The required 2010th digit in the above pattern
is 6.
EXPLANATION:
Let N(i) = # i-digit perfect squares
Then, we must have:
Numbers # digit N(i)
12 to 32 1 3
42 to 92 2 6
102 to 312 3 22
322 to 992 4 68
1002 to 3162 5 217
3172 to 9992 6 683
From the above table, total # digits contained in the squares
of 1 to 999 inclusively
= Σ(i = 1 to 6) i*N(i)
= 1*3+2*6+3*22+4*68+5*217+6*683
= 5536 > 2010, which is a contradiction
Total # digits contained in the squares of 1 to 316 inclusively
= Σ (i = 1 to 5) i*N(i)
= 1*3+2*6+3*22+4*68+5*217
= 1438
Accordingly, we require 2010 - 1438 = 572 more digits.
Now, 572/6 = 95 + 2/6
Since, 316 + 95 = 411, we require 2 digits into the square of 412.
Now, 4122= 169744, whose 2nd digit is 6 and consequently
the 2010th digit in the above pattern is 6. |
