tengaku squared
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About tengaku squared
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- Birthday 03/11/1988
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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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The answer is 6. You took a really roundabout way of getting it, though. Here's how I did it.
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2011 = 11 (mod 1000), ergo 2011^2011 = 11^2011 (mod 1000). So:
11^2011 = (1+10) ^ 2011
= 1 + (2011)*10 + (2011*2010)*100 + ...
You could go on and on, but note that all other terms would have three or more trailing zeros, so we can pretty much disregard the rest and get what we have here.
= 1 + 20110 + 5500
= 25611
= 611 (mod 1000)
The hundreds digit of 11^2011 is 6, so the hundreds digit of 2011^2011 is also 6.