A Tree of Numbers
Bob has the number 2014 written on a chalkboard. He splits the number into a sum of two integers, and writes out the product of the two aside. Then he is left with a sum of two numbers. He then splits both of these two and lists out the product. When he can no longer split a number into a sum of two positive integers, he adds the products he had listed out. As an example, if Bob started with 10, he could split it as \begin{equation*}\begin{aligned} 10 & = 6 + 4 \\ & = (3 + 3) + 4\\ & = ((1+2) + (1+2)) + (2+2)\\ & = ((1 + (1+1)) + (1+(1+1))) + ((1+1) + (1+1)) \end{aligned} \quad \begin{aligned} 24\\ 9\\ 2, 2,4\\ 1,1,1,1\\ \end{aligned}\\ \implies 45 \end{equation*} Find the maximum possible value that Bob can get.
The answer is 2027091.
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1 solution
As you can see that the final sum is always the same no matter in what way you split the number [try by yourself]. So the easiest way to do it is by taking 1 out in each step like this: 10=9+1=(8+1)+1.... and so on. The formula for the final sum will be n*(n+1)/2 ,where n is the given number. For n=2014, the result will be 2027091.