Group Prime-Power Packets into Coprime Composite Numbers
第17讲 · 把质数幂“整包”分配给两两互质的合数
Factor a fixed product, keep every complete prime-power packet together, build three composite factors whose pairs have GCD 1, and then compare every legal grouping to prove the smallest or greatest possible sum.
Separate “composite” from “pairwise coprime”
The worked example requires three numbers that are composite individually and coprime in every pair. Both conditions matter.
Composite number
A composite number is greater than 1 and has more than two positive divisors.
The number 13 is prime, so it cannot stand alone as one of the required factors.
Pairwise coprime
For three numbers A, B, and C:
This is stronger than saying that all three numbers have no single factor in common.
Live GCD audit
Check the definitions
Need a hint? Start here
Check all three pairs separately. For example, 6, 10, 15 have no common factor shared by all three, but each pair shares a factor.
How to check my reasoning
49=7², 52=2²×13, 121=11²; no prime appears in two factors. Every pair has GCD 1; yes; yes.
Factor the worked example product into prime-power packets
All copies of one prime are bundled into one packet. These packets, not the individual prime factors, are the pieces we may assign.
Fixed product
Prime-factor and packet machine
Record the worked example factorization
Need a hint? Start here
Factor the whole product, then collect every copy of the same prime into one power.
How to check my reasoning
1001×28×11=(7×11×13)×(2²×7)×11. Packet values 4,49,121,13; product 308308; prime packet 13.
Keep every prime-power packet whole
Splitting copies of the same prime between two groups would make those two groups share that prime.
Illegal split
Then both A and B are divisible by 7:
Legal packet
Exactly one factor receives the complete 7² packet. The other two factors receive no factor 7.
Packet-rule demonstrator
The exponent counts copies of the prime in a product: 7² means 7×7. In this tool assign all copies to A or B; C receives none of this prime. A common factor shown here comes only from this one prime, not from other packets.
State the packet rule
Need a hint? Start here
Splitting copies of a prime across two groups gives those groups a shared factor.
How to check my reasoning
Use a 7² preset: splitting one 7 into each group makes both divisible by 7. One factor only; GCD at least 7.
Build three legal composite factors
Assign each packet to A, B, or C. The page audits nonempty groups, compositeness, and all three pairwise GCDs.
Complete a legal construction
Assign all four packets to build three legal factors before checking. Keep your valid grouping on screen.
Need a hint? Start here
Every group must be composite and every pair must have GCD 1. A lone packet 13 fails the composite test.
How to check my reasoning
Each group must be nonempty and composite, with no shared prime factor between groups. 13 cannot stand alone because it is prime.
Find the minimum sum in Worked example 6
Four packets must become exactly three nonempty groups, so exactly one pair of packets is formed. Since 13 cannot stand alone, it must be the packet that is paired.
Enumerate every legal partner for 13
| Packet joined to 13 | Three composite factors | Sum |
|---|---|---|
| 4 | 52, 49, 121 | 222 |
| 49 | 4, 637, 121 | 762 |
| 121 | 4, 49, 1573 | 1626 |
Why the first row is best
The three factors are:
They are all composite and pairwise coprime. Their sum 222 is smaller than the other two legal sums, 762 and 1626; comparing all three cases proves the minimum.
Complete original conclusion
The worked example asks for the minimum, not merely one legal construction.
Record the minimum case
Need a hint? Start here
Only the packet 13 needs a partner. Compare its three possible partners to prove the minimum.
How to check my reasoning
13 must join 4,49 or121. The resulting sums are 222,762,1626. Join 4; factors 49,52,121 in any order; minimum 222.
Find the maximum sum in Guided Practice 6
The guided practice uses the same fixed product, but asks for the greatest possible sum.
Why the packet pool is unchanged
To maximize
Pair the prime packet 13 with the largest composite packet, 121:
Maximum sum
All three factors remain composite and pairwise coprime.
Complete the guided practice
Need a hint? Start here
Use the same complete list for the maximum; no new case is needed.
How to check my reasoning
Choose the largest of the same three sums. Join 121; factors 4,49,1573; maximum 1626.
Enumerate all legal packet partitions
A complete search proves that no legal grouping has been missed. The explorer factors a product, builds whole packets, partitions them into three groups, and keeps only composite triples.
Prime-power partition explorer
Analyze the transfer product 1260
Use the 1260 preset in the explorer before checking these answers. Swapping the names A, B and C does not create a new partition.
Its packets are 4, 9, 5, and 7. Since both 5 and 7 are prime packets, they must be paired together to make the third composite factor.
Need a hint? Start here
Each packet goes into exactly one group. Do not split it, repeat it, or leave it unused.
How to check my reasoning
Its prime packets 5 and 7 must join each other, leaving composite packets 4 and 9 alone. One partition; factors 4,9,35; sum 48.
Prove which partner gives the minimum or maximum
When one prime packet p must join one of three composite packets x<y<z, comparing the three sums gives a clean optimization rule.
Comparison rule
If p joins x, the sum is:
If p joins y instead, the increase is:
Because both factors are positive, the sum grows when the partner grows.
Conclusion
Partner comparison laboratory
Use a prime p and three powers of different other primes, each composite. Every input value must be from 2 through 1,000,000,000. All four packets must be pairwise coprime.
Apply the rule to packets 4, 9, 25, and 7
Use the transfer comparison preset before checking these answers.
Need a hint? Start here
Compare px + y with x + py. Their difference is (p − 1)(y − x).
How to check my reasoning
Compare the sum when 7 joins 4,9 or25. Increasing the partner increases the sum by (7−1) times the partner increase. Minimum 62; maximum 188; join 25 for the maximum.
Prime-power packet workshop
Correct all eight questions to complete the workshop.
1. Pairwise coprime means the GCD of every pair is what number?
2. List the four complete prime-power packets of 308308.
3. If one 7 is placed in A and another 7 in B, GCD(A,B) is at least what?
4. How many unordered partitions of 4,49,121,13 form three composite factors?
5. Least sum of three pairwise coprime composite factors with product 308308?
6. Greatest sum of three pairwise coprime composite factors with product 308308?
7. Three pairwise coprime composite factors have product 1260. What is their unique sum?
8. For packets 4, 9, 25, and 7, what is the maximum sum?
Need a hint? Start here
Check composite, pairwise coprime, and product conditions before comparing sums.
How to check my reasoning
Only compare sums after all grouping conditions pass. 1; 4,49,121,13; 7; 3; 222; 1626; 48; 188.
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. Which of packets 4,49,121,13 is prime and cannot be a composite factor alone?
2. For three pairwise coprime composite factors A,B,C with ABC=308308, what is the least A+B+C?
3. For the same product 308308 and factor rules, what is the greatest A+B+C?
4. What must every pairwise GCD equal?
5. Three pairwise coprime composite numbers have product 900 = 4 × 9 × 25. What is their sum?
Optional reflection — not automatically graded
Certificate of completion
Prime-Power Packet Architect
has completed Lesson 17.6 and demonstrated factorization, pairwise-coprime reasoning, composite-factor construction, complete partition enumeration, and minimum/maximum proof.
Learning notes
- The worked example factors the fixed product into 2², 7², 11², and 13, then groups these factors into three pairwise-coprime composite numbers.
- The worked example minimum construction is 121, 49, and 52, with sum 222.
- The guided practice uses the same product because 11011 = 1001 × 11; the maximum construction is 4, 49, and 1573, with sum 1626.
- The packet language, GCD audits, exhaustive partition explorer, comparison proof, transfer examples, workshop, and exit ticket are added instructional scaffolds.
Need a hint? Start here
In the new product there are three packets and three groups. Each group must receive exactly one packet.
How to check my reasoning
Explain your method in the reflection box. The checker grades the answer; your explanation is for you to review.