Asked by Jason
Suppose A,B, and C are positive integers such that 24/5 = A + 1/B+1/C+1
The value of 1A + 2B + 3C equals:
(A) 9 (B) 12 (C) 15
(D) 17 (E) 20
hey i am trying to solve the problem, is it written like
A+1
_____
B+1/C+1
A + 1
______
B + 1
____
C + 1
sorry it sort of doesnt make sense i am really good at math too... so it is written like
a+1
________
B+1
__
C+1
from your equation we can say
A + 1/B + 1/C = 19/5 (subtract 1 from both sides)
since the common denominator BC is 5
B=1 and C=5 or B=5 and C=1
neither combination produces your stated result,
but when A=3, B=-5 and C=1
we get 3 -1/5 + 1 = 19/5
we get the same result when A=3, B=1 and C=-5
I actually ran a computer program in QuickBasic which gave me the following results, which all work in your original equation
A,B,C
4,-5,1
4,1,-5
5,-30,-6
5,-10,-10
6,-1,-5
THERE IS NO SOLUTION FOR ALL POSITIVE INTEGERS
The value of 1A + 2B + 3C equals:
(A) 9 (B) 12 (C) 15
(D) 17 (E) 20
hey i am trying to solve the problem, is it written like
A+1
_____
B+1/C+1
A + 1
______
B + 1
____
C + 1
sorry it sort of doesnt make sense i am really good at math too... so it is written like
a+1
________
B+1
__
C+1
from your equation we can say
A + 1/B + 1/C = 19/5 (subtract 1 from both sides)
since the common denominator BC is 5
B=1 and C=5 or B=5 and C=1
neither combination produces your stated result,
but when A=3, B=-5 and C=1
we get 3 -1/5 + 1 = 19/5
we get the same result when A=3, B=1 and C=-5
I actually ran a computer program in QuickBasic which gave me the following results, which all work in your original equation
A,B,C
4,-5,1
4,1,-5
5,-30,-6
5,-10,-10
6,-1,-5
THERE IS NO SOLUTION FOR ALL POSITIVE INTEGERS
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