All the positive perfect squares 1, 4, 9, 16, 25.... are written in strictly ascending order of magnitude and without the commas and spaces, resulting in the following infinite string:
149162536496481100121144......
Reading left to right, what is the 2010th digit in the above pattern?
There are 3 1-digit squares, 6 2-digit squares, etc., per the following table:
digits total digits
digits number accounted accounted for
for thus far
1 3 3 3
2 6 12 15
3 22 66 81
4 68 272 353
5 217 1085 1438
6 683 4098 5536
7 2163 15141 20677
8 6837 54696 75373
9 21623 194607 269980
10 68377 683770 953750
So for example, the six 2-digit squares account for 12 digits in all, and added to the 3 digits from the single-digit squares, the result is 15, representing the total of the numbers of digits in all squares of two or fewer digits.
By the table, the squares with 5 or fewer digits account for 1438 digits, so we need to go the the (2010 - 1438)th or 572nd digit thereafter, among the 6-digit squares, making that the second digit (ten thousands position) in the 96th 6-digit square.
The last 5-digit square is the square of 3+6+22+68+217, or 316^2 = 99856, so the 96th 6-digit square is the square of 316 + 96, or 412^2 = 169744.
The digit in question, therefore, is 6.
For the above table:
DEFDBL A-Z
pwr = 10
FOR n = 1 TO 10
num = INT(SQR(pwr - 1))
numDiff = num - prevNum
prevNum = num
pwr = pwr * 10
totDig = totDig + n * numDiff
PRINT USING "############"; n; numDiff; n * numDiff; totDig
NEXT
Computer (brute force) verification:
list
5 S1="":S2=""
10 while len(S1)+len(S2)<2010
20 I=I+1
30 if len(S1)<1050 then S1=S1+cutspc(str(I*I)):else S2=S2+cutspc(str(I*I))
40 wend
60 print mid(S2,2010-len(S1),1)
OK
run
6
OK
?i*i
169744
OK
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Posted by Charlie
on 2010-01-28 14:18:33 |
solution
