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)z=222(x+y)^{\displaystyle z} = 2^{2^{2}}, what is the maximum value of xyzxyz?

2 - Full Circle

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

3 - Citamotuautomatic

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

#FunctionalEquations #Exponential #WholeNumbers #DelaCorte

Note by Guilherme Dela Corte
7 years, 2 months ago

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Comments

Answer 1

Alternative representations for 222 2^{2^2} on whole numbers include 161,  42, 16^1, \; 4^2, and 24 2^4 , which leads straightforward to z=1,  2,  4 z = 1, \; 2, \; 4 . Respectively, by AM-GM, we have that x+y=16max(xy)=64,  x+y=4max(xy)=4 x+y=16 \to \max(xy) = 64, \; x+y=4 \to \max(xy)=4 and x+y=2max(xy)=1 x+y=2 \to max(xy)=1 . In a clear way, our desired answer is max(xyz)=64. \max(xyz) = \boxed{64.} .

Answer 2

Multiplying abbcacab \cdot bc \cdot ac yields (abc)2=602 (abc)^2 = 60^2 . Since we are talking about whole numbers, it is clear that abc=60>0 abc = 60 > 0 , thus. Comparing the product to the missing multiplying variables on 12,1512, 15 and 2020 , we can see that c=5c = 5 , b=3b=3, a=4a=4, which means f(x)=(x3)(x4)(x5) f(x) = (x-3)(x-4)(x-5) . This leads to f(6)=123=6. f(6)=1\cdot2\cdot3=\boxed{6.}

Answer 3

Let xy=22=41 x^y = 2^2 = 4^1 . From the function's definition, it follows that h(22)=2+2=4 h(2^2) = 2 + 2 = 4 and h(41)=4+1=5 h(4^1) = 4 + 1 = 5 . However, 22=41 2^2 = 4^1 , and our result shows us that h(22)h(41) h(2^2) \neq h(4^1) , which means h(x) h(x) is not a function. QED     \text{QED} \; \; \blacksquare

Guilherme Dela Corte - 6 years, 5 months ago
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