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# On Whole Numbers

Hello, buddies! In this note you will find a few atypical exercises I've made about the whole numbers. They should not seem too hard, but if they do, don't be afraid to search the topics and ask for help.

That is how I intend to start my series of notes. Keep an eye for more pastimes!

1 - Lift Me Up

If $$(x+y)^{\displaystyle z} = 2^{2^{2}}$$, what is the maximum value of $$xyz$$?

2 - Full Circle

The numbers $$a, b, c$$ are the roots of the monic third degree polynomial $$f(x)$$, such that $$ab=12, bc=15, ac=20$$. Evaluate $$f(6)$$.

3 - Citamotuautomatic

Prove that no $$\; \mathbb{Z_{+}} \rightarrow \mathbb{Z_{+}}$$ functions $$h(x)$$ exist such that $$h(x^y) = x + y$$.

Note by Guilherme Dela Corte
2 years, 7 months ago

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Alternative representations for $$2^{2^2}$$ on whole numbers include $$16^1, \; 4^2,$$ and $$2^4$$, which leads straightforward to $$z = 1, \; 2, \; 4$$. Respectively, by AM-GM, we have that $$x+y=16 \to \max(xy) = 64, \; x+y=4 \to \max(xy)=4$$ and $$x+y=2 \to max(xy)=1$$. In a clear way, our desired answer is $$\max(xyz) = \boxed{64.}$$.

Multiplying $$ab \cdot bc \cdot ac$$ yields $$(abc)^2 = 60^2$$. Since we are talking about whole numbers, it is clear that $$abc = 60 > 0$$, thus. Comparing the product to the missing multiplying variables on $$12, 15$$ and $$20$$, we can see that $$c = 5$$, $$b=3$$, $$a=4$$, which means $$f(x) = (x-3)(x-4)(x-5)$$. This leads to $$f(6)=1\cdot2\cdot3=\boxed{6.}$$

Let $$x^y = 2^2 = 4^1$$. From the function's definition, it follows that $$h(2^2) = 2 + 2 = 4$$ and $$h(4^1) = 4 + 1 = 5$$. However, $$2^2 = 4^1$$, and our result shows us that $$h(2^2) \neq h(4^1)$$, which means $$h(x)$$ is not a function. $$\text{QED} \; \; \blacksquare$$ · 1 year, 10 months ago