1995 AMC 8 Problem 15

Attempt Problem 15 of the 1995 AMC 8 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 1995 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

15.

What is the 100100th digit to the right of the decimal point in the decimal form of 437?\dfrac{4}{37}?
00
11
22
77
88
Answer: B
Concepts:repeating decimalmodular arithmetic
Small hint:
Write 437\dfrac{4}{37} as a repeating decimal and find the length of the repeating block
Big hint:
The block has length 3;3; use the remainder of 100100 divided by 33
Solution:
437=0.108‾,\dfrac{4}{37} = 0.\overline{108}, repeating with block length 3.3. The digits in positions 3,3, 6,6, 9,9, …\ldots (multiples of 33) are 8.8.
Since 9999 is a multiple of 3,3, the 9999th digit is 8,8, so the 100100th digit starts the next block: it is 1.1.
Thus, the correct answer is B .
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.