Recover Primes from Equations and Reciprocal Sums
第17讲 · 从方程与倒数和中恢复质数
Factor a denominator, translate each unit fraction, use parity and prime factors to shrink the search, and then verify the exact equation. This lesson includes the complete original Example 5 and Guided Practice 5.
Turn unit fractions into one common denominator
A reciprocal such as 1/7 is a unit fraction. To add several unit fractions exactly, rewrite them with one common denominator.
Unit-fraction model
The least common denominator is the least common multiple of 7, 11, and 13.
Common-denominator contributions
Live unit-fraction builder
Record the worked example values
Need a hint? Start here
For unit fractions with common denominator D, the numerator contributed by 1/p is D ÷ p.
How to check my reasoning
LCM(7,11,13)=1001; each numerator is 1001 divided by its original denominator. 1001; 143; final numerator 311.
Use denominator factors as clues — not as a substitute for proof
The worked example factors 1001 and 1986 to suggest the prime denominators. The final equality still has to be checked.
Factor the displayed denominator
These three prime factors are natural candidates for the three unit-fraction denominators.
Why exact substitution matters
If repeated primes are allowed, cancellation can hide a denominator:
The reduced denominator is 3 even though the prime 2 was used. Therefore “factor the denominator” is a powerful original clue, but not a universal proof by itself.
Prime-factor explorer
“Square-free” means no prime factor occurs more than once. For example, 42=2×3×7 is square-free, but 12=2²×3 is not.
Use the clue responsibly
Need a hint? Start here
Factor the denominator with small prime trials; divide out each successful factor completely.
How to check my reasoning
Repeated denominators may cancel during reduction, so always verify the exact fraction. 7,11,13 in any order; no, the factor list alone is not the full check.
Solve Worked example 5
Three primes have reciprocal sum 311/1001. Recover the primes and their sum.
Complete practice problem
Complete the worked example result
Reveal all four original steps before checking these answers.
Need a hint? Start here
Add the common-denominator contributions and reduce the fraction before comparing.
How to check my reasoning
143+91+77=311 verifies the reciprocal sum. Primes 7,11,13; sum 31.
Verify a proposed prime triple exactly
A candidate is accepted only when all denominators are prime and the cross-multiplied fraction equals the target.
Exact reciprocal verifier
Name the proof step
Run both verifier presets with target 311/1001: accept 7,11,13 and reject 7,11,17 before checking the questions. This checks whether a proposed triple works; the uniqueness proof comes in Mission 6.
Need a hint? Start here
Substitution proves a candidate works. To prove it is the only candidate, account for all possible denominators.
How to check my reasoning
All denominators must be prime and the exact fractions must agree. Substitution proves a candidate works, not that it is unique. Exact substitution; no, 7,11,17 does not match.
Complete Guided Practice 5
The same worked solution works with a denominator containing one much larger prime factor.
Three primes have reciprocal sum 1661/1986
Exact check
Prime total
331 is prime: test possible prime divisors 2,3,5,7,11,13,17. None divides 331, and the next prime, 19, has 19²=361>331.
Record the guided-practice result
Need a hint? Start here
Test 331 using primes 2, 3, 5, 7, 11, 13, 17, since 19² is too large.
How to check my reasoning
1986=2×3×331; contributions 993,662,6 total 1661. 2,3,331; largest 331; sum 336.
Explore triples, then prove uniqueness
The worked example exhibits a working triple. The explorer checks only primes within your chosen limit. A separate argument below proves uniqueness without relying on that limit.
Why these two triples are the only possibilities
- Start with a reduced fraction. For example, adding fractions whose denominators use only primes 2 and 3 cannot create a factor 5 in the reduced denominator. Cancelling factors can only remove them.
- For three prime denominators p, q, r, a common denominator is p × q × r. Every prime factor of the reduced denominator must therefore be one of p, q, r.
- 311/1001 is reduced and 1001 = 7 × 11 × 13. Those three different primes must all appear among the three denominators. This forces exactly 7, 11, 13.
- 1661/1986 is reduced and 1986 = 2 × 3 × 331. This forces exactly 2, 3, 331. The exact substitutions in Missions 3–5 verify both triples.
Repetition cannot help here: repeating a denominator would leave room for at most two different primes. This argument depends on the reduced denominator having three different prime factors; do not assume it works for every reciprocal equation.
Exact reciprocal-triple search
Record the counts justified by the proof
Run both target presets before checking. Count unordered triples: reordering the same three primes does not create another solution. The proof applies beyond the search limit.
Need a hint? Start here
A search limit is not a proof. Use the reduced-denominator argument above for these two targets.
How to check my reasoning
Each reduced denominator has three different prime factors. All three must occur among the three prime denominators. One unordered triple for each target; no, a search up to 20 cannot rule out larger primes.
Recover prime pairs from a linear equation
Parity often forces one prime to be 2 before any long search begins.
Two primes satisfy 3p + 2q = 12
Because 2q and 12 are even, 3p must be even. Therefore p must be the only even prime.
Prime-pair equation explorer
The equation is A×p+B×q=T. Pairs are ordered: p fills the first position and q the second. Both primes must be within the chosen limit; absence of a result rules out only that range.
Complete the test problem
Run the explorer with A=3, B=2 and T=12 before checking.
Need a hint? Start here
Use parity first in 3p + 2q = 12; an even prime must equal 2.
How to check my reasoning
Parity forces p=2; substitute to find q. p=2; q=3; product 6.
Turn a product–sum equation into factor pairs
A prime factor on one side can force one of the unknown primes, and then a simple factorization finishes the search.
Three primes satisfy pqr = 7(p + q + r)
The right side is divisible by 7, so the prime product pqr is divisible by 7. At least one of the three primes must be 7. Call it r.
Factor-pair audit for 8
Prime-triple search within the chosen limit
Every prime in a triple must be at most the displayed limit. Repeated primes are allowed; rearrangements count once. The factor-pair argument above proves the worked example triple is unique without a search limit.
Record the unique original-test triple
Run the product–sum search with k=7 before checking. List the three primes in any order.
Need a hint? Start here
If a prime divides a product, it divides one factor. Substitute the forced prime and factor what remains.
How to check my reasoning
Set one prime to 7. Factor pairs of 8 give (p,q)=(2,9),(3,5),(5,3),(9,2); only the middle pairs are prime. Unordered triple 3,5,7; sum 15; product 105.
Prime-equation and reciprocal-sum workshop
Correct all eight questions to complete the workshop.
1. In the worked example, what is 1001 ÷ 7?
2. Three primes have reciprocal sum 311/1001. List them.
3. What is their sum?
4. List the three prime factors of 1986.
5. Three primes have reciprocal sum 1661/1986. What is their sum?
6. For primes p,q with 3p+2q=12, what is pq?
7. For primes with pqr=7(p+q+r), set r=7. What does (p−1)(q−1) equal?
8. List the three primes satisfying pqr = 7(p+q+r).
Need a hint? Start here
Separate finding candidates, checking equality, and proving uniqueness.
How to check my reasoning
The reduced denominator can contain only prime factors from the original denominators. 143; 7,11,13; 31; 2,3,331; 336; 6; 8; 3,5,7.
Exit ticket
Complete Missions 1–8, earn 8/8 in the workshop, and earn 5/5 on this exit ticket to unlock the certificate. The reflection is optional and not automatically graded.
1. Three primes have reciprocal sum 311/1001. What is their sum?
2. Three primes have reciprocal sum 1661/1986. What is their sum?
3. In 3p + 2q = 12, what is q?
4. Product of the primes in that linear equation?
5. Write 1/2 + 1/3 + 1/5 in lowest terms. What is its numerator?
Optional reflection — not automatically graded
Certificate of completion
Prime Equation & Reciprocal-Sum Solver
has completed Lesson 17.5 and demonstrated exact fraction verification, prime-factor reasoning, parity, complete enumeration, and equation solving.
Learning notes
- The worked example factors each displayed denominator and verifies the proposed unit fractions exactly.
- The repeated-prime cancellation example, exhaustive uniqueness searches, configurable laboratories, feedback, workshop, and exit ticket are added teaching scaffolds.
- The exhaustive search allows repeated primes unless the student selects “require distinct primes.”
Need a hint? Start here
Put the three unit fractions over denominator 30. Add the three contributions, then reduce.
How to check my reasoning
Explain your method in the reflection box. The checker grades the answer; your explanation is for you to review.