<p>Okay so basically I'm really bad at questions like this:</p>
<p>how many integers between 1 and 100000 will have exactly one digit equal to 3? Two digit? three digit?</p>
<p>How do you solve it? Is there a formula?</p>
<p>These are counting problems. And even the first one you stated is already at the edge of maximum SAT difficulty. To restate it: Of the first 1000000 positive integers, how many contain exactly one digit which is a 3?</p>
<p>This is one where I think it is easier to calculate how many do NOT contain a 3. Well, there’s 1000000 itself. Then, all the other numbers have 5 or fewer digits. Imagine that you have 5 slots to fill – the digits of the numbers. Instead of 10 choices for each one, you have only 9 choices since you can’t use 3’s. So there are 9x9x9x9x9 numbers with 5 digits (or fewer: zero can show up in the lead digits) that contain no 3’s. But wait: that includes 0 0 0 0 0 – which represents 0 and is not positive. We didn’t want ot count that one but we did want to count 1000000 itself so it all washes out. So 9^5 of the first million integers do not contain a 3. Subtract to get the ones that do…</p>
<p>OK, even that one was too hard to be on the SAT. Exactly two 3’s or exactly three makes it even harder. And you have to either break it down into multiple cases based on where the 3’s are placed, or you have to use combinatorics:</p>
<p>So say it was three 3’s you wanted. Of the 5 available digits, you would have to choose 3 places to put the 3’s. That’s 5C3. Then, for each of the remaining two digits you would have 9 choices ('cause now you don’t want even more threes!) So I believe that you are loooking at 9^2 times 5C3.</p>
<p>As I said, this is way beyond SAT counting problems. But still a cool question…</p>
<p>The “exactly” one 3 is not too hard to count. Out of the set {1,2,…,99999} we can count the number of numbers without a 3 and subtract from 99999:</p>
<p>99999 - 9^5 = 40950 (this assumes 427 = 00427, for example)</p>
<p>Add 1 to account for “100,000” so the answer is 40951.</p>
<p>The “exactly two digits” and “exactly three digits” can be done in a similar way, but it’s a bit trickier. It’ll more likely appear on an AMC, not an SAT.</p>
<p>^ Isn’t that wrong? 99,999 - 9^5 counts a number like 33333. But the question asks for only “exactly” one digit equal to 3.</p>
<p>Wouldn’t it just be (9^4)(5) = 32805 ?</p>
<p>^ I think you are right. Both rpsence and I seem to have solved for “at least one” rather than “exactly one” 3. </p>
<p>For exactly one, you could use the same approach as the one I outlined for exactly 2 or exactly 3: </p>
<p>You have 5 digits to choose from to place the one 3. That’s obviously 5 – or 5C1 if you prefer. Then for the remaining 4 digits, you have 9 options for each, where again we are allowing leading zeroes so as to generate the fewer-digit numbers as well. So its 5x9x9x9x9 numbers with exactly one 3.</p>
<p>@adityap, whoops, you’re right. It’s 5(9^4). </p>
<p>Then for two 3’s, there are 5C2 positions for the 3’s to go, and 9^3 ways of selecting the other three digits.</p>