Problem 6 · 2005 AMC 8
Easy
Fractions, Decimals & Percents
place-value-comparison
Suppose d is a digit. For how many values of d is 2.00d5 > 2.005?
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Answer: C — 5 values.
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Hint 1 of 2
Both numbers start 2.00…, so everything up to there is a tie. Only the digits after that point can break it.
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Hint 2 of 2
When numbers share a long matching prefix, peel it off — compare just the part where they actually differ.
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Approach: strip the matching prefix, then compare
- Write the right side as 2.0050 so both have the same length. They agree through 2.00, so subtract that common part away — the contest is now 0.00d5 vs 0.0050.
- The first place they can differ is the thousandths: d vs 5. If d > 5, left wins. If d < 5, right wins. If d = 5, they're still tied (0.0055 vs 0.0050) — and the next place, 5 vs 0, tips it to the left. So d = 5 works too.
- Conclusion: d ≥ 5, i.e. d = 5, 6, 7, 8, 9 ⇒ 5 values.
- Why this transfers: to compare decimals, ignore the shared leading digits and judge at the first place they split — matching length first (pad with a 0) keeps you from misreading place values.
Another way — algebra on the gap:
- Subtract 2.005 from both sides: the inequality becomes 0.00d5 − 0.0050 > 0, i.e. (d − 5)·0.001 + 0.00005 > 0.
- That holds exactly when d − 5 ≥ 0, so d ≥ 5 — the 5 digits 5 through 9.
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