AMC 8 · 2005 · #6
Grade 5 arithmeticPick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question asks "for how many values," so the goal is to count digits that pass a test (Tool #13). The test itself is the decimal comparison 2.00d5 > 2.005. To run the test cleanly, split the 10 digits into three cases by how d compares to 5 in the thousandths place (Tool #7). The case d=5 needs an extra look at the ten-thousandths place, so isolating it as its own subproblem keeps the reasoning honest.
Line up the place values: rewrite 2.005 as 2.0050 so it has the same four decimal places as 2.00d5.
Comparing decimals is a Grade 5 skill: same place values lined up, then compare digit by digit from the left.
5.NBT.A.3Identify SubproblemsThe ones, tenths, and hundredths all match, so the whole comparison is decided at the thousandths place.
Whichever number wins the first place where they differ wins the whole comparison.
5.NBT.A.3Identify SubproblemsCase A, d > 5: 2.00d5 already wins at the thousandths place, so d = 6, 7, 8, 9 all work — four values.
A bigger digit in the first differing place makes the whole decimal bigger.
5.NBT.A.3Identify SubproblemsCase B, d < 5: 2.00d5 loses at the thousandths place, so d = 0, 1, 2, 3, 4 all fail.
A smaller digit in the first differing place makes the whole decimal smaller.
5.NBT.A.3Identify SubproblemsCase C, d = 5: the thousandths tie, so 2.0055 beats 2.0050 at the ten-thousandths place — so d = 5 counts.
If the deciding place ties, slide one more place to the right and compare there.
5.NBT.A.3Identify SubproblemsCollect the winners 5, 6, 7, 8, 9 — that is five digits, so the answer is (C).
Tool #13 in action: sweep through all 10 digits, count the ones that pass the test.
5.NBT.A.3Convert To AlgebraStack the decimals, find the first place where they differ, and count the digits that come out on top — that turns a tricky-looking inequality into a quick Grade 5 check.