Nara-chan
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Content Count
2948 -
Joined
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Last visited
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Days Won
7
Status Replies posted by Nara-chan
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Damn, I eat a lot.
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But there is something called "Internet anonymity". And that would be a compliment to you.
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Damn, I eat a lot.
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Do you force people to find out your name?
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Damn, I eat a lot.
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Then I'm worse than you
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Damn, I eat a lot.
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*hands lead out* But you reveal your name through the forum? Wow, you're just the oposite of me.
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Damn, I eat a lot.
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I am not a kid. And it's not like a younger member of DCW will randomly open KtPT's status, especially a newbie. And *into
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Damn, I eat a lot.
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Do they punish you?
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Hi guys...join TROPANG WALEY in facebook...
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Damn, I eat a lot.
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Do you get fat?
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Ran's hair makes her look like a unicorn... or is that just me?
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Ran's hair makes her look like a unicorn... or is that just me?
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RT @YoMARCIAlaw: FACT: 99% ng kabataan ngayon ay namo-mroblema dahil sa concert ni Daniel Padilla.
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No... DCTP has stopped subbing the latest ep..... SO Sad... Because of those Funimation... Grrr!!
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I'm bored
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If there was anything you were curious about me, here's your chance to ask. I'll answer any questions in reply to this status.
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What is your favorite straight ship in DC?
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If there was anything you were curious about me, here's your chance to ask. I'll answer any questions in reply to this status.
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Question: What percent of your posts are from forum games?
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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2011), making 201 numbers.
We would add 201x46 to 0, the initial term
201X46+0=9246, which ends in 6.
Therefore, my answer is 6.
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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11^2-121 HD 1 (0+1)
11^3-331 HD 3 (1+2)
11^4 641 HD 6 (3+3)
and so on.l
But when you get to the point where you add 9, you have to carry 1, so the sum becomes 1 more.
So instead of 45 (1+2+3+....+9), we add 46 to the first term to get the 11th.
But we have 2011 numbrs to add (counted 11^0)
2011/10 (0,1,2,3,4,,5,6,..,9)=201 r.1
But 11^10 is the 11th term, so 46 would be added every 10 terms starting from 11.
2011 is counted in the list (11,21,31,......
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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@Moho answer is 6.
2011^2011's hundreds digit is the same as 11^2011.
Through observation,
Pattern of ones: They are always 1
Pattern of tens: 1,2,3,4,5...
Through more observation:
11^1- 11- Hundreds digit 0
11
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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@Moho Of course it's a one-digit number! A place value can only have one digit!
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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@Moho Obviously
@AL the love expert, philosopher and mathematician It's complicated. But I found a not-really-a-pattern pattern.
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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Thank you for teaching me ow to multiply trinomials, and that it always ends in 1 and the tens are in a pattern of 1,2,3,4,5,6.... BUT THERE IS NO PATTERN IN THE HUNDREDS >____<
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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Clue 1-Duh.
Clue 2- *starts analyzing*
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Nara's problem is boring. Here's a more challenging problem: What is the hundreds digit of 2011^2011?
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Clue, Moho? I just found out that it's the same as 11^2011, but I don't really see a pattern in them, unlike in last digitsl
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I'm at that point on an all-nighter where regret and drowsiness sink in and you realize even if you fall asleep now you'll be woken in a matter of tens of minutes. -.- Internet...
