Chapter 14 · Lesson 6 · Grade 5

Solve Number Networks with Global Totals and Constraints

填数阵图：用整体总和与条件破解数阵

Learn to combine local sum rules, the total of the whole number pool, upper and lower bounds, an unused-number ledger, and systematic case testing. Every diagram and clue needed to solve the tasks is included here.

Solve the whole network before chasing every single circle.

10 interactive missionsWorked example 6“王” practiceCube-face transfer

Answers save in this browser

Mission 1

Read the complete network before calculating

The diagram uses thirteen different labels—the numbers 1 through 13. The purple circle marked 43 is not another label; it records the required sum of the four lowest numbered circles.

Not complete

unknown label given label total, not a label

Three kinds of information

Number pool: use 1 through 13 exactly once.

Local rule: each lower connected circle equals the sum of the two circles feeding it.

Global rule:

D + F + H + K = 43

Do not place 43 in the number pool. It is a total outside the thirteen labels.

Check what the diagram contains

How many labels from the pool must be placed?How many local “lower = sum of two above” rules are present?Is 43 one of the labels 1–13?

Need a next step?

Hint

Separate labels from the total outside the pool.

Reasoning and worked explanation

The thirteen placed labels are 1–13. The 43 below them is a total of D,F,H,K, not a fourteenth label.

Mission 2

Translate every connection into an equation

A network becomes manageable when every local picture is rewritten as a short equation. Select a relation to highlight the circles that create it.

Not complete

Choose a relation to see which circles it uses.

All six original relations

C = A + B

D = 4 + C

F = B + E

H = 3 + G

J = G + I

K = J + I

Read the equations

Which expression equals C?Which expression equals K?Which four circles have total 43?

Need a next step?

Hint

Trace the two parents of each lower node.

Reasoning and worked explanation

For example C=A+B, and K=J+I. The second equation uses I a second time when J is replaced by G+I.

Mission 3

Use the whole number pool to find the uncounted circle

The key original idea is to count two complete groups with known total 43. Every label except A belongs to exactly one of those groups.

Not complete

All thirteen labels

1 + 2 + ··· + 13 = 91

This is the total value available in the complete pool.

The four bottom circles

DFHK

D + F + H + K = 43

The eight circles that produce them

4CBE3GJI

(4+C)+(B+E)+(3+G)+(J+I)=43

Only A is absent from both groups. The two 43-groups contain the other twelve labels exactly once.

A = 91 − 43 − 43 = 5

Rebuild the global-total argument

What is 1 + 2 + ··· + 13?What is the total of the eight producer circles?What must A equal?

Need a next step?

Hint

Add the bottom four labels; then add their eight different parents.

Reasoning and worked explanation

Both groups total 43. They cover twelve labels with no overlap, leaving A=91−43−43=5.

Mission 4

Use bounds and the unused-number pool

After finding A, the network equations create tight upper bounds. Bounds are most useful when they are combined with the numbers already used.

Not complete

Bound B from the left branch

C = B + 5

D = 4 + C = B + 9

Every label is at most 13, so:

B + 9 ≤ 13 ⟹ B ≤ 4

Bound I from the right branch

J = G + I

K = J + I = G + 2I

Since G is at least 1 and K ≤ 13, G + 2I ≤ 13 gives I ≤ 6. If I were 6, these bounds force G = 1. Then H = 3 + G = 4, duplicating the already used 4.

I < 6

Now use the number pool

A, 3, and 4 already use the labels 5, 3, and 4. B is a positive unused value no larger than 4. I is a positive unused value below 6. B and I must also be different.

{B, I} = {1, 2}

Check the bounds

Largest possible value of B before using the unused-number ruleWhole-number upper bound for I after rejecting I = 6, before removing the used labels 3, 4, 5Which two labels must B and I use?

Need a next step?

Hint

Every placed value is at most 13.

Reasoning and worked explanation

D=B+9 gives B≤4. K=G+2I gives I≤6; I=6 forces G=1 and repeats the fixed 4 at H. After removing used 3,4,5, B and I must be 1 and 2.

Mission 5

Test both remaining cases and prove uniqueness

There are only two possible orders for B and I. Test both; one completes the network, while the other demands four consecutive unused labels that do not exist.

Not complete

Case A

B = 1, I = 2

Run the case to reveal its forced values.

Case B

B = 2, I = 1

Run the case to reveal its contradiction.

State the surviving case

Which order survives?Why does the other order fail?

Need a next step?

Hint

After testing B and I, list the remaining pool before choosing G.

Reasoning and worked explanation

B=1,I=2 forces C=6,D=10. The remaining labels are 7,8,9,11,12,13. The group G,G+2,G+3,G+4 fits only at G=9, giving J=11,H=12,K=13. Then E=7,F=8. The other case needs four consecutive remaining labels, which do not exist.

Mission 6

Complete and audit the worked example network

Enter values for A through K. The checker verifies the number pool, all six local equations, and the bottom total. It does not compare your work with a picture; it checks the mathematics.

Not complete

Need a next step?

Hint

Read the first failing audit and check its two parent values.

Reasoning and worked explanation

The completed A–K row is 5,1,6,10,7,8,9,12,2,11,13. Each label 1–13 appears once including the fixed 3 and 4, and 10+8+12+13=43. After studying it, complete the independent prediction too.

Your independent check

You may study a model. To complete this mission, also answer this new prediction and press the mission’s Check button.

In a new network branch, C=A+B and D=C+4. Increase B by 2 while A stays fixed. How much must D increase?

Explain this prediction

The increase passes through C and then D unchanged: two more at both nodes.

Mission 7

Build the worked example’s 王-shaped equal-sum array

Place 1 through 11 in the eleven cells so the three horizontal rows and the central vertical column each total 18. The worked example gives the challenge; the global-total analysis below makes it approachable.

Not complete

Count the horizontal rows first

1 + 2 + ··· + 11 = 66

3 horizontal rows = 3 × 18 = 54

The two bridge-only cells are not in any horizontal row, so together they total:

66 − 54 = 12

Now use the vertical column

The vertical column contains the two bridge cells plus the three row-center cells.

row centers + 12 = 18

row-center total = 6

The only three distinct positive labels totaling 6 are 1, 2, and 3.

Derive and validate

Sum of the two bridge-only cellsSum of the three row centersWhich labels must occupy the row centers?

Need a next step?

Hint

The three horizontal rows leave only the bridge cells uncounted.

Reasoning and worked explanation

66−54=12 for the bridges; vertical 18 leaves 6 for the centers, forcing 1,2,3. One construction has horizontal rows 6,1,11 / 7,2,9 / 5,3,10 and bridges 4,8.

Your independent check

You may study a model. To complete this mission, also answer this new prediction and press the mission’s Check button.

A Wang-shaped diagram uses 1–11 once but its three horizontal rows each total 19. Find the total of the two bridge-only cells.

Explain this prediction

The pool 66 minus three row totals 57 leaves 9.

Mission 8

Transfer weighted counting to equal cube-face sums

Place eight different whole numbers chosen from 1 through 9 on the eight cube vertices (corners), one per vertex. Every face has the same four-vertex sum k. Dividing k by the one unused number must leave a nonzero remainder.

Not complete

Count every vertex by its face participation

Each cube vertex belongs to exactly three faces. Let the unused label be u.

used-label total = 45 − u

sum of all 6 face totals = 6k

the same count by vertices = 3(45 − u)

k = (45 − u) ÷ 2

Test the possible unused odd labels (45 − u must be even because it equals 2k)

u	k	Does u divide k?	Keep?

Verify that a construction exists

This worked face explorer shows one arrangement using every label except 7. Inspect all six faces to verify the common sum k.

Inspect all six faces using the face buttons to complete this mission. Each highlighted face uses four corners, including corners hidden behind the front face.

Complete the weighted-counting proof

How many faces use each vertex?Which number is unused?What is the common face sum k?

Need a next step?

Hint

Count each cube vertex once for each face using it.

Reasoning and worked explanation

Six face totals count eight vertices three times each: 6k=3(45−u). Thus k=(45−u)÷2. Test odd u=1,3,5,7,9. Only u=7 makes k not divisible by u, giving k=19.

Explore further: line totals and integer neighbor averages

Use a line’s remainder. If a three-node line totals 23 and two labels are 6 and 9, the third is 23−6−9=8. When two lines share a node, both equations must give the same value. A bent path is not a straight line unless the problem explicitly counts it. Mark each required line before adding totals.

Averages need divisibility. A node replaced by the average of three neighbors gets a whole number only if their total is divisible by 3. Neighbors 1,4,7 give (1+4+7)÷3=4; neighbors 1,4,8 give 13÷3, which is not a whole number. A useful construction is to use distinct labels that are all 1 more than a multiple of 3, such as 1,4,7,10,13,16,19,22. Any three then total a multiple of 3, so every three-neighbor average is an integer. Apply the averages simultaneously to the original labels, not one update at a time.

Mission 9

Fresh strategy workshop

Try these new situations before opening the explanations. Correct all eight answers to complete this workshop.

Not complete

1. A nine-label network uses 1–9 once. Two groups with no shared position each total 20 and leave only A uncounted. Find A.

Worked explanation — try first

The pool totals 45; A=45−20−20=5.

2. A branch has D=B+8 and every label is at most 12. What is the largest possible B?

Worked explanation — try first

B+8≤12 gives B≤4.

3. If G=5 and I=3, and J=G+I and K=J+I, find K.

Worked explanation — try first

J=8; K=8+3=11, or G+2I=11.

4. Labels 1–11 total 66. Three horizontal rows with no shared position each total 19 and leave two bridge cells outside the rows. Find the bridge total.

Worked explanation — try first

66−3×19=9.

5. The same diagram has a vertical total of 19 using those bridges and three row centers. Find the row-center total.

Worked explanation — try first

19−9=10. Unlike a total of 6, this does not uniquely determine the three centers.

6. A cube uses eight different labels totaling 52. All face sums are equal. Find the common face sum.

Worked explanation — try first

Six faces count each vertex three times: 6k=3×52, so k=26.

7. A lower node is fed by values 7 and 4. What value must it contain?

Worked explanation — try first

The local addition rule gives 7+4=11.

8. An 8-vertex cube has how many total vertex appearances across its six faces?

Worked explanation — try first

Six faces × four vertices = 24; also eight vertices × three faces = 24.

0 / 8

Mission 10

Independent exit ticket

Try these new situations before opening the explanations. All five correct completes this mission. The certificate requires all ten missions.

Not complete

1. A network uses 1–9 once. Two groups with no shared position total 18 and 23, leaving only the label at position A outside both groups. Find A.

Worked explanation — try first

45−18−23=4.

2. D=B+7, with D at most 15. What is the largest possible B?

Worked explanation — try first

B≤15−7=8.

3. G=6, I=2, J=G+I and K=J+I. Find K.

Worked explanation — try first

6+2×2=10.

4. A cube uses eight different labels with total 60. All six face sums agree. Find each face sum.

Worked explanation — try first

6k=3×60, hence k=30.

5. Three centers must be distinct positive integers and total 6. What is the largest center?

Worked explanation — try first

The smallest possible distinct triple is 1,2,3, already totaling 6. Thus its largest is 3.

0 / 5

Optional reflection — saved, not automatically graded

Describe a strategy you can now use on a new puzzle. Explain one check that prevents a mistake.

✓

Number-Network Constraint Architect

This certifies that

Grade 5 Mathematician

used local equations, global totals, bounds, unused-number ledgers, and complete constraint audits to solve number networks.

Teaching notes

The interactive validators, face explorer, partial starters, workshop, and exit ticket are added teaching scaffolds.

The worked example network is unique under its stated conditions. The 王-shaped practice has multiple valid arrangements, so its checker validates the conditions rather than demanding one specific pattern.