Lesson 9.4 — Three Types and Integer Constraints

Mission 1

Why do three types need a third clue?

A total count and a weighted total give two relationships. Three unknown counts can still have more than one whole-number solution.

Not complete

A small example

Six boxes hold either 1, 2, or 3 tokens. Together they hold 12 tokens. Every type must appear at least once.

x + y + z = 61x + 2y + 3z = 12

These two arrangements both work:

Candidate A

(1, 4, 1)

1 + 4 + 1 = 6
1 + 8 + 3 = 12

Candidate B

(2, 2, 2)

2 + 2 + 2 = 6
2 + 4 + 6 = 12

The third clue chooses

Suppose the problem adds:

Type 2 is strictly more numerous than either other type.

Candidate A works because 4 is greater than both 1 and 1. Candidate B does not because all three counts are tied.

Strictly greatest means no tie.

Three-type problems often need:

a ratio between groups;
at least one of each type;
a largest or smallest count;
an even/odd condition;
or another independent relationship.

Check the idea

Why are the two totals not enough here?

In general, what kind of numbers can object counts be?

What does “at least one of each type” mean?

Which candidate has Type 2 strictly greatest?

Mission 2

Build a three-type constraint table

Write what each type contributes, then record every independent condition before trying values.

Not complete

General model

Type	Count	Contribution each	Subtotal
A	x	a	ax
B	y	b	by
C	z	c	cz

x + y + z = Nax + by + cz = T

Use a low-type baseline

Choose Type A to have the smallest contribution: a < b and a < c.

If every object were Type A, the baseline would be:

aN

Every Type B object adds b − a. Every Type C object adds c − a.

(b − a)y + (c − a)z = T − aN

This removes one unknown count, but a third condition may still be needed.

1

Count conditionAll type counts add to the total number of objects.

2

Contribution conditionThe weighted subtotals add to the stated total.

3

Extra conditionUse the ratio, positivity, parity, ordering, maximum, or minimum statement.

4

Integer checkCounts must be whole and cannot be negative. “Positive” means at least 1; “nonnegative” allows 0.

Match each relationship

Which equation records the total number of objects?

Which equation records the weighted total?

What does an all-Type-A baseline equal?

What should be checked last?

Mission 3

Split the mentor groups using a ratio

A factory has 27 mentors supervising 40 apprentices. Each apprentice is assigned to exactly one mentor. Each mentor supervises 1, 2, or 3 apprentices. The number supervising 1 apprentice is twice the combined number supervising 2 or 3 apprentices.

Not complete

Translate the ratio

1-apprentice mentors : other mentors = 2 : 1

The 27 mentors form three equal ratio units:

27 ÷ (2 + 1) = 9 mentors per unit2 × 9 = 18 one-apprentice mentors1 × 9 = 9 other mentors

Remove what is already known

The 18 one-apprentice mentors supervise:

18 × 1 = 18 apprentices

The remaining 9 mentors must supervise:

40 − 18 = 22 apprentices

Now the problem has only two types left: mentors supervising 2 or 3 apprentices.

Two ratio units · 18 mentors supervise 1 each

One ratio unit · 9 mentors supervise 2 or 3 each

Complete the first stage

Mentors supervising 1 apprentice

Mentors remaining in the 2-or-3 group

Apprentices supervised by the first group

Apprentices supervised by the remaining group

Mission 4

Use a baseline inside the remaining group

Nine mentors supervise 22 apprentices. Each of these mentors supervises either 2 or 3 apprentices.

Not complete

Change some 2-apprentice mentors into 3-apprentice mentors

Mentors supervising 34

5supervise 2

4supervise 3

22apprentices

0difference from 22

Baseline-and-extra reasoning

Assume all nine mentors supervise 2 apprentices:

9 × 2 = 18

The actual total is 22, so there are:

22 − 18 = 4 extra apprentices

Each 3-apprentice mentor adds exactly one extra apprentice. Therefore 4 mentors supervise 3 apprentices, and the other 5 supervise 2.

Complete the second stage

All-2 baseline

Extra apprentices above the baseline

Mentors supervising 3 apprentices

Mentors supervising 2 apprentices

Mission 5

Validate the complete integer triple

A candidate is not a solution until it satisfies every original condition.

Not complete

Enter your three mentor counts

Supervise 1 apprentice

Supervise 2 apprentices

Supervise 3 apprentices

Why check all conditions?

Four checks

1

Total mentorsCounts must add to 27.

2

Total apprentices1x + 2y + 3z must equal 40.

3

Ratio cluex must equal 2(y + z).

4

Integer countsEvery count must be a nonnegative whole number.

Verification matters. A neat-looking triple can still have the wrong total, wrong weighted total, or wrong ratio.

Mission 6

Build the dormitory baseline

A school has 12 completely full dorm rooms holding exactly 80 students. There is at least one room of each size: small rooms hold 5, medium rooms hold 7, and large rooms hold 8. The number of medium rooms is strictly greater than either other count. Find the greatest possible number of medium rooms.

Not complete

Assume all rooms are small

12 × 5 = 60 students

The real arrangement holds 80, so the room upgrades must add:

80 − 60 = 20 extra places

Changing one small room to medium adds:

7 − 5 = 2

Changing one small room to large adds:

8 − 5 = 3

Write the upgrade equation

Let L be the large-room count and M the medium-room count.

3L + 2M = 20

Then the small-room count is:

S = 12 − L − M

Additional constraints:

L, M, S ≥ 1 because every size occurs.
M > L and M > S.
All three counts are whole numbers.

Complete the model

All-small baseline

Extra places needed

Extra places from one large room

Extra places from one medium room

Which upgrade equation is correct?

Mission 7

List the integer candidates, then apply the ordering clue

The upgrade equation creates a short candidate list. Positivity and “strictly greatest” decide which candidate works.

Not complete

Explore possible large-room counts

Large rooms L2

2large

7medium

3small

80students

This candidate has medium rooms strictly greatest.

Why is L even?

2M = 20 − 3L

The left side is even, so the right side must be even. Since 20 is even and 3 has the same parity as L, L must be even.

At least one of every room size leaves:

L = 2, 4, or 6

L = 0 is not allowed because the problem says all three room sizes occur.

Large L	Medium M	Small S	Total rooms	Total students	Medium strictly greatest?

Use the candidate list

How many positive whole-number triples fit both totals before applying the “medium strictly greatest” clue?

Smallest possible positive large-room count

Greatest possible medium-room count

Small-room count in the winning candidate

Why is the candidate (4, 4, 4) rejected?

Mission 8

Use an integer-candidate laboratory

Search small whole-number triples systematically instead of guessing. The laboratory starts with a new practice problem.

Not complete

Practice preset

Ten packages are Type A, B, or C. They contribute 1, 2, or 4 points each, for 20 points altogether. Every type appears. Type B is strictly the most numerous.

Total objects N

Type A contribution

Type B contribution

Type C contribution

Weighted total T

Display rule

Require at least one of every type

The laboratory will list every candidate.

Type A	Type B	Type C	Count check	Weighted check	Rule match

Systematic search is not random guessing. Start at 0 or 1, move one count at a time, calculate the remaining count, and reject any triple that violates a condition.

Read the practice candidate list

Use the fixed preset: 10 objects, contributions 1, 2, and 4, total 20, every type present, and Type B strictly greatest. Editing the laboratory does not change these questions.

Positive integer candidates before the ordering rule

Type A count in the rule-matching candidate

Type B count in the rule-matching candidate

Type C count in the rule-matching candidate

Why is the ordering clue necessary?

Mission 9

Three-type workshop

Solve at least six of the eight questions. Enter only the requested number or choice.

Not complete

1. Twelve packages weigh 1, 3, or 5 kg each. Their total mass is 28 kg, and at least one of each type is present. What is the all-1-kg baseline, in kg?

Hint

Use all twelve packages.

Worked solution — open after trying

12 × 1 = 12 kg.

2. For those twelve packages, how much extra mass above the baseline must the heavier types explain, in kg?

Hint

Compare the real total with the baseline.

Worked solution — open after trying

28 − 12 = 16 kg.

3. Let A, B, C count the 1-, 3-, and 5-kg packages. Dividing the extra-mass equation by 2 gives B + 2C = 8. If C = 1, what is B?

Hint

Subtract 2C from 8.

Worked solution — open after trying

B = 8 − 2 × 1 = 6.

4. For C = 1 and your B count, what is A?

Hint

All three counts add to 12.

Worked solution — open after trying

A = 12 − 6 − 1 = 5.

5. How many positive whole-number candidates (A,B,C) satisfy both totals? Try C = 1, 2, 3. Enter only the candidate count, then explain to yourself why C = 4 is excluded.

Hint

B = 8 − 2C must be positive.

Worked solution — open after trying

The candidates are (5,6,1), (6,4,2), (7,2,3). C = 4 would make B = 0, which is forbidden. There are 3 candidates.

6. An extra clue says there are twice as many 3-kg packages as 5-kg packages: B = 2C. What is C?

Hint

Combine B = 2C with B + 2C = 8.

Worked solution — open after trying

4C = 8, so C = 2. This extra clue chooses just one candidate.

7. With B = 2C, what is B?

Hint

Double the C count.

Worked solution — open after trying

B = 2 × 2 = 4.

8. With that extra clue, what is A? Check both totals on paper.

Hint

Subtract B and C from 12.

Worked solution — open after trying

A = 12 − 4 − 2 = 6. Count: 6 + 4 + 2 = 12. Mass: 6 + 12 + 10 = 28 kg.

Mission 10

Try a fresh challenge

Try these new questions before opening help. Earn your certificate by completing Missions 2, 4, and 9 and answering all five exit questions correctly. Optional reflections and extra exploration do not affect your score.

Not complete

1. Nine packages each weigh 1, 2, or 4 kg, with at least one of each type. Total mass is 19 kg. What is the all-1-kg baseline, in kg?

Hint

Keep all nine packages in the baseline.

Worked solution — open after trying

9 × 1 = 9 kg.

2. Let B count 2-kg packages and C count 4-kg packages. In B + 3C = ___, what is the missing extra mass?

Hint

Subtract the baseline from 19 kg.

Worked solution — open after trying

19 − 9 = 10 kg; B + 3C = 10.

3. The positive candidates (A, B, C) are (1,7,1), (3,4,2), and (5,1,3). An extra clue says B = 2C. What is C?

Hint

Test the extra clue for every row.

Worked solution — open after trying

Only (3,4,2) has B = 2C, so C = 2. The other rows have 7 ≠ 2 and 1 ≠ 6.

4. For the candidate satisfying B = 2C, what is A?

Hint

Use A + B + C = 9.

Worked solution — open after trying

A = 9 − 4 − 2 = 3.

5. Verify that candidate: what total mass does A + 2B + 4C give, in kg?

Hint

Substitute all three counts, then check positivity too.

Worked solution — open after trying

3 + 2 × 4 + 4 × 2 = 19 kg. All counts are positive integers, and 3 + 4 + 2 = 9.

Optional explanation: Why does your method work? Check both totals.

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Three-Type Constraint Solver

Lesson checkpoints completed

You practised baselines, whole-number candidates, positivity and strict ordering. Revisit any steps for which you needed solution help.

Lesson 9.4 complete

Teaching notes

The worked solution treats all three room sizes as present and rejects a tie for “most.” To make the child-facing lesson unambiguous, this page explicitly states “at least one room of each size” and “medium rooms are strictly more numerous than either other type.

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