<p>This is from blue book math level1,</p>
<p>If the positive integers, starting with 1, are written consecutively, what will be the 90th digit written?
(A)0 (B)1 (C)5 (D)8 (E)9 </p>
<p>the answer is (c)5</p>
<p>I read the explanation, but still don't understand.
Can somebody explain this?
Thanx in advance!!</p>
<p>I think this is what it is saying.... 1,2,3,4,5,6,7,8,9,10,11,12.. so 0 (in the 10), would be the eleventh digit written. Since there are 9 digit numbers, there are 81 remaining digits to be written, coming out to be 40.5 two digit numbers, so the .5 signifies that it will be the first digit in the next two digit number. Since you are starting from ten, the number will be 10+(40.5-1) (for example, the second digit starting from the first number 0 would be 1, not two, thus you subtract 1) =49.5, since there is a .5, it will carry on to the first digit in the next number, which is the 5 in 50.</p>
<p>I work really fast on the math sections, so for problems like this I take my time and write it out (yes all 90 digits). It takes a LOT of time, so if you are pressed for time, don't do it the hard way, but personally, problems like these are where i make the most mistakes. If you're not comfortable with your solution do it the hard way. (not sure if you were asking how to do the problem or what the problem was asking). Just my .02$</p>
<p>As Christm4sKitty said, there are 81 places left for 2-digit numbers.
10,11,12,13,14,15,16,17,18, and 19 take 2x10=20 places
20 through 29 ---- another 20 places, 20+20=40 altogether from 10 to 29.
30->39 ---- 20 more, 60 places total
40->49 ---- 80 total.
On 81st place is the first digit of 50 (50 follows 49, right?).
Answer: 5.</p>
<p>Faster:
81/20 = 4 remainder 1:
there are four complete ten-sets (from 10-19 to 40-49) and one place after.</p>