less than 5. because if any of the number a,b,c becomes 5. then the value of a+b+c will be greater than 7. thus three numbers will be less than 5. so possible values available for a,b,c are 1,2,3,4. of these possible values only 1,2,4 results in the some of 7. thus product of a,b,c will be 1*2*4=8 st.2 again, if any of the number a,b,c becomes 5 or bigger then the sum of ab+bc+ca will be greater than 5. hence a,b,c will be less than 5. from the possible set of values of a,b,c only 1,2,4 results in the sum of 14. hence product of a,b,c will be 1*2*4 =8 as statement alone is sufficient to answer the question. hence answer should be D.



Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1348
Location: Malaysia

Re: a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
13 Feb 2017, 19:39
anceer wrote: a, b, and c are three distinct positive integers. What is the product abc?
(1) a + b + c = 7 (2) ab + bc + ca = 14 Official solution from Veritas Prep. D For S1, let's try a few possibilities. 1 + 2 + 3 = 6 is the smallest, and that doesn't work. 1 + 2 + 4 does, so that's a possibility. 1 + 2 + 5 is too big, so that (or anything 1 + 2 + a number bigger than 5, for that matter) is out. The smallest possibility without 1 involved is 2 + 3 + 4 = 9, which is also too big ... so we must have 1, 2, and 4. Hence abc = 1 * 2 * 4 = 8, so S1 is SUFFICIENT. For S2, algebra doesn't seem to help, so let's use number properties instead. ab + bc + ca = 14, so ab + bc + ca = even. We can only have an EVEN sum if AT LEAST one of ab, bc, and ca is even. (If they're all odd, we'd have odd + odd + odd, which is NOT even.) If at least one of ab, bc, and ca is even, then at least one of a, b, and c is even. Now let's suppose that ONLY one of them is even: in other words, that a is even, b is odd, and c is odd. Then ab + bc + ca = evenodd + oddodd + odd*even = even + odd + even = odd. So this isn't possible either! Hence, we know that at least TWO of a, b, and c are even. Now let's suppose that ALL THREE are even. Using the SMALLEST possible even a, b, and c, we'd have 2 * 4 + 2 * 6 + 4 * 6, which is too big ... so they can't all be even! So we know we have two even and one odd. 1 * 2 + 1 * 4 + 2 * 4 = 14 works, so we could have a * b * c = 1 * 2 * 4 2 * 3 + 2 * 4 + 3 * 4 = 26, which is too big ... so anything other than 1, 2, and 4 is too big. Hence a * b * c = 8, and S2 is ALSO sufficient.
_________________
"Be challenged at EVERY MOMENT."“Strength doesn’t come from what you can do. It comes from overcoming the things you once thought you couldn’t.”"Each stage of the journey is crucial to attaining new heights of knowledge."Rules for posting in verbal forum  Please DO NOT post short answer in your post! Advanced Search : https://gmatclub.com/forum/advancedsearch/



Intern
Joined: 05 Aug 2012
Posts: 18
WE: Project Management (Pharmaceuticals and Biotech)

Re: a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
14 Feb 2017, 19:24
anceer wrote: a, b, and c are three distinct positive integers. What is the product abc?
(1) a + b + c = 7 (2) ab + bc + ca = 14 Can someone suggest a less time consuming way to evaluate statement 2?



Current Student
Joined: 29 Nov 2016
Posts: 14
Location: United States (TX)
GPA: 3.58

a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
03 Aug 2017, 05:48
raeann105 wrote: anceer wrote: a, b, and c are three distinct positive integers. What is the product abc?
(1) a + b + c = 7 (2) ab + bc + ca = 14 Can someone suggest a less time consuming way to evaluate statement 2? (2) ab + bc + ca = 14 > 2(a + b + c) = 14 > a + b + c = 7 Please let me know if this solution is incorrect.



Intern
Joined: 14 Dec 2016
Posts: 11

Re: a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
19 Mar 2018, 10:11
hazelnut wrote: anceer wrote: a, b, and c are three distinct positive integers. What is the product abc?
(1) a + b + c = 7 (2) ab + bc + ca = 14 Official solution from Veritas Prep. D For S1, let's try a few possibilities. 1 + 2 + 3 = 6 is the smallest, and that doesn't work. 1 + 2 + 4 does, so that's a possibility. 1 + 2 + 5 is too big, so that (or anything 1 + 2 + a number bigger than 5, for that matter) is out. The smallest possibility without 1 involved is 2 + 3 + 4 = 9, which is also too big ... so we must have 1, 2, and 4. Hence abc = 1 * 2 * 4 = 8, so S1 is SUFFICIENT. For S2, algebra doesn't seem to help, so let's use number properties instead. ab + bc + ca = 14, so ab + bc + ca = even. We can only have an EVEN sum if AT LEAST one of ab, bc, and ca is even. (If they're all odd, we'd have odd + odd + odd, which is NOT even.) If at least one of ab, bc, and ca is even, then at least one of a, b, and c is even. Now let's suppose that ONLY one of them is even: in other words, that a is even, b is odd, and c is odd. Then ab + bc + ca = evenodd + oddodd + odd*even = even + odd + even = odd. So this isn't possible either! Hence, we know that at least TWO of a, b, and c are even. Now let's suppose that ALL THREE are even. Using the SMALLEST possible even a, b, and c, we'd have 2 * 4 + 2 * 6 + 4 * 6, which is too big ... so they can't all be even! So we know we have two even and one odd. 1 * 2 + 1 * 4 + 2 * 4 = 14 works, so we could have a * b * c = 1 * 2 * 4 2 * 3 + 2 * 4 + 3 * 4 = 26, which is too big ... so anything other than 1, 2, and 4 is too big. Hence a * b * c = 8, and S2 is ALSO sufficient. why arent we considering zero here. zero is also an integer and ques doesnt say anything about numbers being greater than zero



Intern
Joined: 14 Dec 2016
Posts: 11

Re: a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
19 Mar 2018, 10:12
qazi11 wrote: hazelnut wrote: anceer wrote: a, b, and c are three distinct positive integers. What is the product abc?
(1) a + b + c = 7 (2) ab + bc + ca = 14 Official solution from Veritas Prep. D For S1, let's try a few possibilities. 1 + 2 + 3 = 6 is the smallest, and that doesn't work. 1 + 2 + 4 does, so that's a possibility. 1 + 2 + 5 is too big, so that (or anything 1 + 2 + a number bigger than 5, for that matter) is out. The smallest possibility without 1 involved is 2 + 3 + 4 = 9, which is also too big ... so we must have 1, 2, and 4. Hence abc = 1 * 2 * 4 = 8, so S1 is SUFFICIENT. For S2, algebra doesn't seem to help, so let's use number properties instead. ab + bc + ca = 14, so ab + bc + ca = even. We can only have an EVEN sum if AT LEAST one of ab, bc, and ca is even. (If they're all odd, we'd have odd + odd + odd, which is NOT even.) If at least one of ab, bc, and ca is even, then at least one of a, b, and c is even. Now let's suppose that ONLY one of them is even: in other words, that a is even, b is odd, and c is odd. Then ab + bc + ca = evenodd + oddodd + odd*even = even + odd + even = odd. So this isn't possible either! Hence, we know that at least TWO of a, b, and c are even. Now let's suppose that ALL THREE are even. Using the SMALLEST possible even a, b, and c, we'd have 2 * 4 + 2 * 6 + 4 * 6, which is too big ... so they can't all be even! So we know we have two even and one odd. 1 * 2 + 1 * 4 + 2 * 4 = 14 works, so we could have a * b * c = 1 * 2 * 4 2 * 3 + 2 * 4 + 3 * 4 = 26, which is too big ... so anything other than 1, 2, and 4 is too big. Hence a * b * c = 8, and S2 is ALSO sufficient. why arent we considering zero here. zero is also an integer and ques doesnt say anything about numbers being greater than zero my bad ,it says positive intergers..



Intern
Joined: 27 Jun 2019
Posts: 1

Re: a, b, and c are three distinct positive integers. What is the product
[#permalink]
Show Tags
01 Jul 2019, 02:09
manpreetsingh86 wrote: anceer wrote: a, b, and c are three distinct positive integers. What is the product abc?
1. a + b + c = 7 2. ab + bc + ca = 14 st.1 as a,b,c>0 thus all three of them must be less than 5. because if any of the number a,b,c becomes 5. then the value of a+b+c will be greater than 7. thus three numbers will be less than 5. so possible values available for a,b,c are 1,2,3,4. of these possible values only 1,2,4 results in the some of 7. thus product of a,b,c will be 1*2*4=8 st.2 again, if any of the number a,b,c becomes 5 or bigger then the sum of ab+bc+ca will be greater than 5. hence a,b,c will be less than 5. from the possible set of values of a,b,c only 1,2,4 results in the sum of 14. hence product of a,b,c will be 1*2*4 =8 as statement alone is sufficient to answer the question. hence answer should be D. st.2 again, if any of the number a,b,c becomes 5or bigger and and others will be 1 and 2 minimum then the sum of ab+bc+ca will be greater or at least than 17. hence a,b,c will be less than 5.




Re: a, b, and c are three distinct positive integers. What is the product
[#permalink]
01 Jul 2019, 02:09






