AMC 10 · 2016 · #22
Grade 7 number-theoryPick an answer.
A repeating decimal is a disguised fraction, so Tool #13 (Convert to Algebra) is the way in: 0.abcdef = abcdef/999999 turns the decimal statement about 1/n into the divisibility statement n ∣ 999999, and likewise (n+6) ∣ 9999. That converts a question about digits into a question about factors. Tool #7 (Identify Subproblems) says to work the 4-digit condition first, because 9999 is much smaller and factors into 99 × 101 with the prime 101 doing all the work. That prime forces 101 ∣ (n+6), which cuts the search to a handful of numbers. Tool #2 (Make a Systematic List) writes those few candidates out, and Tool #3 (Eliminate Possibilities) knocks out all but one using the 999999 condition. The whole solve is a funnel: two divisibility facts, one prime, three candidates, one survivor.
Turn repeating decimals into fractions
Each repeating decimal becomes a divisibility fact.
Multiplying 0.abcdef by 10⁶ shifts it exactly one block, so subtracting the original leaves a whole number over 999999.
Multiplying by a power of ten shifts the decimal exactly one block, so subtracting the original leaves a whole number.
▸ Why?
Multiplying by a power of ten slides every digit along by that many places without changing any of them.
▸ Why?
The two endless tails are identical, so the subtraction removes them entirely.
Make the period exactly right
Exactly means two conditions apiece.
Dividing a shorter string of nines is exactly what a shorter period looks like, so those divisors must be forbidden.
4.OA.B.4Eliminate PossibilitiesFactor 9999 and find the forced prime
Factoring forces one particular prime.
The only thing 9999 has that 99 lacks is the prime 101, so escaping period 2 means carrying that prime.
6.NS.B.4Identify SubproblemsList every candidate for n
That leaves only three candidates.
Once 101 is forced, n+6 is 101 times a divisor of 99, and the bound n ＜ 1000 leaves only three.
6.EE.B.6Make A Systematic ListScreen the three against 999999
The other condition screens them.
A number divides 999999 only if its whole prime recipe is already inside 3³ × 7 × 11 × 13 × 37.
6.NS.B.2Eliminate PossibilitiesVerify both periods, then read the interval
One survivor names the interval, choice (C).
Writing out the two decimals is the final receipt that the divisor reasoning matched the digits.
6.NS.B.3Eliminate PossibilitiesA decimal that repeats in blocks of k digits is a fraction over k nines, so 'period 6' means n divides 999999 and 'period 4' means n+6 divides 9999; the prime 101 hiding in 9999 pins n+6 down to a few values, and only n = 297 survives.
- Turn repeating decimals into fractions
- Make the period exactly right
- Factor 9999 and find the forced prime
- List every candidate for n
- Screen the three against 999999
- Verify both periods, then read the interval