<p>[2008</a> AMC 12A Problems/Problem 15 - AoPSWiki](<a href=“http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_12A_Problems/Problem_15]2008”>Art of Problem Solving)</p>
<p>I tried looking for a better explanation, but even so I could not comprehend. How do they get 8 squared increased by 2^4, and 4+6? How does modulus even apply to this problem…?</p>
<p>You’re probably better off asking these questions on the AoPS forum. But I’ll see if I can help. Do you understand mods? Because all we’re looking for is the last digit, and mods are clearly useful for that. Is “how do they get 8 squared increased by 2^4, and 4 + 6” your only question, or is there something else you don’t understand?</p>
<p>Okay so I’ll try and explain the solution (I don’t think that any “better” solution exists).</p>
<p>Obviously k (congruent) (2008)^2 + 2^(2008) Now the last digit of (2008)^2 is just the last digit of 8^2 = 4. And the last digit of powers of 2 (2^n) cycle through (2,4,8,6,2,4,8,6…) And 2008 is divisible by 4, thus the last digit of 2^(2008) = 6. Thus k (congruent) 10 (congruent) 0 (mod 10). Do you need the rest of the solution explained?</p>
<p>Yes, it’d definitely be a lot helpful if the rest of the solution was explained.</p>
<p>I get how you got 8^2=4, but after that you lost me on everything. Yeah, I get how the powers of 2 go through the cycles 2,4,8 but what does 2008 being divisible by 4 have to do with anything? It’s also divisible by 2 and 8. Maybe I just need to read a detailed explanation on mods to understand this better… I was actually never exposed to them till this problem. And I still do not get how mods apply to this problem… ;_;</p>
<p>Thanks spratley!</p>
<p>Yeah, they don’t take long to learn, I think AoPS has a page in their wiki on mods. So the last digit of 2^(n), for natural numbers n, repeats (2,4,8,6,…) Thus we divide 2008/4 and we have remainder 0. So 2008 has the same last digit as the last number in the cycle of powers of 2 (i.e 6). This is very similar to find i^(n) - if you have experience doing that. Do you understand so far? If not you probably should read up on mods…</p>
<p>Yes, I will indeed locate a modulus article of proper decency to scrutinize. It puzzles me as to why you divided 2008 by 4, and not 2 or 8, which will also leave a remainder of 0. Nonetheless, I go onward to exploring modulus.</p>
<p>We divide 2008 by 4 because powers of 2 repeat in cycles of 4. Its just like finding i^(2008) - if you know how to do that… but yes mods are very important in contest math - questions like this one, and #1 on this year’s AIME I, are really easy if you know mods, but quite difficult otherwise. AoPS’s Introduction to Number Theory covers mods and much more number theory in a very intuitive, and yet rigorous manner, so you might want to think about getting it…</p>
<p>^^^ yes, I remember doing this problem for practice and the quickest way to do it is using modular arithmetic. AOPS volume 1 does a nice job of covering this technique and how it cna be used to in solving competition problems.</p>
<p>Well AoPS Vol I does introduce mods, but I’d say its coverage of mods is still rather skimpy. Their Introduction to Number Theory gives a much more comprehensive introduction to mods, and number theory in general - so I’d probably recommend it to the OP.</p>