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This is a really good question!(Everyone has encountered a question which makes them love math, this is mine:)

We write a series of numbers $$1,\frac12,\frac13,..........,\frac{1}{2010}$$

Now we can pick any two numbers $x$ and $y$ and we replace these two numbers by just one number $x+y+xy$

This process is repeated until only one number is left, find the last number.

This has just baffled me! Seriously having no idea how to proceed. Isn't it interesting that we end up with the same number despite where we start? All Hints are welcome on how to solve it

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3 Answers 3

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This is an invariant question : imagine a function $f(x_1,...,x_m)$ (where $m$ is a certain number of arguments and $x_i$ are all real numbers) with the following property : $f(x_1,...,x_m)$ doesn't change when you take any two of the these $x_i,x_j$ and replace them by just $x_i+x_j+x_ix_j$.

Then what happens? If there's just one number $N$ on the board left after all that, then $f(x_1,...,x_m) = f(N)$, so $N = f^{-1}(f(x_1,...,x_m))$ provided that $f(x_1,...,x_m)$ has exactly one preimage.

A hint for this function $f$ comes from $(1+x)(1+y)=1+(x+y+xy)$, so something like : add $1$ to all the numbers you have, and multiply these results together?

It is obvious that such a function does the job! In which case, we must add $1$ to each of the numbers, and multiply them all. That's like multiplying $\frac{2}{1}, \frac 32, \frac 43 ,...\frac {2011}{2010}$, which is just $2011$.

Now, whatever last number is on the board, one plus that is $2011$, so it is $2010$.

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The operation $x*y=x+y+xy=(x+1)(y+1)-1$ on real numbers is associative so the result does not depend on the order of steps and is equal to $$(1+1)(1+1/2)...(1+1/2010)-1=2011!/2010!-1=2010$$

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Suppose you choose $\frac1m$ and $\frac1n$ in the first turn, replace them by $\left(\frac{m+1}m\frac{n+1}n\right)-1$

(note that $x+y+xy=(x+1)(y+1)-1$)

In the next turn, you might choose two numbers $\frac1a$ and $\frac1b$, and the replaced number will look just as above, with $a,b$ replacing $m,n$. However, if you choose the new number obtained in the previous step, i.e. $\left(\frac{m+1}m\frac{n+1}n\right)-1$ and one of the original numbers $\frac1a$, then you replace them by $\left(\frac{m+1}m\frac{n+1}n\frac{a+1}a\right)-1$.

Fill in the intermediate steps to show by induction that the replaced number in any step will look like $\left(\prod_j\frac{a_j+1}{a_j}-1\right)$, so that the final answer will be $$\dfrac{2011}{2010}\dfrac{2010}{2009}\cdots \dfrac{2}{1}-1=2010$$.

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