AMC 10 · 2013 · #15
Grade 6 number-theoryPick an answer.
The phrase 'as small as possible' points straight at the Extreme Principle: chase the smallest a₁ and the smallest b₁ that can still work. The lever is prime factors. A factorial n! contains a prime p only if n reaches p, so the largest prime inside 2013 forces a₁, and the largest unwanted prime dragged along forces b₁. Break the problem into 'how small can a₁ be' and 'how small can b₁ be', then build one explicit expression to prove those minimums are reachable.
Factor 2013 into primes
The number splits into three primes.
Splitting into primes shows exactly which building blocks the factorials must supply.
4.OA.B.4Identify SubproblemsThe prime 61 forces a₁ = 61
The largest one forces the top factorial.
A prime can only enter a factorial once the count climbs up to that prime.
A prime can only enter a factorial once the count has climbed all the way up to that prime.
▸ Why?
Every number has one prime recipe, so a prime factor cannot appear from any smaller number.
▸ Why?
So a factorial below that prime simply does not contain it, which forces the smallest allowed choice.
61! drags in 59, forcing b₁ = 59
That factorial drags in an unwanted prime.
Whatever unwanted prime the top factorial pulls in must be matched by a factorial on the bottom to erase it.
6.NS.B.4Extreme PrincipleBuild an expression that hits 120
A real expression reaches that minimum.
Showing one working expression proves the smallest a₁ and b₁ are really possible, not just wished for.
5.OA.A.1Guess And CheckTake the difference
The difference is 2, choice (B).
The two forced anchors sit just two apart because 59 is the prime right below 61.
4.NBT.B.4Identify SubproblemsThe biggest prime inside a number decides the top factorial, and the biggest unwanted prime it drags along decides the bottom one.
- Factor 2013 into primes
- The prime 61 forces a₁ = 61
- 61! drags in 59, forcing b₁ = 59
- Build an expression that hits 120
- Take the difference