RiemannIntegralzzz

Sorry, the answer was actually 3, which I solved myself, after it said your solution was incorrect. Thank you for trying however! Here is the solution:

Call the two integers a and b. So, $lcm[a,b]*gcd[a,b]=ab$, so $gcd[a,b]=\frac{ab}{lcm[a,b]}$.

We know that $a<10^6$ and $b<10^6$, so $ab<10^12$. We also know that $lcm[a,b] \geq 10^9$. Therefore, $gcd[a,b]<10^3$, so it has at most $\boxed{3}$ digits.

Ah, I am sorry Guest! I realized my mistake, and I found the same question asked with a correct answer here: https://web2.0calc.com/questions/paint-a-sphere/new?

Hope this helps!

Sorry for edits, I realized my LaTeX wasn't working, so I had to redo it.

The surface area of the sphere is given by $4*\pi*r^2$, so we have the surface area is $4*\pi*10^2=400*\pi$.

To find the amount of paint as a volume, multiply it by $0.001$ meters to get about $400*0.001*\pi=\frac{4*\pi}{10}=\frac{2*\pi}{5} \approx \frac{2*3}{5} \approx \boxed{\frac{6}{5}m^3}$ of paint

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If you use the square of a binomial, you get $(1^2+1*2+2^2)+(2^2+2*3+3^2)+...+(99^2+99*100+100^2)=(1+2)^2+(2+3)^2+(3+4)^2+...+(99+100)^2=3^2+5^2+7^2+...+197^2+199^2$. Now, this is equivalent to the sum of the first odd squares, minus $1^2$. The formula for the sum of the first n odd perfect squares is $\frac{n(2n+1)(2n-1)}{3}$. Now, let's say we include the $1^2$ so we can use the formula, just remember to subtract $1^2$ at the end. This is then the first $100$ perfect squares. Using the formula, we get $\frac{100(201)(199)}{3}=\frac{3999900}{3}=1333300$. But wait! We need to subtract $1^2$, because we added it to use the formula. So our final answer is $1333300-1^2=\boxed{1333299}$

Oh my god, thank you so much, I just realized where I went wrong. I had a brain freeze, on the last step, where I divide by 6, you get $x>=\frac{-10}{6}=\frac{-5}{3}$, not $x>=-6$. I'm really sorry.

Hello! Completing the square is not used for linear equations, it is a method of turning the quadratic $ax^2+bx+c=0$ into the form $a(x+b)^2+c=0$, which is not applicable for linear equations.

Out of a $10$-sided die, there are $4$ prime numbers: $2,3,5,7$

The proability of rolling a prime number is $\frac{4}{10}=\frac{2}{5}$.

So the probability of rolling 3 prime numbers is $(\frac{2}{5})^{3}=\frac{8}{125}$.

But wait! The problem says EXACTLY three, so we can't roll a prime number on the fourth roll.

So the probability of rolling EXACTLY 3 prime numbers is $\frac{8}{125}*\frac{6}{10}=\frac{8}{125}*\frac{3}{5}=\frac{24}{625}$.

But wait! There are ${4 \choose 3}=4$ ways to choose 3 of the four dice to roll prime numbers on, so the final probability is $\frac{24}{625}*4=\boxed{\frac{96}{625}}$