fun binomial question

Question

imqwerty · Answer

Let $\large x=\left(3 \sqrt{6}+7\right)^{89}$ . If  $\{x\}$ denotes the fractional part of $'x'$ then find the remainder when $x \{x\}+(x\{x\})^2+(x\{x\})^3$ is divided by 31.

empty · Answer

Wait, since x is irrational, there is no fractional part, therefore {x}=0 and when $x \{x\}+(x\{x\})^2+(x\{x\})^3$ is divided by 31 you get 0. Done, HA!

ganeshie8 · Answer

Don't you mean the expression is
$$x -\{x\}+(x-\{x\})^2+(x-\{x\})^3$$

?

ganeshie8 · Answer

Let $y = (3\sqrt{6}-7)^{89}$

Since $x-y $ is an integer and $0\lt y \lt 1$, it is easy to conclude that $\{x\}=y$.

ganeshie8 · Answer

$$x - \{x\} = x - y  = (3\sqrt{6}+7)^{89} -(3\sqrt{6}-7)^{89} $$

imqwerty · Answer

the fractional part is not equal to 0

the binomial expansion of $(3\sqrt{6}+7)^{89}$  will have some irrational terms like these -

$~^{89}C_1(3\sqrt{6})^1(7)^{88}~,~^{89}C_3(3\sqrt{6})^3(7)^{86}~,~^{89}C_5(3\sqrt{6})^5(7)^{84}~~...$ 

so all these terms will have $\sqrt{6}$ in them  
and then when they are added there is no way that the fractional part will become 0

imqwerty · Answer

and the expression is this-> $x \{x\}+(x\{x\})^2+(x\{x\})^3$

ganeshie8 · Answer

what do you mean by $x\{x\}$ ?

are you multiplying x by its fractional part or what ?
that expression is not so clear

alivejeremy · Answer

That's some hard math qwerty

hari5719 · Answer

that aint fun bro

dan815 · Answer

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