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10 Apr 2020, 00:03
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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:
85%
(hard)
Question Stats:
49% (02:25) correct
51%
(02:18)
wrong
based on 196
sessions
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Not Attempted Yet
If a,b and c are all positive integers satisfying \(a*b*c= 7(a+b+c)\), what is the smallest possible value of \((a+b+c)\)?
A. 12
B. 15
C. 18
D. 21
E. 24
A. 12
B. 15
C. 18
D. 21
E. 24
09 Oct 2020, 23:52
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a*b*c= 7(a+b+c)
So one of a,b and c must be a multiple of 7. Since we have to minimize a+b+c, one of them is equal to 7(minimum positive multple of 7).
Consider c=7
a*b=a+b+7
a*b-a-b= 7
a*b-a-b+1 = 7+1
a(b-1)-1(b-1)=8
(a-1)(b-1) = 1*8 or 2*4
a=2 and b=9 or a=3 and b=5
To minimize a+b+c, consider 2nd case; a=3 and b=5.
So one of a,b and c must be a multiple of 7. Since we have to minimize a+b+c, one of them is equal to 7(minimum positive multple of 7).
Consider c=7
a*b=a+b+7
a*b-a-b= 7
a*b-a-b+1 = 7+1
a(b-1)-1(b-1)=8
(a-1)(b-1) = 1*8 or 2*4
a=2 and b=9 or a=3 and b=5
To minimize a+b+c, consider 2nd case; a=3 and b=5.
KeyurJoshi
Is there any other way besides brute force method for this problem?
General Discussion
10 Apr 2020, 00:29
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If a,b and c are all positive integers satisfying a∗b∗c=7(a+b+c), what is the smallest possible value of (a+b+c)?
here, a∗b∗c=7(a+b+c)
already one factor 7 . let , a=7
so, (a+b+c) must be a number of two factors so that b,c can be positive integers but keeping one condition in mind that is,
(a+b+c) = bc
question wants to find smallest number .so starting from choice a,
(a+b+c) = 12 =2*6
so, b=2, c=6
and a= 7 but (a+b+c) = 7+2+6 = 15 not 12 (rejected )
(a+b+c) = 12 = 3*4
so, b=3, c=4
and a= 7 but (a+b+c) = 7+3+4 = 14 not 12 (rejected )
(a+b+c) = 15 =5*3
so, b=5, c=3
and a= 7 but (a+b+c) = 7+5+3 = 15 not 15 (accepted )
correct answer B
here, a∗b∗c=7(a+b+c)
already one factor 7 . let , a=7
so, (a+b+c) must be a number of two factors so that b,c can be positive integers but keeping one condition in mind that is,
(a+b+c) = bc
question wants to find smallest number .so starting from choice a,
(a+b+c) = 12 =2*6
so, b=2, c=6
and a= 7 but (a+b+c) = 7+2+6 = 15 not 12 (rejected )
(a+b+c) = 12 = 3*4
so, b=3, c=4
and a= 7 but (a+b+c) = 7+3+4 = 14 not 12 (rejected )
(a+b+c) = 15 =5*3
so, b=5, c=3
and a= 7 but (a+b+c) = 7+5+3 = 15 not 15 (accepted )
correct answer B
