18.6Math Adventure • Prime Factorization
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Chapter 18 • Lesson 6

Use Factor Pairs, Bounds, and Shared Divisors to Recover Hidden Counts

用因数对、范围与公因数还原隐藏数量

All people, item counts and equal shares in this lesson are positive whole numbers unless a laboratory explicitly allows zero adults or zero leftovers. A hidden number of children, students, winners, or items is often a factor of a known total. Build every factor candidate, apply the story’s bounds and remainder rules, then state whether the answer is unique or whether several cases survive.

Factor → filter → assign roles → verify
Grade 510 interactive missionsSelf-contained
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Mission 1

Build a factor-pair staircase

Factor pairs come in a predictable order: the first factor rises while the partner falls. Once the factors cross near the square root, every pair has already appeared.

Not complete

One product, many pairings

If ab = N, then a and b form a factor pair of N. For 420, the staircase begins:

1 × 420, 2 × 210, 3 × 140, …

Stop after the smaller factor passes:

√420 ≈ 20.49
The final pair before crossing is 20 × 21. Reversing it would not create a new unordered pair.

Factor-pair laboratory

Read the 420 staircase

Need a hint? Start here

A factor pair records both the number of people and the amount each gets.

How can I check my reasoning?

Stop testing smaller factors after √420. There are 12 unordered pairs. The pair with the smallest difference is 20 and 21.

Mission 2

Use neighboring factor pairs in the 420-apple story

This is Chapter 18, Example 5. The total stays 420 while one factor rises by 1 and the other falls by 2.

Not complete

original problem

A kindergarten teacher divides 420 apples equally among the children. If one more child joins, each child receives 2 fewer apples. How many children were there originally?

Original arrangementn × a = 420n children, a apples each
One child joins(n + 1)(a − 2) = 420same total

Audit neighboring rows of the factor staircase

The surviving transition is 14 × 30 → 15 × 28. Both products are 420, the number of children rises by 1, and the share falls by 2.

Complete the worked solution

Need a hint? Start here

Keep the total fixed when comparing old and new shares.

How can I check my reasoning?

na=(n+1)(a−2) gives a=2n+2, so the share is larger than the people count. The staircase orientation is valid here. 14×30=15×28=420. Original children 14; original share 30; new share 28.

Mission 3

Use a general factor-pair transition laboratory

The same idea works whenever a fixed total is shared by a changed number of people and each share changes by a known amount.

Not complete

Challenge: T = 330, one person joins, each share falls by 3

Need a hint? Start here

Compare two factor pairs with the same total: people increase by 1 while the share decreases by 3.

How can I check my reasoning?

Try 10×33 and then add one person while subtracting three from the share. 10×33=11×30=330. Original people 10; original share 33; new share 30.

Mission 4

Subtract leftovers, then find a shared divisor

This follows Guided Practice 5. The class size divides every amount actually distributed, and it must be larger than every leftover.

Not complete
Laboratory assumption: Every recipient gets a positive whole-number share of every kind of item. Therefore each leftover is smaller than its original total, as well as smaller than the group size. Enter whole-number lists separated by commas.

original problem

A teacher has 40 oranges, 200 biscuits, and 120 candies. The teacher gives out as many complete rounds as possible, giving every child one item of that kind per round. Afterward, 4 oranges, 20 biscuits, and 12 candies remain. There are too few of each leftover item to give one more to every child. How many children are in the class?

Oranges

40 − 4 = 36

Biscuits

200 − 20 = 180

Candies

120 − 12 = 108

Shared-divisor step

gcd(36, 180, 108) = 36

Every possible class size must be a divisor of 36.

Remainder bound

A remainder must be smaller than the divisor. Because the largest leftover is 20:

class size > 20

Only one common divisor survives.

Complete the worked example practice

Need a hint? Start here

Subtract each leftover first. The class size divides every distributed amount and is greater than every leftover.

How can I check my reasoning?

Distributed amounts are 36,180,108. The class size must divide each and exceed 20, the largest leftover. Their GCD is 36. Its only divisor greater than 20 is 36, so there are 36 children.

Mission 5

Use a general equal-distribution laboratory

A common divisor creates candidates. The largest remainder filters them. Sometimes one answer survives, sometimes several survive, and sometimes the data are impossible.

Not complete

Analyze totals 50,100,150 with leftovers 2,4,6

These questions use the multiple-candidate preset, even if you change the laboratory. Enter all candidate group sizes separated by commas, in any order.

Need a hint? Start here

Keep every common divisor greater than the largest leftover; more than one survivor means the size is not uniquely determined.

How can I check my reasoning?

The distributed amounts are 48,96,144, with GCD 48. Keep its divisors greater than 6. Five candidates: 8,12,16,24,48. Classification: multiple.

Mission 6

Use repeated count labels to recover the circle size

This follows Chapter 18, Example 6. If the same student says two count labels, the number of students divides the difference between those labels.

Not complete

original problem

More than 30 but fewer than 50 students stand in a circle and take turns, one student at a time in the same direction. They say consecutive whole-number labels without skipping or restarting. The same student says 30 and 198. How many students are there?

Subtract the labels

198 − 30 = 168

The class size must divide 168.

See the full laps

30
+42 →
72
+42 →
114
+42 →
156
+42 →
198

Four complete circuits of 42 students separate the two labels.

Same-speaker laboratory

Complete the worked example

Need a hint? Start here

Subtract the labels of the same speaker. The circle size divides that difference and must satisfy both strict bounds.

How can I check my reasoning?

198−30=168; keep divisors strictly greater than 30 and strictly less than 50. Only 42 works. 168÷42=4 complete circuits.

Mission 7

Assign factor roles using a grouping condition

This follows Guided Practice 6. A product’s factors can represent “number of people” and “trees per person,” but the group structure decides which factor plays which role.

Not complete

original problem

Students are divided equally into three groups. The teacher plants the same number of trees as each student. Altogether they plant 1,073 trees. How many trees does each person plant?

Factor the total

1,073 = 29 × 37

If there are 37 people including the teacher, then there are:

37 − 1 = 36 students

And:

36 ÷ 3 = 12 students per group

Role audit

Total people37
Students36
Per group12
Trees each29

Factor-role laboratory

Every adult and student contributes the same positive whole-number amount. Only students form the equal groups, and each group must contain at least one student. Adults may be zero in this laboratory.

Complete the worked example practice

Need a hint? Start here

The people count includes the teacher. Subtract one before checking whether the students form three equal groups.

How can I check my reasoning?

1073=29×37. Test every people divisor: people minus 1 must be positive and divisible by 3. 37 people, 36 students, 29 trees each, 12 students per group.

Mission 8

Transfer the method and report every valid case

The chapter exercises combine factor roles, square relations, and complete factor enumeration. One problem is unique; another has two valid triples.

Not complete

original Exercise 10: prize winners

750 yuan is divided equally among n winners. The amount each person receives, measured in jiao, is 12 times the number of winners.

1 yuan = 10 jiao, so total = 7,500 jiao
7,500 ÷ n = 12n
n² = 625

Because n is positive:

n = 25

original Exercise 13: sum many products at once

Choose one one-digit prime and one one-digit composite. Multiply each of the 4 × 4 = 16 possible pairs once and add the results.

Try recalling the shortcut, then reveal the worked solution
Primes: 2 + 3 + 5 + 7 = 17
Composites: 4 + 6 + 8 + 9 = 27
17 × 27 = 459

original Exercise 12: enumerate the two-digit triples

Find all two-digit integers a < b < c with even sum and:

a × b × c = 3,960
original-condition note: the printed data allow two valid triples. This page does not add an unstated uniqueness condition, and it accepts both cases.

State what each complete search proves

Need a hint? Start here

Use jiao in the prize equation, list increasing two-digit factor triples, and use two sums for the prime–composite products.

How can I check my reasoning?

For triples, a,b,c are three separate two-digit numbers multiplied together, with a<b<c. 25 winners. Triples (10,18,22) and (11,15,24) both total 50. The 16 prime–composite products total 17×27=459.

Mission 9

Factor-pair and hidden-count workshop

Correct all eight questions to complete the workshop.

Not complete
Need a hint? Start here

Choose the strategy first: factor, apply the conditions, and multiply back. Revisit the relevant local hint above.

How can I check my reasoning?

For leftovers, subtract first, then use a divisor larger than every remainder. Answers: 12; 14; 30; 36; 5; 42; 29; 2.

Mission 10

Exit ticket

Complete all ten missions, including 5 out of 5 on this exit ticket, to earn the certificate.

Not complete
18.6

Certificate of completion

Hidden-Count Factor Detective

Awarded to Student for building factor pairs, applying bounds, filtering shared divisors, and reporting every valid case.

Need a hint? Start here

Work without the laboratories first. Explain which prime copies or factor pairs your answer uses, then verify the result.

How can I check my reasoning?

For the final product sum, use (2+3)(4+6). Answers: 10; 36; 4; 12; 50.