Hoàng Nguyễn

2022-07-03

11, Find the least integer n such that f(x) is 0( nx ) for each
of these functions.
a) f(x) = 2x + (logx)^10
b) f(x) = (x^4 + 5logx) / (x^4 + 10)

12/A sequence of pseudorandom numbers is generated as follows

x0  = 4

x_i  = ( 6x_i–1  + 5 ) mod 13 if i > 0

Find x6

user_27qwe

11. Let's solve the given functions one by one to find the least integer n for which f(x) is 0 (mod nx).
a) $f\left(x\right)=2x+\left(\mathrm{log}x{\right)}^{10}$
To find the least integer n such that f(x) is 0 (mod nx), we need to find the least value of n such that f(x) is divisible by nx for all values of x.
To solve this, let's set up the congruence equation:
$2x+\left(\mathrm{log}x{\right)}^{10}\equiv 0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}nx\right)$
We can rewrite the equation as:
$2x\equiv -\left(\mathrm{log}x{\right)}^{10}\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}nx\right)$
To simplify the equation, let's assume x > 0. Taking the logarithm on both sides:
$\mathrm{log}\left(2x\right)\equiv \mathrm{log}\left[-\left(\mathrm{log}x{\right)}^{10}\right]\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}n\right)$
Now, let's apply properties of logarithms:
$\mathrm{log}\left(2x\right)\equiv 10\mathrm{log}\left(-\mathrm{log}x\right)\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}n\right)$
To eliminate the logarithms, we can exponentiate both sides with a suitable base. Let's choose the base 10:
${10}^{\mathrm{log}\left(2x\right)}\equiv {10}^{10\mathrm{log}\left(-\mathrm{log}x\right)}\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}n\right)$
Simplifying further:
$2x\equiv \left(-\mathrm{log}x{\right)}^{10}\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}n\right)$
Now, we have a simplified congruence equation. We can use this equation to find the least integer n.
b) $f\left(x\right)=\frac{{x}^{4}+5\mathrm{log}x}{{x}^{4}+10}$
Similar to the previous function, we need to find the least integer n such that f(x) is 0 (mod nx). Let's set up the congruence equation:
$\frac{{x}^{4}+5\mathrm{log}x}{{x}^{4}+10}\equiv 0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}nx\right)$
Multiplying both sides by $\left({x}^{4}+10\right)$:
${x}^{4}+5\mathrm{log}x\equiv 0\phantom{\rule{1em}{0ex}}\left(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}nx\right)$
Now, we have a congruence equation for this function as well. We can use this equation to find the least integer n.

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