When I take the 5 root of a big number, js gives me a wrong answer   var pow = 5; var A = 133; var B = 110; var C = 84; var D = 27; var E;  E = Math.pow((Math.pow(A, pow) + Math.pow(B, pow) + Math.pow(C, pow) + Math.pow(D, pow)), 1/pow);  console.log(E)    This should give 144 but gives 144.00000000000003 5 root of 61917364224 = 144
27
Consider the following code:   0.1 + 0.2 == 0.3  -&gt;  false   0.1 + 0.2         -&gt;  0.30000000000000004   Why do these inaccuracies happen?
83643
I need to find the last digit of $103^{103^{103^{103^{103}}}}$ so the value in $\mod10$. I know  \begin{align} 103^{103^{103^{103^{103}}}}&amp;=(100+3)^{103^{103^{103^{103}}}}\\ &amp;=100\cdot(stuff)+3^{103^{103^{103^{103}}}}\\ &amp;=3^{103^{103^{103^{103}}}} \mod10 \end{align}  And then I am stuck. I have observed that the last digit of a power of 3 is cyclic: $$3 \to 9 \to 7 \to 1 \to 3 \to 9 \to 7 \to\cdots$$  This means it depends on $103^{103^{103^{103}}}\mod4$.  Is this the right method? do I just need to keep doing this or is there a simpler "trick"?
7879
I'm trying to solve this equation:  $$2-x=-\sqrt{x}$$ Multiply by $(-1)$: $$\sqrt{x}=x-2$$ power of $2$: $$x=\left(x-2\right)^2$$ then: $$x^2-5x+4=0$$ and that means: $$x=1, x=4$$    But $x=1$ is not a correct solution to the original equation.  Why have I got it? I've never got a wrong solution to an equation before. What is so special here?
103775
I always forget this, when a trig function has an exponent does that mean multiply itself or apply itself to the result recursivly?  e.g. does $\sin(x)^2=\sin(x)\sin(x)$ or $=\sin(\sin(x))$? What about $\sin^2x$?
87190
Does math have an incorrect notation / syntax? I don't mean writing misaligned notation (google), but when you take something like a number to powers to powers to powers, $${{2^2}^2}^3$$ (I was told this is incorrect notation by a teacher). Is it really incorrect, or does it just need to be simplified with parentheses? Do people write maths like this?  a radical expression with the root being a radical expression? $$\sqrt[\sqrt{2^3}]{2}$$
20436
$$\sqrt[4]{161-72 \sqrt{5}}$$  I tried to solve this as follows:  the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:  $$a^4+30 a^2 b^2+25 b^4=161$$ $$4 a^3 b+20 a b^3=-72$$  In an attempt to solve this, I first tried to factor stuff and rewrite it as:  $$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$ $$4 a b \left(a^2+5 b^2\right)=-72$$  Then letting $p = a^2 + 5b^2$ and $q = ab$ you get  $$4 p q=-72$$ $$p^2+10 q^2=161$$  However, solving this yields messy roots. Am I going on the right path?
72082
I have tried this on my Windows 7 calculator with $\sqrt9 -3 $ it too gives some weird answer- ie$1.1546388020691628168216106791278e-37$. And so for any $n$(positive) $\sqrt n^2-n= wierd_ .answer$ Why does this happen?
21441
Some one please help me with this question. I am getting -10.5 as the wrong final answer.I found the formula in the following slide.Since the result was coming out in complex number i got its modulus.That is how i ended up with a 10.5 (2 * 5.25).
113604
I was messing around with my TI-84 Plus Silver Edition calculator and discovered that it will actually give me values when taking the factorial of any number $n/2$ where $n$ is any integer greater than $-2$. Why does this happen? I thought factorials were only defined for positive integers and $0$, so what is my calculator doing to get the answer $3.32335097$ when I enter $2.5!$? Is there actually a definition of $2.5!$ or is my calculator just being weird? How is the factorial function implemented?  I understand the binary implications of $2.5$, so that could possibly have something to do with it. I get a domain error when trying to take the factorial of $-1, 2.3, e$, and any number that is not of the form $n/2$ where $n$ is any integer greater than $-2$.
43762
I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
51377
I am looking over , and I came across a solution containing this:  Problem Bernardo chooses a three-digit positive integer $N$ and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer $S$. For example, if $N = 749$, Bernardo writes the numbers $10,\!444$ and $3,\!245$, and LeRoy obtains the sum $S = 13,\!689$. For how many choices of $N$ are the two rightmost digits of $S$, in order, the same as those of $2N$? $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 25$ Solution 1 First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities. Say that $N \equiv a \pmod{6}$ also that $N \equiv b \pmod{5}$ Substituting these equations into the question and setting the units digits of $2N$ and $S$ equal to each other, it can be seen that $b &lt; 5$ (because otherwise $a$ and $b$ will have different parities), and thus $a=b$. $N \equiv a \pmod{6}$, $N \equiv  a \pmod{5}$, $\implies N=a \pmod{30}$, $0 \le a \le 4$ Therefore, $N$ can be written as $30x+y$ and $2N$ can be written as $60x+2y$ Just keep in mind that $y$ can be one of five choices: $0, 1, 2, 3,$ or $4$, ; Also, we have already found which digits of $y$ will add up into the units digits of $2N$. Now, examine the tens digit, $x$ by using $\mod{25}$ and $\mod{36}$ to find the tens digit (units digits can be disregarded because $y=0,1,2,3,4$ will always work) Then we take $N=30x+y$ $\mod{25}$ and $\mod{36}$ to find the last two digits in the base $5$ and $6$ representation.[N \equiv 30x \pmod{36}][N \equiv 30x \equiv 5x \pmod{25}]Both of those must add up to[2N\equiv60x \pmod{100}] $N\equiv{30x\mod{36}}\\N\equiv{30x}\equiv{5x\mod{25}}$ Both of these must add up to $2N\equiv{60x\mod{100}}$ [...]  I'm not really sure which modular arithmetic rule to apply in the final three congruences, or where to start. How would I go about understand this?
86248
Why is any number (other than zero) to the power of zero equal to one? Please include in your answer an explanation of why $0^0$ should be undefined.
9763
I noticed relation between modulo operation and number which is power of two  Example I have to calculate $ 3431242341 \mod 2^5 $, which is $ 5 $ but it is equivalent to  $ ( 3431242341 \mod 2^9 ) \mod 2^5 $   I tried many examples and it seems to be true in general, and I am not sure if it is a coincidence or true in general that I can use first modulo operation ( greater number) and the result will be the same.
13150
My teachers have gone over rules for dealing with fractional exponents. I was just wondering how someone would compute say: $$(-5)^{2/3}$$ I have tried a couple ways to simplify this and I am not sure if the number stays negative or turns into a positive. I know that if a negative number is raised to an odd power it is negative, but fractional powers are neither odd or even. Is there a general rule for dealing with these types of problems?
80740
Using 5000-digit precision in PARI/GP, I discovered that the fractional part of $(\sqrt{2}+\sqrt{3})^{2008}$ is extremely small, less than $10^{-999}$.  Is there a simple explanation for this fact ?  This looks like a Pisot number issue (similar questions have already been studied on MSE, see for example ), but it’s a more complicated situation.   Related :
59458
Not sure what to do / how to start this... I have equcation of 504 is: $2 \cdot2 \cdot 2 \cdot 3 \cdot 3 \cdot 7$
64155
I was working on a problem that resulted in the calculation: $20 \equiv 10x \pmod{11}$.  I got the answer $x \equiv 2 \pmod{11}$ with the thought process: Since $10$ is a factor of $20$ I can rewrite $20 \pmod{11}$ as $10x \pmod{11}$ with $x=2$. But isn't this basically the same as using division, which we aren't supposed to do in modular spaces?  If I solve by multiplying $20$ by the multiplicative inverse of $10 \pmod{11}$ which is $10$, I get $200 \pmod{11}$ which simplifies to $2 \pmod{11}$ anyway. Did these numbers just happen to be the same or is my original method valid?