View source for Talk:2011 AMC 10B Problems/Problem 23

I don't get the Pascal's Triangle solution. I did it using Pascal's triangle, but I had more steps, I didn't do it like that. How did you ignore the tens digit and get the right answer?

== Pascal's Triangle Solution ==

Be aware that the 2000 in 2011 doesn't contribute to the hundreds digit of the power. The question is then simplified into finding the hundreds digit of /Users/hourosemary/Desktop/截屏2025-08-31 20.47.37.png. Then, we need to find out the last 3 numbers in the 2012th row of the Pascal's triangle.
By the Pascal's triangle, we have:


/Users/hourosemary/Desktop/截屏2025-08-31 20.33.01.png



By this we know that the last three numbers of the 2012th row are /Users/hourosemary/Desktop/截屏2025-08-31,  20.36.05.png/Users/hourosemary/Desktop/截屏2025-08-31 20.36.56.png, and 1. The calculations are 2021055, 2011, and 1. We need the units digit of the first number and the tens digit of the second number, so we do 5+1= /Users/hourosemary/Desktop/2c024459b191f13b58f34b5e366d4e49e32cad7c.png.