AMC 10 · 2003 · #18
Grade 8 number-theoryPick an answer.
The numbers here are far too big to write out, but only their prime exponents matter, so Tool #4 (Introduce a Variable) names those exponents: let c be how many 7s are in x and d how many 11s. Tool #7 (Identify Subproblems) then splits one huge equation into one small equation per prime, because unique factorization forces the exponents to match prime by prime. Tool #14 (Extreme Principle) is what the word "minimum" is asking for: each exponent has a smallest legal value, and pushing every one of them down as far as it can go is exactly how x is made smallest. Tool #6 (Guess and Check) finishes each small equation — 5c+1=13m has so few candidates that walking m=1,2,3,… settles it in seconds.
Compare one prime at a time
Unique factorization lets each prime be matched separately, splitting one equation into many small ones.
Two numbers are equal only if they are built from the same primes in the same amounts, so the primes can be checked one by one.
Two numbers are equal only if they are built from the same primes in the same amounts, so the primes can be checked one at a time.
▸ Why?
Every number breaks into primes in exactly one way, so its prime counts are a fingerprint nothing else shares.
▸ Why?
For one prime, two equal powers of it must carry equal exponents, so each prime gives its own plain equation.
Primes other than 7 and 11 cost too much
Any other prime would need a huge exponent, so the smallest x uses only the two named primes.
An extra prime can only enter in blocks of thirteen, and thirteen copies of anything is a huge price to pay for no gain.
4.OA.B.4Extreme PrincipleWrite both sides in exponent form
Writing both sides as powers and matching gives two congruences.
Multiplying powers of the same prime just adds their exponents, so the whole equation becomes bookkeeping on two counters.
8.EE.A.1Introduce A VariableSmallest c with 5c+1 a multiple of 13
Walking upward, the smallest exponent of the first prime is 5.
One equation with two whole-number unknowns is solved by stepping the smaller multiplier up until the other side finally divides evenly.
8.EE.C.8Guess And CheckSmallest d with 5d-1 a multiple of 13
The same walk gives the second exponent as 8.
The same stepping search works for the second prime; only the leftover +1 moves to the other side.
8.EE.C.8Guess And CheckConfirm the minimum, then add
Exhibiting a matching partner confirms it, and the four numbers add to 31, choice (B).
A lower bound that is actually reached is a minimum, so showing one working pair finishes the job.
6.EE.A.1Extreme PrincipleWhen two products of primes are equal, count one prime at a time; the smallest number is the one where every exponent is pushed down to its lowest legal value.
- Compare one prime at a time
- Primes other than 7 and 11 cost too much
- Write both sides in exponent form
- Smallest c with 5c+1 a multiple of 13
- Smallest d with 5d-1 a multiple of 13
- Confirm the minimum, then add