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# a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b +

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Joined: 02 Sep 2009
Posts: 55271
a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b +  [#permalink]

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05 Jul 2018, 04:41
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Difficulty:

45% (medium)

Question Stats:

70% (02:21) correct 30% (02:37) wrong based on 43 sessions

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a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b + c = 6. What is the value of abc ?

A. 2
B. 6
C. 9
D. 12
E. 18

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Posts: 173
Concentration: Technology, Nonprofit
WE: Analyst (Non-Profit and Government)
Re: a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b +  [#permalink]

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05 Jul 2018, 04:59
1
Given:
ab + c = 7 -- (eq. 1)
ac + b = 5 -- (eq. 2)
a + b + c = 6 -- (eq. 3)

(a+1)(b+c) = 12
Substitute b+c = 6 - a from eq. 3
(a+1)(6-a) = 12
This gives a = 2, 3

Pick a value. Say a = 3
Substitute in eq. 1 and 2
3b + c = 7
3c + b = 5
==> b = 2 and c = 1
a*b*c = 3*2*1 = 6

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Re: a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b +  [#permalink]

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05 Jul 2018, 05:12
Bunuel wrote:
a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b + c = 6. What is the value of abc ?

A. 2
B. 6
C. 9
D. 12
E. 18

Going from the answer options can be advantageous in this kind of problem.

A. 2 = 2*1 | a = 2,b = c = 1
Whichever way we go about the numbers, we can't get a sum s.t a + b + c = 6

B. 6 = 2*3 | a = 2,b = 3,c = 1
In this case a + b + c = 6. Also, ab + c = 6 +1 = 7 and ac + b = 2 + 3 = 5.

Therefore, Option B(6) is the value of abc.
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a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b +  [#permalink]

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16 Mar 2019, 15:40
Given...

ab + c = 7... c = 7 - ab
ac + b = 5... b = 5 - ac
a + b + c = 6

From the third eq...

a + 5 - ac + 7 - ab = 6

$$c + b = \frac{6+a}{a}$$

So from a + b + c = 6

$$a^2 - 5a + 6 =0$$
$$(a-2)(a-3)$$

If A1 = 2 then b +c could be (2,2) and (3,1)
If A2 = 3 then b + c could be (2,1)

abc...

(1)(2)(2) = 8 or (2)(3)(1) = 6 (A1)
(3)(2)(1) = 6 (A2)

B
a, b, and c are integers such that ab + c = 7, ac + b = 5, and a + b +   [#permalink] 16 Mar 2019, 15:40
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