ΣPMath Education · Chapter 17
Lesson progress0 / 10 missions
Chapter 17 · Lesson 17.3

Decompose Numbers into Distinct Prime Sums

把一个数拆成不同质数之和

Bound the number of terms, search in a fixed order, record each unordered decomposition once, and optimize only after the complete list is known.

Bound first. Enumerate once. Optimize last.
10 interactive missionsGrade 5No this lesson requiredCondition-based checking
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Mission 1

Know what “different primes” means

In this lesson’s distinct-prime problems, a decomposition uses at least two prime addends. No prime may repeat, and changing the order does not create a new decomposition. Later activities explicitly allow repeats when needed.

Not complete
Prime addendsEvery addend must be prime.
No repeats“Different” means distinct.
Order ignored3 + 5 + 29 is the same set as 29 + 5 + 3.
Exact totalThe addends must sum to the target.

Representation normalizer

Enter two comma-separated lists, each containing 2–20 prime numbers from 2 through 1,000,000,000. The tool sorts each list, checks primality and repeats, and decides whether they represent the same decomposition.

Normalized A
Normalized B
Same decomposition?
All primes distinct?
Sum of A
Product of A

Quick check

Need a hint? Start here

Rearranging a sum does not make a new decomposition. Sort both lists and check every entry is a distinct prime.

How to check my reasoning

Order does not matter; sort both lists to compare them. No; no; four distinct primes.

Mission 2

Bound the number of terms before searching

The worked example first proves that 37 cannot be split into five or more different primes. This turns an open-ended search into a finite one.

Not complete
Worked example 3

How many different prime addends can 37 have?

Size bound

2 + 3 + 5 + 7 + 11 = 28 < 37
2 + 3 + 5 + 7 + 11 + 13 = 41 > 37

Size alone says there can be at most five terms.

Parity sharpens the bound

Five different primes totaling the odd number 37 cannot include 2, because 2 plus four odd primes is even.

3 + 5 + 7 + 11 + 13 = 39 > 37

Therefore there can be at most four terms.

Term-bound explorer

Terms5
Smallest possible sum
Smallest all-odd sum
Can total 37?

Record the bound

Need a hint? Start here

Add the smallest possible distinct primes to find when a longer sum becomes too large.

How to check my reasoning

Six smallest total 41. Five terms for odd total 37 cannot include 2, and five smallest odd primes total 39. At most four; 39; 41.

Mission 3

Search by the largest prime

Choose one order and keep it. The worked example fixes the largest prime, works downward, and decomposes only the remaining amount using smaller primes.

Not complete
Largest prime31Remainder to split: 6
Branch count0
Smaller primes available
Target37
OrderDescending

Complete branch audit

Run all five branches before checking the questions. A one-term expression such as 37 is excluded: a decomposition here needs at least two addends. Two terms also fail: parity would force 2+35, and 35 is composite.

Read the completed audit

Need a hint? Start here

Fix the largest prime and subtract it from 37; split only the remainder.

How to check my reasoning

Exclude one term; two terms would require composite 35. Branch 31 leaves 6, whose prime pair 3+3 repeats. Largest prime 19; four decompositions; ten overall; remainder 6 has no distinct-prime decomposition.

Mission 4

Audit the complete list of ten decompositions

A proof of completeness needs more than several successful examples. The search must cover every possible largest prime and count each unordered set once.

Not complete
Why the search stops at 17: if every prime were at most 13, even the four largest available distinct primes would total only 13 + 11 + 7 + 5 = 36.
Complete original search
Displayed0
All decompositions
Three-term
Four-term
Run the complete search to generate the worked example list.

Completeness check

Generate the complete list before checking. Answer using the total counts across all rows, even if a display filter hides some rows.

Need a hint? Start here

A complete search needs every possible largest prime and every possible term count.

How to check my reasoning

With at most four addends, using only primes at most 13 cannot reach 37: 13+11+7+5=36. Ten decompositions; five with three terms; five with four.

Mission 5

Optimize only after the list is complete

The worked example next multiplies the primes in every decomposition and asks for the smallest product.

Not complete

Correct order of work

1
List every valid decomposition.
Otherwise a better case may be missed.
2
Calculate one product per decomposition.
Order does not affect the product.
3
Compare the complete product list.
Then the minimum is proved.

original conclusion

The smallest product comes from:

29 × 5 × 3 = 435

The chart below makes every competing product visible.

Product comparison

Identify the minimum

Need a hint? Start here

Compare products only after listing all legal decompositions. A small sum does not mean a small product.

How to check my reasoning

The smallest product uses the three primes 29,5,3. 435; 29; 5; 3.

Mission 6

Use a general distinct-prime decomposition laboratory

The laboratory searches in nondecreasing order, so each unordered decomposition appears exactly once.

Not complete
Decompositions
Minimum product
Maximum product
Shortest length
Longest length

29 as exactly three different primes

Set the laboratory to target 29, no repeats, exactly 3 terms.

Need a hint? Start here

Check the repeat setting and exact term count before reading the search results.

How to check my reasoning

The lists are 3+7+19, 5+7+17, and 5+11+13. Three decompositions; minimum product 399; maximum product 715.

Mission 7

When repeats are allowed, minimize the largest prime

A practice problem asks for 60 as a sum of ten primes and minimizes the largest prime. It does not say “different,” so repeated primes are allowed.

Not complete
original Exercise 8

Lower bound

If every prime were at most 5, the total could be at most:

10 × 5 = 50 < 60

Therefore the largest prime must be at least 7.

Attainability

To prove the minimum, build ten primes totaling 60 while using no prime larger than 7.

number of terms = 10
sum = 60
largest prime = 7

Count-builder

Choose how many copies of 2, 3, 5, and 7 to use.

Terms0
Sum0
Largest used
Valid optimum?No

Prove the optimum

Build ten primes totaling 60 in the count-builder and run the complete enumeration before checking. Count different multisets of ten prime addends: changing their order does not create a new decomposition.

Need a hint? Start here

If every term were at most 5, ten terms could total at most 50. A construction using 7 proves the next bound is attainable.

How to check my reasoning

Ten primes all at most 5 total at most 50. Five copies of 5 and five copies of 7 total 60. Minimum largest prime 7. Four unordered ten-prime decompositions attain that optimum.

Mission 8

Optimization studio: products and equal pair sums

Two original exercises show why optimization should follow a complete enumeration.

Not complete
original Exercise 12

Split 33 into different primes and maximize the product

Decompositions
Maximum product
Number of primes
original Exercise 11

Four different two-digit primes form two equal pair sums

Maximize the common sum:

? + ?
=
? + ?

Record both optima

Run both searches before checking. For the equal-pair problem, use each of the four distinct two-digit primes once, with two primes on each side.

Need a hint? Start here

For 33, first enumerate the legal sums. For equal pair sums, no prime may be used twice.

How to check my reasoning

33=2+7+11+13. For the pair problem, 71+97=79+89. Maximum product 2002; primes 2,7,11,13 in any order; maximum common sum 168.

Mission 9

Distinct-prime decomposition workshop

Correct all eight questions to complete the workshop.

Not complete

1. Does changing the order of the same prime addends create a new decomposition?

2. What is the maximum number of different primes that can sum to 37?

3. How many unordered distinct-prime decompositions of 37 have largest prime 23?

4. How many different-prime decompositions of 37 are there?

5. What is the minimum product among those ten decompositions?

6. In exactly three different prime terms, how many decompositions does 29 have?

7. When 60 is written as ten primes, allowing repeats, what is the minimum possible largest prime?

8. What is the maximum product when 33 is split into different primes?

Workshop score0 / 8
Need a hint? Start here

Use sorted lists to avoid duplicate cases and verify each total.

How to check my reasoning

Check whether repeats and one-term expressions are allowed before counting. No; 4; 2; 10; 435; 3; 7; 2002.

Mission 10

Exit ticket

Complete Missions 1–8, earn 8/8 in the workshop, and earn 5/5 here to unlock your certificate. The optional reflection is not automatically graded.

Not complete

1. What is 29 × 5 × 3?

2. How many of the 37 decompositions use four primes?

3. For unordered distinct-prime sums of 37, which largest-prime branch contains four decompositions?

4. Allowing repeated primes, what is the minimum possible largest prime in a ten-prime sum of 60?

5. How many unordered pairs of distinct primes total 18?

Exit score0 / 5

Optional reflection — not automatically graded

Certificate of completion

Distinct Prime-Sum Enumerator

Grade 5 Mathematician

has completed Lesson 17.3 and demonstrated accurate term bounds, duplicate-free enumeration, completeness checks, and optimization over prime-sum decompositions.

Teaching notes

The normalizer, term-bound explorer, complete enumerators, product chart, feedback, workshop, and exit ticket are added instructional scaffolds.

Need a hint? Start here

For a two-prime sum, list pairs in increasing first-prime order and stop when the order reverses.

How to check my reasoning

Explain your method in the reflection box. The checker grades the answer; your explanation is for you to review.