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The second statement is almost there

EDIT: If you want to figure it out yourself, DO NOT OPEN THE SPOILER BOX:

IN one hour, together they would finish 1/6.25 of the work, or 1/6.25 x. This is 28 less than the sum of their rates, (1/11 x+1/13 x) 24/143 x.

Equating:

1/6.25 x - 24/143 x =28

28/3575 x = 28

x = 3575 copies

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Nope

EDIT: Hello! This is the creator of this thread. Since I'll be leaving soon, I need some volunteers to contribute to this thread when I'm gone. Please post here if you want :)

And maybe some people are confused with some of the problems we've posted. Here's an easier one:

What is the smallest number you can multiply to 126 to get a perfect square?

 

I'd be happy to contribute. If I had time :/

14.

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Hi folks! Back to:

A. Give answers to Sammie19 (no one's answering her!)

B. Post a question here

C. Do other stuff

And now, here is the question for the Math Thread!

1/1x2x3 + 1/2x3x4 + 1/3x4x5 + ... 1/98x99x100.

Normal calculators can't help you here. There is a need for a certain formula that enables cancellation, but that can be discovered :)

@Rye- hey how was Nationals? Probably saw you there. Everyone who won first was from NCR-B! And almost everyone that one was a boy! I think the 3rd placer from Grade 6 Individuals with a weird name and one of the team members of the team that won 2nd place in Grade 6 with a long name were the only girls...That has me thinking on why Math is a "boy subject"....

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Hi folks! Back to:

A. Give answers to Sammie19 (no one's answering her!)

B. Post a question here

C. Do other stuff

And now, here is the question for the Math Thread!

1/1x2x3 + 1/2x3x4 + 1/3x4x5 + ... 1/98x99x100.

Normal calculators can't help you here. There is a need for a certain formula that enables cancellation, but that can be discovered :)

@Rye- hey how was Nationals? Probably saw you there. Everyone who won first was from NCR-B! And almost everyone that one was a boy! I think the 3rd placer from Grade 6 Individuals with a weird name and one of the team members of the team that won 2nd place in Grade 6 with a long name were the only girls...That has me thinking on why Math is a "boy subject"....

A. ? Where? I'm assuming some names changed or something...

B. Is there a level restriction?

C. Other stuff... Well...

Challenge accepted.

First of all.... Just for laughs.

I don't know why, but I feel stupid for making this even more complicated by thinking this first...

5507e8b28d944ab79538ea77f791187b.png

..... which does give same result but....

...let's make it simple...

I'm pretty confident about my answer, but I can't say for sure about any typos I made....

Took me a while to type them up in Microsoft Word...

So If anyone see the typo, point it out and I'll.. uh... try to fix it. :P

And just in case, I also put extra steps and colored some to help the you guide through the problem with extra detail as possible.

So... feel free to ask if you don't understand something.

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8cb538788499daa5982a65ce5e408c3c.png

14b4aeb1c3f29838db861375c213ecc5.png

afab67da64457173eff0693b08c77ab1.png

445f49523811dac90e509cff791c5339.png

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a5e905df5a22312b3e5582909475d0e2.png

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If you already got the answer then consider this. Especially if you don't know what LIMITS are. It might help you understand a bit since this is a good example. You might have seen this in Algebra 2 and above.... well, I did at least...

That's really close to 0.25. In fact, the infinite sum would result in 0.25.

0a15d0285bcadd17c72f5b4dec2cb0fb.png

In case you don't know limits, they just mean "as the variable in the function(equation) reaches for certain number (in this case, infinity)"

The reason why I have 0 there is because any number divided by ridiculously large number will give you a number very close to 0. Therefore, 1/infinity will give you a number so small, we consider them as 0. and we can prove that already by looking at the original sum from n=2 to 100, which is very close to 0.25, but not close enough. if you try to do this with larger number other than 100, then you'll see that the number will never pass 0.25. That is the LIMIT of the equation.

Now, if you like to think a bit deeper,

Consider that

S=1/1x2x3 + 1/2x3x4 + 1/3x4x5 + ... 1/98x99x100+.... 1/(n-1)n(n+1)

Also, S=s+r where s is sum from n=2 to 100 and r is the remaining sum (from n=101 to infinity)

You will see that the remainder will be just r=S-s.

Why is this important when it is obvious?

It tells us more that what we just see. But I'll just say that we can find an infinite sum that doesn't add up to infinity, as the limit tells us.

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Ah crap, 4th time ever. But someone has to freaking confirm, right?

First of all: o_O how did infinite sums get here?

Second of all: It's actually quite simple. Sorry for typos. I'm on a tablet.

1/n(n+1)(n+2) = 1/2(1/(n)(n+1)- 1/(n+1)(n+2))

Proof: common denominator- n(n+1)(n+2)

So, the first fraction would have a numerator of n+2, and the second one, n. Their diffrerence is 2, and multiplying that by 1/2 would result in 1 over the denominator.

=1/2(1/1x2 - 1/2x3 + 1/2x3 - 1/3x4 + ... + 1/98x99 -1/99x100)

=1/2(1/1x2 - 1/99x100) (all the other terms would be cancelled)

=1/2(1/2 - 1/9900)

=1/2(4949/9900)

=4949/19800

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Ah crap, 4th time ever. But someone has to freaking confirm, right?

First of all: o_O how did infinite sums get here?

Second of all: It's actually quite simple. Sorry for typos. I'm on a tablet.

1/n(n+1)(n+2) = 1/2(1/(n)(n+1)- 1/(n+1)(n+2))

Proof: common denominator- n(n+1)(n+2)

So, the first fraction would have a numerator of n+2, and the second one, n. Their diffrerence is 2, and multiplying that by 1/2 would result in 1 over the denominator.

=1/2(1/1x2 - 1/2x3 + 1/2x3 - 1/3x4 + ... + 1/98x99 -1/99x100)

=1/2(1/1x2 - 1/99x100) (all the other terms would be cancelled)

=1/2(1/2 - 1/9900)

=1/2(4949/9900)

=4949/19800

I noticed my typo in almost every step. it's not sum from n=2 to 100 but n=2 to 99....

Your way makes things much easier... it's basically comes down to the same thing though since the equation from my solution will equal to yours if I did it correctly.

1/4 + (-1)/(2x(x+1))

1/4 - (1/(2x(x+1)))

1/2 (1/2 - 1/ x(x-1)) ----- x = 99 (not 100, which I made a huge mistake in my solution.)

1/2 (1/2 - 1/ 99(100))

1/2( 1/2 - 1/9900) yay we get the same answer!

But then again, I like things complicated for some reason because... uhh... idk

i just like thinking through long steps. I just find them more entertaining. I do wish I can think of such methods quickly though.

I love how math can be manipulated, having multiple ways to solve a problem... or sometimes none, which is such a depressing moment. =_=

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^Hahaha! I learned that method in International Math training when I was Grade 5/6, so I'm not really comfortable using that method because in International Contests, you have to learn a LOT of techniques to shorten solutions.

The level of these problems is around Grade 8-9? I don't know, actually.

This next one is relatively easy. It's about Combinatorics.

Karen has 9 pieces of candy. She wants to distribute it among 5 children such that each one has at least one candy. In haow many ways can she do this?

Oh, Cure-senpai (can I call you that o_O?), you can post a question if you like.

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I don't mind whatever people call me really. You don't have to have honorifics, but you are welcome to do so if that's comfortable.

 

One thing I really hate is word problems. I never really liked them since I was 3rd grade because of the language barrier despite how much I love math,

Truthfully, I don't know. I'm just gonna guess with 126 using combination. 9! / [5!(9-5)!] = 9x2x7 = 126. but that seems really odd. 

 

...oh about giving question... I kinda don't want to. I might give something too confusing....

so let's go with an easy one:

 

I'm thinking about a 3 digit number that can be divided by 11.

If I took that number and added each digits to find that the sum is 21, what must be one of its digit?

 

(... if there's more than one answer than I'm screwed lol...)

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Unfortunately, that's not the answer.

Attempt at solving the above problem:

Let this three digit number be xyz.
Because it is divisible by 11, (x+z)-y is equal to either 0 or 11 (22 is not allowed, because the maximum sum of two one digit numbers is 18, 9+9).
Assume (x+z)-y = 0. So x+z = y. Substitute x+z for y, to let the sum of the digits be 2y =21. But there is no one-digit integer y that satisfies this condition, so (x+z)-y does not equal 0, therefore it must equal 11.
(x+z)-y = 11
x+z = y+11
(y+11)+y = 21
2y +11=21
2y=21-11
2y = 10
y = 5
So definitely, the digit 5 is in the number you are thinking

 

EDIT: Just for fun:

The possible 3-digit numbers you are thinking of are: 957, 858, and 759. All are divisible by 11 (giving results of 87, 78 and 69 respectively) and all have a digit sum of 21.

 

EDIT2: Challenge question:

  Show your solution, please! :)

  

   Find 12 consecutive 6 digit integers such that the 2nd is divisible by 2, the 3rd by 3, and so on until the 12th, which is divisible by 12.

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Gyaaah... I liked this thread... And now its deaaad..dead-onion-head-emoticon.gif

And, the reason is probably the above question... Lol

So lemme post another question until the question poster gives us the answer... :)

Here's a simple quadratic equation questions

If x= √(1+√(1+√(1+√(1...till infinity)

Then find the value of x

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