Hey guys, here is the problem
If a bunch of positive integers adds up to 20, what is the greatest possible product of these numbers?
I think its 1458 from 3*6*2=1458
Some please cheack to seeif it is right thanks
I agree .
If you want a mathematical justification read on. We need a few facts to prove that no matter how hard we try that $\displaystyle 3^6\cdot 2 = 1458$ is the best possible answer.
First, if $\displaystyle x,y\geq 0$ are integers (whole numbers) then $\displaystyle 2x+3y\leq 2^x \cdot 3^y$. You might say the RHS is exponents and LHS is just addition so it makes sense that RHS is much larger than LHS so maybe in fact we should have $\displaystyle <$ instead of $\displaystyle \leq $. But that is not completely true. For small values $\displaystyle x=0,1 \ y=0,1$ we actually have equality.
Second, any integer $\displaystyle n\geq 2$ can be written in the form $\displaystyle n=2x+3y$ where $\displaystyle x,y\geq 0$ are integers. Do you see why this is true? See if you can figure this out.
Third, any representation that has $\displaystyle 1$ amongs its summands (for example $\displaystyle 9+9+1+1=20$) cannot be the maximum value. The reason is the follows. Say $\displaystyle a_1+a_2+...+a_k + 1 = 20$ is a representation then its product is $\displaystyle (a_1)(a_2)...(a_k)(1) = a_1a_2...a_k$. While $\displaystyle a_1+a_2+...+(a_k+1) = 20$ is also a representation and its product is $\displaystyle (a_1)(a_2)...(a_k+1)$. Which one is larger? The second one definitely because in the first product the first $\displaystyle k-1$ numbers are the same as in the second product while in the second product the last number, $\displaystyle a_k+1$, is larger than the last number, $\displaystyle a_k$, in the first product.
Now we are ready to state the main result.
Theorem: The maximum product for the representation $\displaystyle 20$ is achieved where all the summands are $\displaystyle 2$'s and $\displaystyle 3$'s.
Proof: Say that $\displaystyle a_1+a_2+...+a_k = 20$ is a representation. To be maxed we require that each term $\displaystyle \geq 2$, look at the third fact we are using. Then each summand can be written as a sum of twos and threes (because each is $\displaystyle \geq 2$) by the second fact. Meaning $\displaystyle a_1 = 2x_1+3y_1$ and $\displaystyle a_2 = 2x_2+3y_2$ ... and $\displaystyle a_k = 2x_k + 3y_k$. Now $\displaystyle a_1a_2...a_k \leq 2^{x_1}3^{y_1}\cdot 2^{x_2}3^{y_2}\cdot ... \cdot 2^{x_k}3^{y_k}$ by the first fact we are using. But the point is we decomposed each $\displaystyle a_1,...,a_k$ into $\displaystyle 2$'s and $\displaystyle 3$'s. Meaning $\displaystyle 20 = a_1 + ... +a_k = 2x_1+3y_1+...+2x_k+3y_k = \underbrace{(2+...+2)}_{x_1}+\underbrace{(3+...+3) }_{y_1}+...$, in fact the number of times $\displaystyle 2$ appears is $\displaystyle x_1+...+x_k$ times and $\displaystyle 3$ appears $\displaystyle y_1+...+y_k$ times. Which means this new representation is at least as big as the previous one. Thus, any representation which is maxed must be made out of $\displaystyle 2$'s and $\displaystyle 3$'s.
Finally we can solve the problem. Say that you have $\displaystyle x$ twos and $\displaystyle y$ threes then $\displaystyle 2x+3y = 20$ and their product is $\displaystyle 2^x3^y$. Thus, we want to maximize $\displaystyle 2^x3^y$ subject to $\displaystyle 2x+3y=20$. But this is simple. The only possibilities are: $\displaystyle (10,0),(7,2),(4,4),(1,6)$ and its respective sums are: $\displaystyle 1024,1152,1296,1458$. Thus, the max product is $\displaystyle 1458$.
We can in fact generalize this result. Say $\displaystyle N\geq 2$.
If $\displaystyle N\equiv 0(\bmod 3)$ the max product is $\displaystyle 3^{N/3}$.
If $\displaystyle N\equiv 1(\bmod 3)$ the max product is $\displaystyle 2^2\cdot 3^{(N-4)/3}$
If $\displaystyle N\equiv 2(\bmod 3)$ the max product is $\displaystyle 2\cdot 3^{(N-2)/3}$.
(The idea is to get as many three's as possible).
You can prove these results by inducting on each of the three cases.
1)Why would the product of $\displaystyle e$'s make the highest number possible?
2)It is not possible to have $\displaystyle e+e+...+e = N$ because $\displaystyle e$ is irrational.
3)I have an idea of what you might mean, look below.
Given a number $\displaystyle N\geq 2$. And let $\displaystyle x_1+x_2+...+x_k = N$ be positive real numbers. Then the maximum product which can be formed is when all are the same, so that product is $\displaystyle ( \tfrac{N}{k})^k$. What is the value of $\displaystyle k$ which makes this the maximum value? We will solve this problem by first defining the function $\displaystyle f(x) = (\tfrac{N}{x})^x = \tfrac{N^x}{x^x}$ (where $\displaystyle x>0$). The maximum occurs when $\displaystyle f'(x) = 0$ (drawing a graph for various values of $\displaystyle N$ is very helpful). Solving this equation we get $\displaystyle x=\tfrac{N}{e}$. Let $\displaystyle [ ~ ]$ be the 'nearest integer function'. Then by the shape of the curve we have that the best integer value is $\displaystyle [ \tfrac{N}{e} ]$. For example, if we want to maximize the product of the sum for $\displaystyle N=10$ then $\displaystyle k = [\tfrac{10}{e}] = 4$. But then the maximized product is $\displaystyle (\tfrac{10}{4})^4 = 39.0625$.