- #1
Albert1
- 1,221
- 0
$a,b,c \in N$
(1) $1<a<b<c$
(2)$(ab-1)(bc-1)(ca-1) \,\, mod \,\, (abc)=0$
$find :a,b,c$
(1) $1<a<b<c$
(2)$(ab-1)(bc-1)(ca-1) \,\, mod \,\, (abc)=0$
$find :a,b,c$
What are the values of a, b, and c in the following equations?
- MHB
- Thread starter Albert1
- Start date
In summary, the conversation discusses the values of a, b, and c as natural numbers and the condition that a<b<c. The problem to be solved is to find the values of a, b, and c that satisfy the equation (ab-1)(bc-1)(ca-1) mod (abc) = 0. After some algebraic manipulation, it is determined that there are no such values of a, b, and c. However, (2,3,5) is later suggested as a possible solution, and it is confirmed to be the only solution through further calculations. The conversation ends by summarizing the solution process and congratulating the person who found the solution.
- #2
mathworker
- 111
- 0
Since modulo is "zero" there is no remainder.
\(\displaystyle \frac{(ab-1)(bc-1)(ca-1)}{abc}\) is not a fraction
\(\displaystyle abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}\) is not a fraction
So the little terms must sum it up to zero or 1 so,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0\)(let)
\(\displaystyle \frac{ab+bc+ca-1}{abc}=0\)
\(\displaystyle ab+bc+ca=1\) which certainly can't be true
If,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1\)
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)
But as \(\displaystyle a,b,c>1\) \(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) has a maxmum value of 1 so there are no solutions to this one either so I arrived at "there are no such a,b,c"
Did I do something wrong?
\(\displaystyle \frac{(ab-1)(bc-1)(ca-1)}{abc}\) is not a fraction
\(\displaystyle abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}\) is not a fraction
So the little terms must sum it up to zero or 1 so,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0\)(let)
\(\displaystyle \frac{ab+bc+ca-1}{abc}=0\)
\(\displaystyle ab+bc+ca=1\) which certainly can't be true
If,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=1\)
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)
But as \(\displaystyle a,b,c>1\) \(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\) has a maxmum value of 1 so there are no solutions to this one either so I arrived at "there are no such a,b,c"
Did I do something wrong?
- #3
mathbalarka
- 456
- 0
I am just going to leave this one here : (a, b, c) = (2, 3, 5). (*) Can you see where you went wrong now?
(*) Note that this doesn't mean there are no other. I haven't even revealed half of the solution yet. Keep trying!
(*) Note that this doesn't mean there are no other. I haven't even revealed half of the solution yet. Keep trying!
- #4
mathworker
- 111
- 0
yeah I got it
- - - Updated - - -
But,I guess (2,3,5) is the only solution...(Smirk)
max value is not 1 \(\displaystyle its \frac{13}{12}\)
- - - Updated - - -
But,I guess (2,3,5) is the only solution...(Smirk)
- #5
Albert1
- 1,221
- 0
yes (2,3,5) is the only solution..,but how to get the answer ?mathworker said:
- #6
mathworker
- 111
- 0
In,
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
\(\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}\)
\(\displaystyle c=5\)
\(\displaystyle \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1+\frac{1}{abc}\)
L.H.S to be greater than one (a,b) should be (2,3) substituting them rest is linear equation
\(\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{c}=1+\frac{1}{6c}\)
\(\displaystyle c=5\)
- #7
Albert1
- 1,221
- 0
very nice solution(Yes)mathworker said:
Related to What are the values of a, b, and c in the following equations?
1. How do you find the values of a, b, and c in an equation?
To find the values of a, b, and c in an equation, you can use the quadratic formula, which is (-b ± √(b²-4ac))/2a. Plug in the values of a, b, and c from the equation into this formula to solve for the unknowns.
2. Can you find a, b, and c if only 2 out of 3 values are given?
No, you need all three values of a, b, and c to solve for each unknown. If only 2 out of 3 values are given, there are infinitely many solutions for the unknowns.
3. What is the significance of a, b, and c in a quadratic equation?
In a quadratic equation in the form of ax²+bx+c=0, a, b, and c represent the coefficients of the equation. The value of a determines the shape of the parabola, b affects the position of the parabola, and c is the constant term. These values are important in solving and understanding quadratic equations.
4. Is there a specific order in which to find a, b, and c?
No, there is no specific order in which to find a, b, and c. However, it is important to follow the correct steps in solving a quadratic equation, such as simplifying and combining like terms, before using the quadratic formula to find the values of a, b, and c.
5. Can you find a, b, and c using any other method besides the quadratic formula?
Yes, there are other methods for solving quadratic equations, such as factoring, completing the square, and graphing. However, the quadratic formula is the most efficient and accurate method for finding the values of a, b, and c in any quadratic equation.
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