<p>How do you figure out this series problem without taking 5 minutes to write down all the numbers? I know there must be a shortcut or something of the sort.</p>
<p>How many of the first 200 positive integers contain the digit 1 exactly once?</p>
<p>A) 36
B)37
C) 99
D)100
E) 121</p>
<p>Is the answer D?</p>
<p>nope..I'm not going to say until someone gets it right with the explanation..</p>
<p>aaaahhhh so it is not D....is it C?</p>
<p>Anyway, I don't know right now....but what you should see is that the first positive integers (with the ramifications) are 1,10,12, 13, 14 , 15, 16, 17, 18 , 19 ,21, 31, 41, 51, 61, 71, 81, and 91. Then the 100's begin. The 100's already have a one so anything that adds to have more ones you will subtract those from your 100's list...</p>
<p>anyway, 18 from 0-100....</p>
<p>then in 100-200..you can't use 101....ok so then you have 100-19= 81..then add the other 18, and you get 99.</p>
<p>ok, so I might be wrong</p>
<p>alright, first you have to count all the numbers from 1-100 that fit your criteria (we're still counting, but 1-100 is the easy part.) 1 obviously, then 12-19 (thats eight more) and then 21, 31, 41, 51, 61, 71, 81,91, 100. So from 1-100 you have 18 that work. </p>
<p>Next you subtract: 100-19=81 then add: 81+18=99 </p>
<p>99 is the final answer</p>
<p>Rationale: finding the first 18 is easy and shouldn't take too long. Then, for 101-200, you simply subtract all the ones that worked for the first hundred digits from 200 (because all the ones that worked from 1-100 would have two ones from 101-200 because the hundreds place will always be 1.) (Also, your subtracting 19 instead of 18 because 11 doesn't count for your criteria but still needs to be subtracted as a number that will not work.) All the others from 101-200 will work because there is always a 1 in the hundreds place. </p>
<p>I know it looks complicated when i write it, but if you understand the procedure, the problem won't take you more than 30 seconds.</p>
<p>Thank you BigE......i was actually right for once</p>
<p>I would try to list systematically, looking for a recurring pattern</p>
<p>1 and 10 = +2</p>
<p>then 12-19 = +8</p>
<p>then there is 1 for each ten... eg 21, 31, 41, 51... that's +8
+ 100 = 19 total
then when you get into the 100's, there will be 100 ints that have the digit 1 in them... you must subtract those with more than 1, i.e:
101
112-19
following the same pattern as per 1-100.. that means 100 - 19
= 81
+ 19 you had in 1-100
= 100 </p>
<p>I say D.</p>
<p>Ok...maybe i'm wrong but was right before.......or not......what is the right answer?</p>
<p>Actually BigE is probably right - I neglected to see that the pattern changes from the 1-100 to the 100-200 set - that is, you subtract 111 from the list as well. </p>
<p>The answer is most likely 99.</p>
<p>C is the right answer..Thanks for everyone's help!</p>
<p>hahahaha...i was right after all!!!!!!!!!!!!!!!!</p>
<p>from bigE's post: "1 obviously, then 12-19 (thats eight more) and then 21, 31, 41, 51, 61, 71, 81,91, 100. So from 1-100 you have 18 that work."</p>
<p>what about 10?</p>
<p>you have 19 that work for 1-100 inclusive. these are 1,10,12-19(eight),21,32,41,51,61,71,81,91,100. </p>
<p>then for 101-200 inclusive, its the same thing, subtract
101,110-119(ten more),121,131,141,151,161,171,181,191,200(because unlike the first set, the last term 200 does not work). thats 20. </p>
<p>or also you could reason the 19 don't work for 101-200 inclusive, and in addition 111 does not work. again thats 20 working with those 100 numbers (101-200). </p>
<p>100-20 = 80<br>
80+19 = 99.</p>
<p>yeah i did forget 10. But since i didn't subtract it later on it basically cancelled out.</p>
<p>how is this a series problem. looks more like a number theory problem</p>
<p>yeah thats what happened. good luck for you :)</p>
<p>you could also reason that all the numbers that work in the first one, work in the second one, and that these must equal 100 because they are both sets of 100 numbers (think 20 and then 80, 40 and then 60,etc). except 111 doesn't work, so 100-1=99. i could only think of this after a while, but that would be the most logic intensive and quickest way to solve it, that is if you somehow think of it as your first approach.</p>
<p>Lol heck if I know math whiz..I labeled it a series problem because that is what category sparknotes put it in</p>
<p>Humm. I did not read all the solutions offered, and this could be a repeat. However, this how I would solve it, without listing the numbers. </p>
<p>Full explanation is:
The first two hundred numbers have to contain at least 1. This means that 200 is excluded. I split the numbers in the three categories: hundreds, tens, and single digits. </p>
<ol>
<li>With a 1 in the hundreds, success is the combination of any number in tens and singles, except for 1 => (10-1) * (10-1) = 81</li>
<li>With a 0 in the hundreds, success is the combination of 1 in tens AND any single number except 1 in singles => 1 * 9 = 9<br>
Plus success is the combination of any number except 1 in tens AND 1 in singles => 9 * 1 = 9<br></li>
</ol>
<p>Answer is 81 + 9 + 9 or 99.</p>
<p>Damn you xiggi!!!!!!!!!!!!! you had a better explanation than everyone.</p>
<p>oh well....you are in college :D</p>
<p>:) :D</p>