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a, b, and c are three distinct positive integers. What is the product

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a, b, and c are three distinct positive integers. What is the product [#permalink]

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New post 26 Nov 2014, 01:01
1
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A
B
C
D
E

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a, b, and c are three distinct positive integers. What is the product abc?

(1) a + b + c = 7
(2) ab + bc + ca = 14
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Re: a, b, and c are three distinct positive integers. What is the product [#permalink]

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New post 26 Nov 2014, 02:44
1
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.
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Re: a, b, and c are three distinct positive integers. What is the product [#permalink]

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New post 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.
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Re: a, b, and c are three distinct positive integers. What is the product [#permalink]

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New post 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?
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a, b, and c are three distinct positive integers. What is the product [#permalink]

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New post 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.
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Re: a, b, and c are three distinct positive integers. What is the product [#permalink]

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New post 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
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Re: a, b, and c are three distinct positive integers. What is the product [#permalink]

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New post 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..
Re: a, b, and c are three distinct positive integers. What is the product   [#permalink] 19 Mar 2018, 10:12
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a, b, and c are three distinct positive integers. What is the product

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