Addition Equals Multiplication
There is a triplet of positive numbers such that their sum and their product are equal, for example, ( 1 , 2 , 3 ) : 1 + 2 + 3 = 1 × 2 × 3 .
True or False?
There are infinitely many triplets of positive numbers (not necessarily integers) such that their sum and their product are equal.
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6 solutions
@Alex Li actually there's another number combination that works for integers and that is (-1,-2,-3) the proof goes below :-
Assuming 3 nos to be x,x+1,x+2 Now sum of these = 3(x+1)
Now product of these numbers is (x)(x+1)(x+2) Which after solving will get u to: x^3 + 3x^2 + 2x
Now as product = sum 3(x+1)= x^3 + 3x^2 + 2x And finally after solving it x^3 + 3x^ -x -3=0 (take it as eqn 1)
Now as the sum of the coefficients in above equation is 0 therefore (x+1) is a factor
Now after dividing eqn 1 by x+1, we have the quotient as,
x^2 + 2x -3=0 (x-1)(x+3)=0
Thus x can be either 1 or -3 or -1(-1 because eqn 1 was divisible by x+1)
And thus the 3 numbers could be (1,2,3) (-3,-2,-1) and (-1,0,1)
Unfortunately I mistook the question and solved for consecutive triplets rather than any triplets......thus got the answer wrong lol....well just wanted to let u know in case of integers(consecutive) another two sets are possible....
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That is indeed valid though note that I was only working with positive integers. If you extend the domain to all integers than you could have infinitely many combinations of the form (-X,0,X).
what if the numbers had to be integers? Is there a similar proof?
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It does not work for positive integers.
Say we have xyz = x+y+z. First I will show that at least one letter must be a 1.
Assume that there are no 1s. In this case, the RHS is always smaller than the LHS. This is because multiplication scales much faster than addition, and even with the smallest numbers we can plug in, 2 2 2>2+2+2. Therefore there is at least 1 1 in the equation. (This part is a bit iffy in terms of a proof but I think it's pretty intuitive in general and can't think of a better way to show it)
WLOG we say x=1. Then,
y z = 1 + y + z
y z − z = 1 + y
z ( y − 1 ) = 1 + y
z = y − 1 1 + y
For z to be an integer if y is an integer, 1+y must divide y-1, so it is at least twice as big as y-1. However, we can see that y − 1 1 + y approaches 1 asymptotically, so there are only a finite number of solutions. Really, just y=2, z=3 and y=3, z=2. So indeed the only working integer number combination is (1,2,3)
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It comes down to the wuestion of what constiuted a triplet of positive numbers. If they have to be sequential as in the example (1,2,3). Then the equation would be n + n+1 + n+2 = n(n+1)(n+2). This yields a polynomial equation with no more than three real roots.
Good proof
Clear and elegant proof. Great example. Good work
I should have tried a little longer.
What about -3 -2-1 = ( -3 )x( -2 )x(-1)
Let x = 1 ⟹ 1 + y + z = y z ⟹ y = z − 1 z + 1 , where z = 1
Let P be set of positive numbers. The set M = { ( 1 , z − 1 z + 1 , z ) ∣ ( z ∈ P ) ∧ ( z = 1 ) } is one such infinite set.
Set M is an infinite set yet it does not represent all the possible triplets whose products are equal to their sum because of your assumption that x=1. But because the question only asks us to prove that an infinite number of solutions are possible, which you did by proofing that set M has infinite elements, I believe your solution is very elegant in a weird way. I love it though.
I like your solution. Clean and elegant.
z must not equal (1)\, to avoid division by zero.
Should be obviously true by considering the cubic equation f ( x ) = x 3 + p x 2 + q x + r with roots: a , b , c . We want a + b + c = a b c ⟺ p = r . Then we want the roots to also be real, which can be ensured by considering the function value of the stationary points of f , ensuring they occur at distinct x -values, with maximum and minimum values of opposite sign. This shows us there will be a range of values of p , q , r which ensure these conditions to bet met, and so there will be infinite triplets.
Right. I solved this question using cubic discriminant , which is essentially the same approach as yours. Nicely done!
In a triangle TanA+tanB+tanC=TanA x TanB x TanC, we just need to take an acute triangle.
Oh wow, I didn't thought about using trigonometric identities in this question! Very creative.
Did not have to solve anything. The sum of the integers equals the product of them. The result is a "perfect" number. I just had to remember there were an infinite number of them. Is this cheating?
You've confused me. A perfect number is a positive integer that is the sum of all of its dividers, including 1, but excluding the number itself. How does this help find any sets of three numbers, not necessarily integers, whose sum and product are equal?
Let x=Tan A, y=Tan B and z=Tan C, where A, B, C are angles of a triangle. Then it is true that x+y+z=xyz. and there are infinite possibilities for the values of A,B and C such that A+B+C=180 and hence the result.
Given a triplet of positive numbers ( x , y , z ) , we'd like to satisfy the equation:
x y z = x + y + z
Solve the equation for z :
x y z x y z − z z ( x y − 1 ) z = x + y + z = x + y = x + y = x y − 1 x + y
Now, as long as x y > 1 , you can choose any positive x and y , and then compute the z that will make the equation true. Thus, there are infinite solutions to the equation.
For example, let x = 2 and y = 5 . Then,
z = 2 × 5 − 1 2 + 5 = 9 7
Then we can see that the product of these numbers is the same as the sum:
2 + 5 + 9 7 = 2 × 5 × 9 7