  # Level 4 math problem

<p>i'm stuck on Section 6 Question 6. Level 4 problem according to answer key.</p>

<ol>
<li>n = 1234567891011.......787980</li>
</ol>

<p>The integer n is formed by writing the positive integers
in a row, starting with 1 and ending with 80, as shown
above. Counting from the left, what is the 90th digit
of n ?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5</p>

<p>[1.] The 10th digit is a 1
(Now we're in the teens)</p>

<p>[2.] The 12th, 14th, 16th, 18th, 20th, ..., 28th digits are 1s too.
(Because we're still in the teens, so every other digit is a 1.)</p>

<p>(Now we're in the twenties)
[3.] The 30th digit is a 2. So is the 30th, 32nd, ..., 48th. (Because every other digit is a 2.)</p>

<p>[4.] Similarly, the 50th, 52nd, 54th, ... ,68th digits are all 3s.</p>

<p>[5.] Similarly, the 70th, 72nd, ..., 88th digits are all 4s.</p>

<p>[6.] Similarly, the 90th, 92nd, 94th, ..., 108th digits are all 5s.</p>

<p>The only difficult in this problem is finding the fastest way to do it. 1-9 is nine numbers and only one digit so you have 9. Keep that number aside.</p>

<p>10-19 - each number in this set has two digit's and there are 10 numbers in the set so you have 20 digits. Still no where near close to n.</p>

<p>10-19 - 20
20-29 - 20 in a similar fashion as above
30-39 - 20
40-49 - 20
+____
80</p>

<p>here add those first nine for a grand total of 89. So 9 in the 49 is the 89th digit. That means 5 in 50 is the 90th digit.</p>

<p>^Wow Thanks that was a great explanation!</p>

<p>no prob. if you got more ill be glad to help since I like solving math problems.</p>