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 \(h(x^y) = x + y\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestAnswer 1Alternative 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 \( \max(xyz) = \boxed{64.} \).

Answer 2Multiplying \(ab \cdot bc \cdot ac \) 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 \( f(6)=1\cdot2\cdot3=\boxed{6.} \)

Answer 3Let \( x^y = 2^2 = 4^1 \). 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. \( \text{QED} \; \; \blacksquare \)

Log in to reply