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17 Apr 2020, 21:54
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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34% (02:38) correct
66%
(02:42)
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based on 111
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[GMAT math practice question]
\(a, b\), and \(c\) are positive integers with \(abc + 2ab + 2bc + 2ca + 4a + 4b + 4c = 447.\) What is the value of \(a + b + c\)?
A. \(17\)
B. \(19\)
C. \(21\)
D. \(23\)
E. \(25\)
\(a, b\), and \(c\) are positive integers with \(abc + 2ab + 2bc + 2ca + 4a + 4b + 4c = 447.\) What is the value of \(a + b + c\)?
A. \(17\)
B. \(19\)
C. \(21\)
D. \(23\)
E. \(25\)
yashikaaggarwal
Senior Moderator - Masters Forum
Joined: 19 Jan 2020
Last visit: 17 Jul 2025
Posts: 3,086
Given Kudos: 1,510
Location: India
Schools: Cranfield (A) Warwick MiM "23 (M)
GPA: 4
WE:Analyst (Internet and New Media)
18 Apr 2020, 01:36
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It's surely a trickier one.
ABC+2AB+2BC+2CA+4A+4B+4C=447
Equation is not the problem. Problem is Product of ABC.
Think of a situation where ABC can be put into product.
Let's consider (A+X)(B+X)(C+X)
It's ABC+ ACX + BCX + ABX + AX^2+BX^2+CX^2+ X^3
Now compare this with equation given
You will find X as 2
Only thing we are lacking is X^3=8
Put it in equation
ABC+2AB+2BC+2CA+4A+4B+4C+8-8=447
(A+2)(B+2)(C+2)-8=447
(A+2)(B+2)(C+2)=447+8
(A+2)(B+2)(C+2)=455
Now factors of 455=5*7*13
Which is (A+2)(B+2)(C+2)
Put it in equation.
You will get A=3,B=5,&C=11
Which is 19.
OA is B
Posted from my mobile device
ABC+2AB+2BC+2CA+4A+4B+4C=447
Equation is not the problem. Problem is Product of ABC.
Think of a situation where ABC can be put into product.
Let's consider (A+X)(B+X)(C+X)
It's ABC+ ACX + BCX + ABX + AX^2+BX^2+CX^2+ X^3
Now compare this with equation given
You will find X as 2
Only thing we are lacking is X^3=8
Put it in equation
ABC+2AB+2BC+2CA+4A+4B+4C+8-8=447
(A+2)(B+2)(C+2)-8=447
(A+2)(B+2)(C+2)=447+8
(A+2)(B+2)(C+2)=455
Now factors of 455=5*7*13
Which is (A+2)(B+2)(C+2)
Put it in equation.
You will get A=3,B=5,&C=11
Which is 19.
OA is B
Posted from my mobile device
19 Apr 2020, 20:42
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=>
\(abc + 2ab + 2bc + 2ca + 4a + 4b + 4c + 8 = 455\)
\(a(bc + 2b + 2c + 4) + 2(bc + 2b + 2c + 4) = 455\) (taking out a common fraction of a from the first \(4\) terms and a common factor of \(2\) from the last \(4\) terms)
\(= (a + 2)(bc + 2b + 2c + 4) = 455\) (taking out a common factor of (\(bc + 2b + 2c +4\)))
\((a + 2)[b(c + 2) + 2(c + 2)] = 455\) (taking out common factors of \(b\) and \(2\))
\(= (a + 2)(b + 2)(c + 2) = 455\) (taking out a common factor of (\(c + 2\)))
\(= 3*5*11.\)
\(a, b,\) and \(c\) are interchangeable since the equation \((a + 2)(b + 2)(c + 2) = 5*7*13\) is symmetric.
Then we have \(a + b + c = 3 + 5 + 11 = 19.\)
Therefore, B is the answer.
Answer: B
\(abc + 2ab + 2bc + 2ca + 4a + 4b + 4c + 8 = 455\)
\(a(bc + 2b + 2c + 4) + 2(bc + 2b + 2c + 4) = 455\) (taking out a common fraction of a from the first \(4\) terms and a common factor of \(2\) from the last \(4\) terms)
\(= (a + 2)(bc + 2b + 2c + 4) = 455\) (taking out a common factor of (\(bc + 2b + 2c +4\)))
\((a + 2)[b(c + 2) + 2(c + 2)] = 455\) (taking out common factors of \(b\) and \(2\))
\(= (a + 2)(b + 2)(c + 2) = 455\) (taking out a common factor of (\(c + 2\)))
\(= 3*5*11.\)
\(a, b,\) and \(c\) are interchangeable since the equation \((a + 2)(b + 2)(c + 2) = 5*7*13\) is symmetric.
Then we have \(a + b + c = 3 + 5 + 11 = 19.\)
Therefore, B is the answer.
Answer: B
