SparklingWater2

you can notice that you are adding 6 more to the previous number, starting with 1. if we divide each of these numbers by 6, we will always get a remainder of 1. we can subtract 1 from the last number and divide it by 6, to see how many numbers in total are in the sequence. we get (271-1)/6=270/6=45.

now that we know there are 45 digits and each of the remainders when divided by 6 is 1, we know that the remainder of the sum divided by 6 is 45/6=7 3/2=$\boxed3$

we can say that one side length is a, another b, and the last c. We know that:

a+b>c

a+c>b

c+b>a

a+b+c=15

We also know that order matters. if a=1, b and c must be 7. if a=2, b must be 7, and c must be 6. we can continue counting:

a=1,b=7,c=7

a=2,b=7,c=6

a=3,b=6,c=6

a=3,b=7,c=5

a=4,b=6,c=5

a=4,b=7,c=4

a=5,b=5,c=5

those are all the possible solutions, so $\boxed{7}$ is our answer

a. 1 because the number could be any prime^19, which has 19+1=20 positive divisors.

b. 3. We can start by finding the prime factorization of 20, which is 2^2x5 so there must be 3 primes.

the volume is the 3 sidelengths multiplied together, so:

$V=4\cdot12\cdot7=\boxed{336}$cm$^3$

surface area is each side length times the other multiplied by two:

$SA=2\cdot(4\cdot12+4\cdot7+12\cdot7+7\cdot4)=2\cdot(160)=\boxed{320}$cm$^2$

Technically, there are infinite answers, but the number that is in the middle of 0 and 1 is, $\frac12$

If it is prime, that means that last number has to be odd. this leaves us with 5 digits for the last one: 1,3,5,7, and 9. but becayse the digit in the units place is 3 less than the digit in the tens place that has to be 3,4, or 5, we have only 1 answer: $\boxed{41}$

we can first calculate the rectangle beneath the smicircle's area, which is 20x25=500

we see that the radius of the semicricle is 12.5,  and the area of the circle is $\pi$r$^2$, so the area of the semicircle is:

$\frac{\pi r^2}{2}=\frac{3.14\cdot156.25}2=245.3125$

500+245.3125=745.3125

It is most likely the 4th one.

there are 12 choose 5 = 792 possible ways of getting to point (5,7) from (0,0), and there are 7 choose 3=35 possible ways of getting to point (4,3) from (0,0), so the number of ways to get to (5,7) from (0,0) are

 792-35=$\boxed{757}$

Good idea; each course lasts around 2-3 months(I just did an intro to geom that lasted half a year), and is pretty pricey, a few hundred dollars ;-; 

OMG! I USED TO FEEL LIKE THAT TOO! (Keywords:USED TO)

I was originally doing Kumon, and I felt like I couldn't get any farther with Kumon, so I quit at level H. I then discovered alcumus, like you. My dad encouraged me to do alcumus, and mathcounts trainer as well, and one day, we found out a way to enroll in CLASSES on AOPS. There are SEVERAL different classes, like algebra A and B, intro to Geom, AMC 8, 10, and 12 classes, number theory, probability and counting, etc. I feel like the classes REALLY helped. Tbh, I am only 11, I just REALLY like math, (spend like 1 hour on here a day, 30 minutes school math, another 30 minutes for class homework and alcumus), and I've been improving at an alarming rate. All you have to do is make an account if you haven't already, and look up on your screen, and you will see a few options, hover over online school(If you are on computer. If you are on iOS or android then press it) and select Course catalog. There, you will find a lot of different options of classes. You can press one you feel like doing, and to the right of the summary, there are two links, one to see if you are ready for this class, and one to see if you need it at all. You can keep doing these mini tests, and find one that you are both interested in, and suits you. If there are things you don't understand in the "do I need this" section, then by the end of the course, you will understand it. Hopefully this is helpful, although it is lengthy.