Simple math question.

ultimablah

The sum of four numbers equals the product of four numbers equals 711.

What are the four numbers?

I see 2 equations with 4 unknowns, and no way to do this but brute force it; is there a special circumstance here?

Werg22

The sum and product of the same 4 numbers are equal to 711? Is that what you mean?

The prime decomposition of 711 is 3^2*79. There aren't too many ways to get 4 numbers; you either have 1, 1, 9, 79, or 1, 1, 3, 237 or 1, 3 , 3, 79. None of those sum up to 711. Can you restate the problem more clearly?

davee123

Admittedly, he did say "numbers" and not "integers", so the prime factorization of 711 isn't necessarily relevant, although I agree it's certainly implied. Hmm...

DaveE

ƒ(x)

Hm...Just thinking out loud here. This approach requires making certain assumptions that could prove to be wrong. Please forgive my formatting errors.

711 = a+b+c+d
711 = abcd

The first two assumptions are that a = 711, so:

0 = b+c+d
1 = bcd

The next one is that bc = 1, which means that

d = -1

and

-1 = bc

and that

-1/b = c

which means that

711 = 711-1+(-1/b)+b
0 = -1-(1/b)+b
0 = b^2-b-1
b = [1+-(1-4*-1*1)^1/2]/2
b = [1+-(5)^1/2]/2
b = 1.618 or -.618

Therefore, since -1/b = c

When b = 1.618, c = -.618 and when b = -.618 c = 1.618
And that since addition and multiplication are communicative, one of these solution is redundant.

So, in conclusion (finally )

Since -1 = bc

711 ?= 711*-1*-1
711 = 711

and

711 ?= 711-1-.618+1.618
711= 711

a = 711
b = -.618
c = 1.618
d = -1

Q.E.D.

jimmysnyder

As the op said, these are two equations in 4 unknowns. There are infinitely many solutions. F(x) has added 3 more equations which seems extravagant under the circumstances.

ƒ(x)

Haha. Nevertheless, those three extravagant equations allowed me to find one possible solution.

Oh, and by the way, prove that there are a infinite number of solutions.

davee123

That's pretty easy, no?

w+x+y+z = 711
wxyz = 711
w = 711/xyz

Assume z=1:

w=711/xy
711/xy+x+y+1 = 711
711 + x^2y + xy^2 = 710xy
xy^2+(x^2-710x)y+711 = 0

You can use the quadratic formula (though it's messy) to solve for y, with:

A = x
B = x^2-710x
C = 711

So as long as 2A =/= 0, and B^2 > 4AC, you're good. Specifically:

x =/=0
x^4+2x^2-1420x+504100x^2 > 2840x
x^3+2x+504100x > 4260

So, without bothering to solve it, it's obvious that for x>1, the quadratic will be solvable, because it'll satisfy the above requirements. And obviously there are an infinite number of values for x>1.

DaveE

jimmysnyder

My bad. f(x) only added two equations, the expected number. As not both of the original equations are linear it is not enough to merely count equations and unknowns as I did. However, davee123 has saved the day for me on this point.

ƒ(x)

Haha fine. I'm too tired to look through your post right now, so I'll just agree for the moment.

Just as a side note, my math teacher has a problem that is exactly similar to this one that that he calls the "7-11 Problem" (as in the gas station). When asked, he said that there was only 1 possible solution to this, and I am tempted to agree because the problem combines addition and multiplication. But, he is not infallible.