<p>Is there some very quick shortcut to solving problems similar to the following?</p>
<p>What is the units digit of 2^2012
What is the units digit of 3^999
etc.
If there is not, how do you solve them anyhow?</p>
<p>For the first one note that </p>
<p>2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16
2^5 = 32, 2^6 = 64,…</p>
<p>In other words, we get the repeating sequence 2, 4, 8, 6, 2, 4, 6, 8,… for the units digit. Since 2012 is divisible by 4, the units digit of 2^2012 is the same as the units digit of 2^4. Thus, the answer is 6.</p>
<p>Try the second one yourself. It can be done in a similar way.</p>
<p>I don’t think I’m understanding the concept. So say for example, you had to do 2^7, which is equal to 58. Because 7 is only divisible by 1, then wouldn’t you get the units digit of 2, when it’s supposed to be 8? And would you always take the numbers to the end of the pattern to divide by, because 2012 is also divisible by 2. I don’t understand.</p>
<p>It doesn’t matter what 7 is divisible by. All that matters is what’s the remainder when 7 is divided by 4… because the pattern the units digits make (2, 4, 8, 6, 2, 4, 8, 6…) is 4 “pieces” long. </p>
<p>For example, if they asked: “On a year that begins on a Sunday, the 100th day of the year will be…”, in that case you would only care about the remainer when 100 is divided by 7 because the days of the week form a repeating pattern, 7 pieces long.</p>
<p>In all of these problems, you divide the # of items by the number of pieces in the pattern. Then, the remainder tells you how far into the pattern you go to find the end “piece”.</p>
<p>Thanks, I understand now.</p>