1983 AMC 12 Problem 16

Attempt Problem 16 of the 1983 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1983 AMC 12 solutions, or check the answer key.

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16.

Let x=0.123456789101112=998999, \begin{aligned} x&=0.123456789101112\\ &\phantom{={}}\ldots998999, \end{aligned} where the digits are obtained by writing the integers 11 through 999999 in order. The 19831983rd digit to the right of the decimal point is

22

33

55

77

88

Answer: D
Concepts:digitsplace valuecounting integers in a range
Difficulty rating: 1830
Small Hint:

First count the digits contributed by the one- and two-digit integers

Big Hint:

After 189189 digits, locate the remaining position within the three-digit integers

Solution:

The one-digit integers contribute 99 digits and the two-digit integers contribute 902=180,90\cdot2=180, for 189189 total. Thus the desired digit is the 1983189=17941983-189=1794th digit among the three-digit integers. Since 1794=5983,1794=598\cdot3, it is the last digit of the 598598th three-digit integer, namely 100+597=697.100+597=697. That digit is 7.7.

Therefore, the correct answer is D.

← Problem 15#15
Full Exam

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