<p>On solving this:</p>
<p>In how many of the integers from 1 to 200 does the digit 7 appear at least once?</p>
<p>Seriously, I can't just write out all the numbers (which is what I did). Any easier way?</p>
<p>he|| yeah there is an easier way. Trust me you dont have to ever list numbers.</p>
<p>Just think about it like this. From 1-100, you know that 7 occurs</p>
<p>from 1-10 at 7
from 11-20 at 17
from 21-30 at 27
so on</p>
<p>Know that you list it like that you realize that it occurs 9 times excluding 70-79. Now include 70-79 and you get 19 times. Then its the same process for 100-200. It</p>
<p>100-110 - at 107
111-120 - at 117</p>
<p>so another 19 times. Final answer 19+19 = 38.</p>
<p>On problems like these which do appear on the sat, you know that the digit appears 19 times since you done it before so it becomes really easy.</p>
<p>make sense?</p>
<p>Yeah, I actually did that. But I ran out of space by 99. Then I guessed to multiply with 2. But I thought maybe there was a formula…</p>
<p>But thank you anyway (:</p>
<hr>
<p>Can the first digit ever be 7? No.
Can the second digit ever be 7? Yes, from 70-79 = 10 and 170-179 = 10
Can the third digit ever be 7? Yes, at 7, 17, 27,…,197 = 20</p>
<p>10 + 10 + 20 = 40
Since we counted 77 and 177 twice, we can subtract 2 from 40.
40 - 2 = 38.</p>
<p>You can also solve this with the counting principle. But you have to realize that to get the number that DO contain 7, you have to find the number that don’t:</p>
<p>To generate that number, consider each digit and ask how many choices you have:</p>
<hr>
<p>The first digit can be a 0 or a 1. That’s two choices. The next two digits can be anything but 7. That’s 9 choices each. 2 x 9 x 9 = 162. So of the numbers from 1 - 200, 162 do not contain any 7’s. So 200 - 162 = 38 that contain at least one.</p>
<p>Notes:</p>
<p>Usually, you don’t let the lead digits be zero, but this time we want to count one digit numbers and two digit numbers as well – this method includes them in the count.</p>
<p>Also, the 162 that this method gets us actually includes 000 but that’s ok b/c it doesn’t include 200 so the total is still correct.</p>
<p>Just my 2 cents here: using pckeller’s method here gets the answer very quickly. In fact, with a bit of practice, the solution is as simple as typing 200-2<em>9</em>9 into your calculator. </p>
<p>However, in general for combinatorics problem I would recommend using eagle’s method of “listing.” </p>
<p>In other words, if you have a direct modification of a counting problem you’ve already practiced, go with the counting principle. If you get a counting problem not exactly like one you’ve seen before use the “listing” method.</p>
<p>When practicing these types of problems at home do it both ways.</p>
<p>^ I agree. Actually, as much as I like the counting principle, I always recommend starting by listing. Many times, that’s all you need. I say switch to the counting principle when the numbers get to big or you start to feel that you won’t be able to count them all or keep track of what you have counted and whether you have double counted or missed any.</p>
<p>But apart from the SAT, it’s just a very cool method with lots of connections to other interesting math.</p>
<p>thank you both :D</p>