moneydude242

For the all the units digits under 9, it just won't work. Even if the first digit is divisible by 7, adding one to the units digits will only change it by 1 value, which doesn't make it divisable by seven. If you know modular arithmetic, the first number would be 0 mod 7, and the second 1 mod 7. For the value 9, it gets slightly more tricky. However, if you try a number like 759, adding 1 will give you 760, or -1 mod 7. Adding another 9, like 5599 (it there must be an even amount of 5 and 9s - can you see why?) pulled up to 5600 will give you 11. 555999 gives you 556000, or 16 (doesn't work!). The next one (55559999) is much too big. From now, we can consider numbers that end with 0. For the number 770 (769), it doesn't work. Same for for 7770, 77770, 777770, etc. 700 doesn't work, and 7000 doesn't. But 70000 works! 70000 and 69999 works! Sadly, afterwards no other number works. So your answer is one. (BTW, I'm not very sure if this is correct, because I somewhat brute forced this.)

Also, just a bonus thing I learned, the sum of the digits (if it has a nine - otherwise it doesn't work) has to be divsible by 21 (7*3) - idk who asked though

I would like to draw a diagram for you, but I was too lazy to learn Asymptote 

But I drew a lil' diagram on my paper, and found that 3 adjecent side are the \(10\) sides. Honestly, I don't have a proof for this, but I'm sure that this means the "rectangular" prism is actually a cube. That means that the side lengths are \(\sqrt{10}\), or the remainder is \(10\sqrt{10}\). 

This person isn't very good at combinatorics, but I guess I will give this a try? 

Let's call the American skaters \(a\), the Canadian skaters \(c\), and the other skaters \(o\). 

(Quick note, here are the numbers for \(a,c,\) and \(o \).)

\(a=2\)

\(c=5\)

\(o=12\)

Than, in the top 3 skaters, we want a standing that looks something like this: \((a_1,o_6,c_2)\)

(for this problem, I'm assuming that the Americans or Canadians are different, let me know otherwise...)

We can think just think of one case, since we will be multiplying the probablities, and multiplication is commutative.

(BTW, let me know if there is AT LEAST ONE in the problem, or just one American/Canadian)

I'm just gonna take the case \((a,c,o)\). For the American, there are 2 choices: \(a_1 \) and \(a_2\). For the Canadians, likewise, there are 5 choices. And for the other 12 people, there are 12 choices. We can simply multiply these 3 numbers together and get \(2*5*12 = 120\). We can multiply this number by 6, since there are \(3! \) ways to arrange the standings, so we get \(120*6 = \boxed{720}\)

First off, you want to simplify the equation.
\(3x^2 - 2x^2 = x^2 \)

\(12x+14x = 26x \)

\(4-18 = -14 \)

From this we can write out a more simplifed equation:

\(x^2+26x-14 = \text{__}(x+\text{__})^2 + \text{__}\)

From on here, we can use Simon's Favorite Factoring Trick (SFFT) to make the left side into a similar form as the right.

\((x+13)^2 = x^2 + 26x + 169, \text{ so if the third blank equals x, than } 169+x = -14, \text{or } x = -183\)

Then, you can factor the right hand side as

\((x+13)^2 -183\)

So that gives you your answer: the first blank is 1 (there is no coefficient, so that means it's one), the second is 13, and the third is -183.

OK corrected version:
 

In each of three boxes are four straws with integer lengths 1 cm, 2 cm, 3 cm, and 4 cm. A straw is randomly selected from each box. What is the probability that a triangle can be formed with the three straws chosen?

Could anyone also add an explanation with their answer? Thanks! 

ummm oops changing that now

Thank you very much for your explanation! I got the answer before this, but it is also nice to learn the equation!

I believe this is a typo, because after a bit of thinking, wouldn't the next number be 191?

I could try to help you with this question, but could you please clarify what the blanks are in this problem? What I am referring to is " both of which are greater than  " and "The resulting product is less than  ". Thank you! 

Ohh, ok I understand. The 14 each time carries onto the next digit, which causes you to add 14. Thank you so much for the help, and your patience! 

So I'm not the best at math, but I'll give this a shot. We have the set {1,2,2,4,4,4}. First off, we have 6! = 720, not counting any duplicates. Because there are 3 copies for each time we rearrange the set, (because some of them are us just rearranging the fours), we divide 720 by 3, to get 240. We repeat this process with the two 2s, and get 120. Now, I'm not 100% sure about this answer, so Melody (or anybody else), if you see this and it's wrong, could you please correct me? Thank you!

Ok, I get that, but how did you get 14 as a remainder? I didn't see any division... (sorry, but I really don't understand)

Thank you for telling me how this works, but could you please explain why you multiplied 20 by a 4 and got 140?

Ok, so right now you've got me confused. Do you mean that \(7+77+777+7777 = 8500?\) I believe the sum of those four numbers is 8638... Also, what do you mean that "The sum of each group is 85000000, which has a 7 in the hundreds place"?

Thank you!

Uh, but when you were multiplying the (n-2)(2n-3), wouldn't it multiply/add to -7n?