I have the feeling I am not going to get any more takers on this, so here is the solution.

xy + x + y = (x+1)(y+1) – 1

Now see how this works.

Starting with 2 and 3 we get (2+1)(3+1) – 1, which replaces 2 and 3

Now let’s combine this with 5.

We get:

[(2+1)(3+1)-1 +1]((5+1) – 1 = (2+1)(3+1)(5+1) – 1

Given k terms, the final answer is (n1 + 1)(n2+1)...(nk+1) – 1

As @Jeruba suggested, the answer comes from the properties of arithemtic. Since it does not matter what order the multiplication occurs in, the answer does not depend on the order that the terms are combined. In the example given, the answer is (2+1)(3+1)(5+1)(7+1)(10+1) -1 =6335

I don’t know if anybody is going to read this, but here is a quick lesson on the general principle being used.

Let f(x) = xy = x + y

Then f(x) + 1 = (x+1)(y+1)

Now let T(x) = x+1, and let g(x) = x*y

We have for the general case:

T(f(x) = g(T(x),T(y))

Why is this of any use? Let me show you a way of looking at this that even English majors can understand. Suppose I want to know the value of f(x,y), but I want to have someone else do it. I find an expert on the calculation, but she speaks some foreign language. I then hire a translator. Think of T(x) as translation. The translation of f is g and of x and y are T(x) and T(y). This will work provided that g(T(x),T(y)) represents a translation of the answer, that it is equal to T(f(x,y). The translator can translate in answer given as f(x,y). In other words, the translation process sets up a kind of parallel world. If the translation is easier to work with or provides additional information then it can be very useful.

For example, the use of logarithms.

log(x*y) = log(x) + log(y). Here T(x) = log(x), f(x,y)=(x*y) and g(x,y) = x + y. This can prove useful because addition is easier to work with than multiplication. Mathematical transforms, of which there are many, like Fourier and Laplace transforms, use the same idea.

Consider any mathematical model, like the determination of fuel cost for a trip based on distance, gas mileage and gas cost. The use of any mathematical model is based on translating certain real world measurements into numerical values and then using the model to mimic the way these measured values work together to compute some other real world value. If the model and measurements are accurate, it is this same translation idea.