HAH! This is right, I'm positive: 9867312, and is the largest I can get.

---

"In many ways, the work of a critic is easy. We risk very little, yet enjoy a position over those who offer up their work and their selves to our judgment. We thrive on negative criticism, which is fun to write and to read. But the bitter truth we critics must face, is that in the grand scheme of things, the average piece of junk is probably more meaningful than our criticism designating it so. But there are times when a critic truly risks something, and that is in the discovery and defense of the new. The world is often unkind to new talent, new creations. The new needs friends. Last night, I experienced something new: an extraordinary meal from a singularly unexpected source. To say that both the meal and its maker have challenged my preconceptions about fine cooking is a gross understatement. They have rocked me to my core. In the past, I have made no secret of my disdain for Chef Gusteau's famous motto, "Anyone can cook." But I realize, only now do I truly understand what he meant. Not everyone can become a great artist; but a great artist can come from anywhere."

-Anton Ego/Peter O'Toole (1932-2013)

Hah, didn't even notice there wasn't a topic on this. I'm actually a 'prodigy' in math, so to speak. I'm in 9th grade and taking 11Honors math (Pre-Calc). Now, most people would think, "That's not big of an achievement. I've seen better." So have I. But what makes me different is that while others went to CTY (Center for Talented Youth) and took math courses there, therefore giving them the credit they need to pass a course, or LISG (Long Island School for the Gifted), where they were accelerated multiple times with heavy-duty education and thrown back into the curriculum with accelerated in almost all of their courses, or some other program. I didn't. I did so well in the public school system where I go that I was recommended to go to a higher level, which is why I'm known as a prodigy. Plus, I'm a grade ahead in English. But let's see about that problem... (will be edited with answer that I get eventually)

I had a big essay ready for you, but I'll just say this: if people just put effort into something they like, it shouldn't be too hard to excel past high school mathematics; The motivation/drive/interest has to be there though, glad you did something with it. 9867312? :V Yeah math can be pretty cool.

---

shout out 2 kai-v 2.

The Complete Competitive HMS Resource

GOT IT! Here: 9867312, this is the largest I can get.

It IS indeed divisible by 7, 6, AND 42. I don't know how I skipped this one, but I was pretty close last night. I should've thought about switching the 6 and 7 than switching the 8 and 7 after 7 failed to divide...

But can you explain how you came across this?

Actually, my dad works as a business manager, and grosses about $1million a year, personal salary ... sure he has to do basic arithmetic, but in no way was he ever good at math, he never even took calculus... My godfather, a strong businessman and entrepreneur owns 11 businesses, and one of his best friends and partners is worth $3 billion ... never even took any form of higher math. Same goes for some other family members I have... Doing well in business, being a CEO, business management, etc etc... if you have the right calculator or computer, or if your brain is just programmed to run a business, you really don't need math to succeed. Sure, as I said, its a great thing to have, but the most successful people didn't get to where they are today solely through math... and yes, some did, but its not the only way.

In my honest opinion, being so closed minded as to say math is the only way (or saying any one thing is the only way) is the downfall of any attempts at success. You must be open minded and try EVERY way, be creative, see all the openings, and create something from nothing.

I live where people just think "success and fame" (Beverly Hills) and have so many connections to people worth money that even I can't even comprehend at times... and math certainly has not played a role in all of their successes.

I like math, but I don't think I will use higher math in the business opportunities that I engage in, but I know that I will be as successful, if not more successful, than all of my family members.

I do not think anybody will take your family as the norm though. Doing a few advanced mathematic classes definitely opens a lot more doors in an education system like in Canada. Of course, there are some careers like being a lawyer that do not even require you to take natural sciences courses and can still get you a huge salary. However, grasping such abstract concepts as there can often be in mathematics is never a bad aspect for intelligence in general, and unless you go for some exceptions in society, the average person who wants to get a great, paying job will have to consider mathematics in their scholarship.

I agree that math is great and is amazing for intelligence... thats why I enjoy it haha, but the sad thing is, unless you are going into research and development and the such, most of the highest paying jobs don't really revolve around math... especially with the technology that allows computers to do the math for us.
At least where I live, my family is the norm...

As I've said, I like math, and do believe that if you can master all subjects you will go farther with yourself, but lawyers and businessmen and, of course actors, singers, and entertainers, are 'successful' in a money sense and probably never even took advanced math.

---

~SaMaster14

you're still missing her point completely

Excellent Ga'Hooleone, you are correct in believing that 9867312 is the largest possible number given what I said. Since I planned to have a new problem every Monday, a tournament in July and to have explained how to achieve the conclusion to the problem on weekends, your timing is perfect. I will have a new problem up by Monday, or if preferred, I'll put it up now as an exception to the roster? Ah, I can't help waiting, I'll post it now. On some other note, in regard to opening doors,bit of a cliche overuse, eh?

Bewildered, Forgetful and not so fail, Bob.

Bob has an unreliable memory. When asked for his phone number on six different occasions, it wasn't surprising that he failed to give out his real number on each of his different attempts.

#1: 913916
#2: 917931
#3: 917601
#4: 914544
#5: 919772
#6: 911805

Although, his memory is not a complete failure since:
a) his telephone number does in fact consist of 6 digits.
b) in every attempt, the first two digits are correct. (91)
c) in each of his attempts, he also had exactly one of the last four digits in the correct place.
Question: What is his telephone number provided that the last four digits are all different?

What I have seen is that people consider the guide to be for losers, when in reality, it isnt. I actually encourage you to make use of the guide anyway. If yu can solve it on your own then thats great, but it is better to show working and use the guide rather to prove just the answer, or complete it yourself then answer the questions?

[The guide shall be posted soon enough.]

---

Although I am inactive, check out my YouTube account for existing Skyrim tutorials.
Skyrim Guides:

• One-Handed And Dual Wielding Guide
  
• ETA Next Guide: 8/9/12

in an equilateral triangle each side is 12 cm.if circle is incribed touching the sides.
root 3 value=1.73,pi=3.14
then raduis of circle?
area of incircle?
the area of portion not included in the circle?

---

draw triangle and circle in which circle is incribed in triangle touching three sides.

Excellent Ga'Hooleone, you are correct in believing that 9867312 is the largest possible number given what I said. Since I planned to have a new problem every Monday, a tournament in July and to have explained how to achieve the conclusion to the problem on weekends, your timing is perfect. I will have a new problem up by Monday, or if preferred, I'll put it up now as an exception to the roster? Ah, I can't help waiting, I'll post it now. On some other note, in regard to opening doors,bit of a cliche overuse, eh?

Bewildered, Forgetful and not so fail, Bob.

Bob has an unreliable memory. When asked for his phone number on six different occasions, it wasn't surprising that he failed to give out his real number on each of his different attempts.

#1: 913916
#2: 917931
#3: 917601
#4: 914544
#5: 919772
#6: 911805

Although, his memory is not a complete failure since:
a) his telephone number does in fact consist of 6 digits.
b) in every attempt, the first two digits are correct. (91)
c) in each of his attempts, he also had exactly one of the last four digits in the correct place.
Question: What is his telephone number provided that the last four digits are all different?

What I have seen is that people consider the guide to be for losers, when in reality, it isnt. I actually encourage you to make use of the guide anyway. If yu can solve it on your own then thats great, but it is better to show working and use the guide rather to prove just the answer, or complete it yourself then answer the questions?

[The guide shall be posted soon enough.]

You say that the last four digits are ALL different. Meaning that within that last four, there shouldn't be any repeating numbers. Therefore, your choices are then limited to...

#1: 913916
#2: 917931
#3: 917601
#6: 911805

After that, ONE of the last four digits are in the correct place. Even though that digit is unknown, you can rule actually rule out a set based on the number of times he repeated a number in a row.

#1: 913916
#2: 917931
#3: 917601

Out of the three, #2 is either followed or preceded by a repeating number.

HOWEVER. While I want to stand by my reasoning for 2, the 3rd hint, "in each of his attempts, he also had exactly one of the last four digits in the correct place," throws me off. Set 3 has 2 numbers "in the correct place" (in my opinion). SO.......I'm kind lost.

GOT IT! Here: 9867312, this is the largest I can get.

It IS indeed divisible by 7, 6, AND 42. I don't know how I skipped this one, but I was pretty close last night. I should've thought about switching the 6 and 7 than switching the 8 and 7 after 7 failed to divide...

But can you explain how you came across this?

This required more common sense then anything. If you had a set of numbers, how do you make the largest one? say, 1 4 2 5 3 6? You put them in descending order: 654321. Since you had to account for 7, 2 had to be the last one. So I switched the 2 numbers that would affect the size the least: 1 and 2. So now we have 9876312. However, by putting 8 and 7 next to each other, you get some divisibility problems, which I'm sure most of you noticed. So we have to switch one of the lower numbers (preferably 6 or 3 to maintain size) with 7 (again to maintain size). Now, when you try to get two fractions with different denominators add together, you would take the LCM of the 2 numbers and add. Since the LCM of 8 and 7 is 56, and really the only place value that matters is the ones value (especially since 5 isn't in here), the 6 should be between the 8 and 7 to make it divisible by both, no? Sorry if it's hard to understand, that's how my mind works when it comes to math, and how I reasoned it out. XD But you come up with 9867312, with only a slight variation from the largest number possible with 1, 2, 3, 6, 7, 8, and 9: 9876321. That's how I did it.

EDIT: I just realized that Mushy explained the 8 and 7 rule better upon looking through past posts. Just look at that one for a better explanation if you have trouble understanding.

---

"In many ways, the work of a critic is easy. We risk very little, yet enjoy a position over those who offer up their work and their selves to our judgment. We thrive on negative criticism, which is fun to write and to read. But the bitter truth we critics must face, is that in the grand scheme of things, the average piece of junk is probably more meaningful than our criticism designating it so. But there are times when a critic truly risks something, and that is in the discovery and defense of the new. The world is often unkind to new talent, new creations. The new needs friends. Last night, I experienced something new: an extraordinary meal from a singularly unexpected source. To say that both the meal and its maker have challenged my preconceptions about fine cooking is a gross understatement. They have rocked me to my core. In the past, I have made no secret of my disdain for Chef Gusteau's famous motto, "Anyone can cook." But I realize, only now do I truly understand what he meant. Not everyone can become a great artist; but a great artist can come from anywhere."

-Anton Ego/Peter O'Toole (1932-2013)

I figured, but I think I forgot to switch the 6 and the 7 and went straight to switching the 8 and the 7.

Hah, didn't even notice there wasn't a topic on this. I'm actually a 'prodigy' in math, so to speak. I'm in 9th grade and taking 11Honors math (Pre-Calc). Now, most people would think, "That's not big of an achievement. I've seen better." So have I. But what makes me different is that while others went to CTY (Center for Talented Youth) and took math courses there, therefore giving them the credit they need to pass a course, or LISG (Long Island School for the Gifted), where they were accelerated multiple times with heavy-duty education and thrown back into the curriculum with accelerated in almost all of their courses, or some other program. I didn't. I did so well in the public school system where I go that I was recommended to go to a higher level, which is why I'm known as a prodigy. Plus, I'm a grade ahead in English. But let's see about that problem... (will be edited with answer that I get eventually)

GOT IT! Here: 9867312, this is the largest I can get.

I'm doing the same my math skils scare my friends

Excellent Ga'Hooleone, you are correct in believing that 9867312 is the largest possible number given what I said. Since I planned to have a new problem every Monday, a tournament in July and to have explained how to achieve the conclusion to the problem on weekends, your timing is perfect. I will have a new problem up by Monday, or if preferred, I'll put it up now as an exception to the roster? Ah, I can't help waiting, I'll post it now. On some other note, in regard to opening doors,bit of a cliche overuse, eh?

Bewildered, Forgetful and not so fail, Bob.

Bob has an unreliable memory. When asked for his phone number on six different occasions, it wasn't surprising that he failed to give out his real number on each of his different attempts.

#1: 913916
#2: 917931
#3: 917601
#4: 914544
#5: 919772
#6: 911805

Although, his memory is not a complete failure since:
a) his telephone number does in fact consist of 6 digits.
b) in every attempt, the first two digits are correct. (91)
c) in each of his attempts, he also had exactly one of the last four digits in the correct place.
Question: What is his telephone number provided that the last four digits are all different?

What I have seen is that people consider the guide to be for losers, when in reality, it isnt. I actually encourage you to make use of the guide anyway. If yu can solve it on your own then thats great, but it is better to show working and use the guide rather to prove just the answer, or complete it yourself then answer the questions?

[The guide shall be posted soon enough.]

I got

Spoiler (Click to View)

914902

How I did it

Spoiler (Click to View)

Having the 4th and 5th digit as 9 and 0 respectively covers all the guesses except for 914544 and 919772, so the 3rd and 6th digit must be 4 and 2 respectively.

---

Check out my NEW story! http://worldbeyblade.org/Thread-Mr-N-s-N...lade-Story
I make idiotic posts sometimes, and I apologize for any of them. I'm still trying to be a more constructive poster. Feel free to correct me with any of my posts through PM or in the thread if it's not off-topic.
Please keep in mind when reading my posts that I know little to nothing about the current metagame and Zero-G.

Thats great that we have some early adopters, and those eager to solve the problems but I cannot reveal the answer until Friday. It's still just a concept at this point, but I'm sure it'll come along well. If you are the first to solve a problem correctly, then you are awarded a Math Credit. When you have accumulated enough of these credits, you get rewards, e.g if you obtain 20 credits you will get a free entry to my July tournaments. Depending on how difficult the problem was, you get a certain value.

Easy: 1 Credit
Medium: 2 Credits
Hard: 4 Credits
Pro: 8 Credits

Hopefully, this idea can be implemented well. Ga, you are awarded one math credit. If the WBO staff are willing to co-operate, perhaps we can also have WBO faces for acheiving a certain amount of credits, and winning the July tournaments.

Thats alright Kai, I see your reasoning.I guess we'll stick with Math Credits...

---

Although I am inactive, check out my YouTube account for existing Skyrim tutorials.
Skyrim Guides:

• One-Handed And Dual Wielding Guide
  
• ETA Next Guide: 8/9/12

Hopefully, this idea can be implemented well. Ga, you are awarded one math credit. If the WBO staff are willing to co-operate, perhaps we can also have WBO faces for acheiving a certain amount of credits, and winning the July tournaments.

Unfortunately, that only revolves around one topic, has nothing to do with Beyblade, and does not particularly mean that the person contributed to the community in any way, so no.

Thats fine Kai, I understand your reasoning. I was kind of expecting it, but I'll just satisfy myself with math credits. Those rules are quite interesting, I never thought the WBO only gave faces for contributing to the community. I guess if you look at the faces page you can make out that assumption but I usually don't think that far ahead, as to go to that page without any given hint.

---

Although I am inactive, check out my YouTube account for existing Skyrim tutorials.
Skyrim Guides:

• One-Handed And Dual Wielding Guide
  
• ETA Next Guide: 8/9/12

Speaking of Mathematics, has anyone here ever played Game 24? Really fun game.

---

"Yeah we're all going the same way down. I'm just trying to write it all down." - The Airborne Toxic Event

| Spectrum GFX Team | Beyblade: Redemption | PenPad Forums | Poké-Fusion Lab |

It took a while, but here's a few questions to answer, that will help you reach a conclusion. If you choose to use this method, post your answers to them. I encourage you to use this over simply working it out. This guide is for use with Bewildered Bob, the problem regarding the phone numbers...

Why is 913642 NOT a possibility. Which condition, a, b, c or d is not satisfied?

---

There are six options from Bobs list, but these must cover only four digits. WHat does this imply?

---

Explain what happens when seven is placed as the third digit. hHat digit must then be placed as the last, and what options are left for the fourth and fifth digits.

---

Using your answers above, determine Fred's phone number. Explain which digit you determine first, the implications of this, which digit you determine to be the second and so on.

---

Although I am inactive, check out my YouTube account for existing Skyrim tutorials.
Skyrim Guides:

• One-Handed And Dual Wielding Guide
  
• ETA Next Guide: 8/9/12

Necroing about a month, but this is important.

OK, so I took yet another practice Algebra Regents (a standardized test) and I came across a question that I could just not solve. So here's what it says (or something like it):

There are three hats. Hat A has 5 green marbles and 4 red marbles. Hat B has 6 red marbles and 5 blue marbles. Hat C has 5 green marbles and 5 blue marbles.
Then it says for the second part:
How many marbles can you add to each hat so that there is a 1/2 probability of picking green in every hat.

Ok, so here's some info. You have to add the same amount to every hat, not different amounts. And, considering it's an Algebra test, you probably must solve it algebraically. Also, since this is an 8th/9th grade test, anything learned in higher grade levels probably can't be used. Also, the numbers might not be exactly correct, but I just need to know how to solve this type of question.

So here's what I did:
Let x = the amount of green marbles added
Let y = the total amount of red and blue marbles added
Then, I tried to solve it using an algebraic system, using:
Hat A: 5 + x = 4 + y
Hat B: x = 11 + y
Hat C: 5 + x = 5 + y

But, in the end, I ended up with y - 1 = y + 11, which makes no sense!

---

"Yeah we're all going the same way down. I'm just trying to write it all down." - The Airborne Toxic Event

| Spectrum GFX Team | Beyblade: Redemption | PenPad Forums | Poké-Fusion Lab |

Must the added marbles come from the other hats, or do you have an infinite amount elsewhere ? Because otherwise, you cannot always really "add" an amount to each hat.

Furthermore, does it need to be the same amount of [colour] marble for each hat ? For instance, two more green marbles in each one ? That is what your system seems to suggest.

It should be 3 variables, one for red, one for green and one for blue. That could be why.

---

"In many ways, the work of a critic is easy. We risk very little, yet enjoy a position over those who offer up their work and their selves to our judgment. We thrive on negative criticism, which is fun to write and to read. But the bitter truth we critics must face, is that in the grand scheme of things, the average piece of junk is probably more meaningful than our criticism designating it so. But there are times when a critic truly risks something, and that is in the discovery and defense of the new. The world is often unkind to new talent, new creations. The new needs friends. Last night, I experienced something new: an extraordinary meal from a singularly unexpected source. To say that both the meal and its maker have challenged my preconceptions about fine cooking is a gross understatement. They have rocked me to my core. In the past, I have made no secret of my disdain for Chef Gusteau's famous motto, "Anyone can cook." But I realize, only now do I truly understand what he meant. Not everyone can become a great artist; but a great artist can come from anywhere."

-Anton Ego/Peter O'Toole (1932-2013)

Must the added marbles come from the other hats, or do you have an infinite amount elsewhere ? Because otherwise, you cannot always really "add" an amount to each hat.

Furthermore, does it need to be the same amount of [colour] marble for each hat ? For instance, two more green marbles in each one ? That is what your system seems to suggest.

Infinite amounts of marbles.

Also, no, it doesn't need to be the same amount. But, half the hat has to be green. The amounts of red and blue don't matter, as long as half is green. And in order for it to be half, it would be 'green = red + blue' so that's what I made my equations on.

---

"Yeah we're all going the same way down. I'm just trying to write it all down." - The Airborne Toxic Event

| Spectrum GFX Team | Beyblade: Redemption | PenPad Forums | Poké-Fusion Lab |

EDIT: Ack, NVM. Noodoo typed the same thing while I typed this.

---

"In many ways, the work of a critic is easy. We risk very little, yet enjoy a position over those who offer up their work and their selves to our judgment. We thrive on negative criticism, which is fun to write and to read. But the bitter truth we critics must face, is that in the grand scheme of things, the average piece of junk is probably more meaningful than our criticism designating it so. But there are times when a critic truly risks something, and that is in the discovery and defense of the new. The world is often unkind to new talent, new creations. The new needs friends. Last night, I experienced something new: an extraordinary meal from a singularly unexpected source. To say that both the meal and its maker have challenged my preconceptions about fine cooking is a gross understatement. They have rocked me to my core. In the past, I have made no secret of my disdain for Chef Gusteau's famous motto, "Anyone can cook." But I realize, only now do I truly understand what he meant. Not everyone can become a great artist; but a great artist can come from anywhere."

-Anton Ego/Peter O'Toole (1932-2013)

It should be 3 variables, one for red, one for green and one for blue. That could be why.

Yeah, but it doesn't really matter how much red and blue you have as long as half of the bag is green. So I figured, I could solve my system and then proceed with finding the amount of blue and red marbles (separately) after I did that part.

---

"Yeah we're all going the same way down. I'm just trying to write it all down." - The Airborne Toxic Event

| Spectrum GFX Team | Beyblade: Redemption | PenPad Forums | Poké-Fusion Lab |

Infinite amounts of marbles.

Also, no, it doesn't need to be the same amount. But, half the hat has to be green. The amounts of red and blue don't matter, as long as half is green. And in order for it to be half, it would be 'green = red + blue' so that's what I made my equations on.

OK, but right now you have :
Hat 1 : green + x = red + blue + y
Hat 2 : green2 + x = red2 + blue2 + y
Etc.

It will not be the same x in each case, so unless I am really missing way too much from my good, high school years of mathematics, you cannot blend them and use "x" as if it were always the same. The problem is that x1, x2 and x3 would not really be linked, nor would three y.

It seems easy to solve without variables, so perhaps you could work backwards, hah.

Infinite amounts of marbles.

Also, no, it doesn't need to be the same amount. But, half the hat has to be green. The amounts of red and blue don't matter, as long as half is green. And in order for it to be half, it would be 'green = red + blue' so that's what I made my equations on.

OK, but right now you have :
Hat 1 : green + x = red + blue + y
Hat 2 : green2 + x = red2 + blue2 + y
Etc.

It will not be the same x in each case, so unless I am really missing way too much from my good, high school years of mathematics, you cannot blend them and use "x" as if it were always the same. The problem is that x1, x2 and x3 would not really be linked, nor would three y.

It seems easy to solve without variables, so perhaps you could work backwards, hah.

Well, x would have to be the same number because you have to add the same exact number of each marble into each bag, so if you add 1 green to Hat A, you would have to add it to all the hats. The same with the y. I tried working backwards, but gave up because it was taking waaay too long and then I tried this plan... which ultimately failed.

---

"Yeah we're all going the same way down. I'm just trying to write it all down." - The Airborne Toxic Event

| Spectrum GFX Team | Beyblade: Redemption | PenPad Forums | Poké-Fusion Lab |