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I have 5 questions I don't know how to solve

1. Given that a^{2}- b^{2}= -280 and a+ b = 20, find the value of a-b.

2.Solve for x: (How do I show this?) x square root to the power of x square root to the power of x square root infinitely = 2

3. In what bases, b, does (b+6) divide into (5b+6) without any remainder?

4. Using each of the ten digits exactly once (1,2,3,4,5,6,7,8,9,0), find the 2, five-digit numbers that have the smallest product

5. If 4 numbers from 1 to 1000 are chosen at random, one at a time, find the probability that they will be chosen in decreasing order.

Thanks

Eggshell

Eggshell Feb 24, 2019

#1**+1 **

\(a^2-b^2 = -280,~a+b=20\\ (a-b)(a+b)=-280\\ (a-b)(20) = -280\\ (a-b)=-14\\ (a+b)=20\\ 2a=6\\ a=3,~b=17\)

\(\sqrt{x}^{\sqrt{x}^{\sqrt{x}^{\dots}}}=2\\ (\sqrt{x})^2 = 2\\ x = 2\)

\(\dfrac{5b+6}{b+6} = \dfrac{4b}{b+6}+1 \in \mathbb{N}\\ \dfrac{4b}{b+6}\in \mathbb{N}\\ \dfrac{4}{1+\frac 6 b} \in \mathbb{N}\\ 1+\dfrac 6 b = 4, 2\\ b=2,~6\)

\(\text{To find the two 5 digit numbers with the smallest product simply choose the smallest}\\ \text{available digit alternately forming the two factors}\\ f_1 = 10468\\ f_2 = 23579\)

\(\text{Assuming no replacement, any set of 4 numbers can be arranged into a decreasing sequence}\\ \text{thus there are }\dbinom{1000}{4} \text{ strictly decreasing length 4 sequences using 1-1000}\\ \text{there are }\dbinom{1000}{4}4! \text{ total length 4 sequences}\\ \text{thus the probability of being strictly decreasing is }\\ p = \dfrac{\dbinom{1000}{4}}{\dbinom{1000}{4}4!}= \dfrac{1}{24}\)

.Rom Feb 24, 2019