A2 = (4/5)(18292 + 12982)
in about 3min, because this comes at the end of a rather difficult geometry problem with 6 min for the entire question. (edit: Yes, calculators weren't allowed because it was a competition. I have verified that everyone comes to this step, so I didn't make any mistakes before that.)
The Attempt at a Solution
I'm quite sure there's some algebraic manipulation to help you with this one, but I can't think of it. I did the 18292 and 12982 the 'brute force' way, then added them together for the sum. This step took the longest - more than 3min for me.
Then I converted this to standard form so as to mentally estimate the common logarithm to 3 decimal points. I also know common lg 4 and lg 5 by memory to 3 decimal points - then I took a 'guess' at A:
lg A = (lg 4 + lg(18292+12982) - lg 5)/2
Comes around to 2 * 103~ rather quickly, correct a bit and 'try' a few values, and I found 2006 eventually (edit: , which is the correct answer.)
Also did think of putting it in (1829+1298)2 - 2*1829*1298, but it still takes about as long. Alternatively, I had a suggestion to use 18002 + 292 + 2*1800*29, and similar for 12002 + 982 + 2*1200*98, which takes about as long for me.
Does anyone see an algebraic manipulation to make this easier? Thanks in advance ;)