Arithmagons math worksheet with four sections of number puzzles for mental math practice.

To solve these Arithmagons, we need to understand the rule: The number in a square (box) is the sum of the two numbers in the circles connected to it.

Let's break down each section step-by-step.

Section A

In this section, all three circle numbers are given. We just need to add them up to find the box numbers.

1. First Triangle:
* Circles: 2, 3, 6
* Left Box: $2 + 3 = 5$
* Right Box: $2 + 6 = 8$
* Bottom Box: $3 + 6 = 9$

2. Second Triangle:
* Circles: 7, 10, 4
* Left Box: $7 + 10 = 17$
* Right Box: $7 + 4 = 11$
* Bottom Box: $10 + 4 = 14$

3. Third Triangle:
* Circles: 4, 7, 3
* Left Box: $4 + 7 = 11$
* Right Box: $4 + 3 = 7$
* Bottom Box: $7 + 3 = 10$

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Section B

Here, some circles are missing. We can find them by subtracting known circle numbers from the box numbers.

1. First Triangle:
* Knowns: Top Circle = 7, Left Circle = 3, Right Box = 28.
* Find Right Circle: The Right Box (28) is the sum of Top (7) and Right Circle.
* $28 - 7 = 21$. So, Right Circle = 21.
* Find Left Box: Top (7) + Left (3) = 10.
* Find Bottom Box: Left (3) + Right (21) = 24.

2. Second Triangle:
* Knowns: Top Circle = 12, Left Box = 24, Right Circle = 5.
* Find Left Circle: The Left Box (24) is the sum of Top (12) and Left Circle.
* $24 - 12 = 12$. So, Left Circle = 12.
* Find Right Box: Top (12) + Right (5) = 17.
* Find Bottom Box: Left (12) + Right (5) = 17.

3. Third Triangle:
* Knowns: Top Circle = 8, Right Box = 35, Right Circle = 7.
* *Wait, let's check consistency.* If Right Box is 35 and Right Circle is 7, then Top Circle should be $35 - 7 = 28$. But the Top Circle is given as 8. Let me re-read the diagram carefully.
* Ah, looking at the third triangle in Section B:
* Top Circle: 8
* Right Circle: 7
* Bottom Box: 35
* This implies the Left Circle + Right Circle = 35? No, usually Bottom Box connects Left and Right circles. Let's assume standard positioning: Left Box connects Top/Left, Right Box connects Top/Right, Bottom Box connects Left/Right.
* Given: Top=8, Right=7, Bottom Box=35.
* Bottom Box (35) = Left Circle + Right Circle (7).
* Left Circle = $35 - 7 = 28$.
* Now find Left Box: Top (8) + Left (28) = 36.
* Now find Right Box: Top (8) + Right (7) = 15.

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Section C

This section is trickier. We have box numbers but need to find the circle numbers.
Rule: If you add all three box numbers together, you get twice the sum of the three circle numbers. Or simpler: Circle = (Sum of two adjacent boxes - Opposite Box) / 2.
Even simpler method for students: Pick a circle. Let's call the top one $T$, left $L$, right $R$.
$L + R = \text{Bottom Box}$
$T + L = \text{Left Box}$
$T + R = \text{Right Box}$

1. First Triangle:
* Boxes: Left=24, Right=64, Bottom=36.
* $L + R = 36$
* $T + L = 24$
* $T + R = 64$
* Add all boxes: $24 + 64 + 36 = 124$.
* Sum of circles ($T+L+R$) = $124 / 2 = 62$.
* Top Circle ($T$) = Total Sum (62) - Bottom Box (36) = 26.
* Left Circle ($L$) = Total Sum (62) - Right Box (64)... Wait, $62-64$ is negative. Let me re-check the calculation.
* $24 + 64 + 36 = 124$. Correct.
* $124 / 2 = 62$. Correct.
* $T = 62 - 36 = 26$.
* $L = 62 - 64 = -2$.
* $R = 62 - 24 = 38$.
* Check: $L+R = -2 + 38 = 36$. Correct. $T+L = 26 + (-2) = 24$. Correct. $T+R = 26 + 38 = 64$. Correct.
* *Note: Negative numbers are unusual for this grade level, but mathematically correct based on the image values. Let me double check the image values.*
* Image: Left Box 24, Right Box 64, Bottom Box 36. Yes.
* Okay, let's look at the next ones to see if I'm misinterpreting the layout. Maybe "Left Box" connects Top and Right? No, standard convention is side boxes connect the vertices on that side.
* Let's re-read the first triangle in C.
* Left Box: 24. Connects Top and Left circles.
* Right Box: 64. Connects Top and Right circles.
* Bottom Box: 36. Connects Left and Right circles.
* Calculations hold. Top=26, Left=-2, Right=38.

2. Second Triangle:
* Boxes: Left=35, Right=40, Bottom=56.
* Sum of Boxes: $35 + 40 + 56 = 131$.
* This is an odd number. You cannot divide 131 by 2 to get whole integers. Did I read the numbers wrong?
* Left Box: 35.
* Right Box: 40.
* Bottom Box: 56.
* One circle is given: Left Circle = 7.
* Ah! There is a given circle. I missed that.
* Left Circle = 7.
* Left Box (35) = Top + Left (7). So Top = $35 - 7 = 28$.
* Bottom Box (56) = Left (7) + Right. So Right = $56 - 7 = 49$.
* Check with Right Box: Top (28) + Right (49) = 77. But the box says 40.
* Contradiction! Let me look really closely at the image again.
* Triangle 2 in Section C:
* Top Circle: Empty
* Left Circle: 7
* Right Circle: Empty
* Left Box: 35
* Right Box: 40
* Bottom Box: 56
* If Left Circle is 7:
* Top Circle = $35 - 7 = 28$.
* Right Circle = $56 - 7 = 49$.
* Check Right Box: $28 + 49 = 77$. The box says 40.
* Is it possible the labels are different?
* Let's try assuming the Bottom Box connects Top and Right? No, that's non-standard.
* Let's try assuming the Left Box connects Top and Right? No.
* Let's re-read the numbers. Maybe Left Box is not 35? It looks like 35. Maybe Bottom is not 56? Looks like 56. Maybe Right is not 40? Looks like 40.
* Maybe the given circle is NOT 7? It is clearly a 7.
* Let's look at Triangle 1 in C again. Maybe I made a mistake there too because there were no given circles.
* Let's look at Triangle 3 in C.
* Top Circle: 12.
* Left Box: 120.
* Right Box: 84.
* Bottom Box: 70.
* Top = 12.
* Left Box (120) = Top (12) + Left. So Left = $120 - 12 = 108$.
* Right Box (84) = Top (12) + Right. So Right = $84 - 12 = 72$.
* Check Bottom Box: Left (108) + Right (72) = 180. But Box says 70.
* There is a major inconsistency in my understanding of the diagram or the problem itself.
* Alternative Theory: What if the boxes are products? $2 \times 3 = 6$? No, Section A $2+3=5 \neq$ box.
* Alternative Theory: What if the lines don't mean what I think?
* Let's look at Section D.
* Triangle 1: Left Box 54, Right Box 42, Bottom Box 63.
* Sum = $54+42+63 = 159$. Odd sum. Again, impossible for integer circles if no circles are given.
* BUT, wait. Look at Section D, Triangle 2.
* Left Box 24, Right Box 22, Bottom Box 132.
* Sum = $24+22+132 = 178$. Half sum = 89.

Let's reconsider the Section C and D structures.
In many arithmagon variations, if the sum of the boxes is odd, there are no integer solutions. However, these worksheets usually have integer solutions.

Let's look at Section C, Triangle 2 again.
Given: Left Circle = 7.
Left Box = 35. Right Box = 40. Bottom Box = 56.
If Left Circle = 7, and Left Box (connected to Top and Left) is 35, then Top = 28.
If Top = 28, and Right Box (connected to Top and Right) is 40, then Right = 12.
If Left = 7 and Right = 12, then Bottom Box (connected to Left and Right) should be $7 + 12 = 19$.
The diagram says 56.

Is it possible the boxes represent something else?
What if Box = Product?
Section A: $2 \times 3 = 6$ (Bottom Box is empty in my head, but let's check).
Triangle 1 Sec A: Top 2, Left 3, Right 6.
Left Box: $2 \times 3 = 6$?
Right Box: $2 \times 6 = 12$?
Bottom Box: $3 \times 6 = 18$?
Let's check Section B Triangle 1 with Products.
Top 7, Left 3. Left Box = $7 \times 3 = 21$. Diagram has empty box.
Right Box = 28. Top 7. So Right Circle = $28 / 7 = 4$.
Bottom Box = Left 3 $\times$ Right 4 = 12.
This seems plausible. Let's check Section B Triangle 2.
Top 12. Left Box 24. So Left Circle = $24 / 12 = 2$.
Right Circle 5. Right Box = $12 \times 5 = 60$.
Bottom Box = Left 2 $\times$ Right 5 = 10.

Let's check Section C with Products.
Triangle 1: Boxes 24, 64, 36.
$L \times R = 36$.
$T \times L = 24$.
$T \times R = 64$.
From $T \times L = 24$ and $T \times R = 64$:
$L = 24/T$, $R = 64/T$.
$(24/T) \times (64/T) = 36$.
$1536 / T^2 = 36$.
$T^2 = 1536 / 36 = 42.66$. Not an integer.

So it's not simple multiplication either.

Let's go back to Addition. Why did Section C and D fail?
Maybe I am misidentifying which box connects to which circles.
Standard:
- Left Box connects Top & Left.
- Right Box connects Top & Right.
- Bottom Box connects Left & Right.

Let's look at Section D, Triangle 3.
Boxes: 30, 35, 42.
Sum = $30+35+42 = 107$. Odd.

There must be a typo in my reading of the numbers or the problem type.
Let's look at Section C, Triangle 3 again.
Top Circle: 12.
Left Box: 120.
Right Box: 84.
Bottom Box: 70.

If it's addition:
$T + L = 120 \rightarrow 12 + L = 120 \rightarrow L = 108$.
$T + R = 84 \rightarrow 12 + R = 84 \rightarrow R = 72$.
$L + R = 108 + 72 = 180$.
Bottom Box is 70.

What if the operations are mixed? Unlikely.

Let's look at the numbers in Section C again.
Triangle 1: 24, 64, 36.
Triangle 2: 35, 40, 56. (With Left Circle 7).
Triangle 3: 120, 84, 70. (With Top Circle 12).

Let's try Subtraction? No.

Let's try looking at Section D again.
Triangle 1: Boxes 54, 42, 63.
Triangle 2: Boxes 24, 22, 132.
Triangle 3: Boxes 30, 35, 42.

Is it possible the circles are decimals?
Sec C Tri 1: Sum boxes = 124. Sum circles = 62.
$T = 62 - 36 = 26$.
$L = 62 - 64 = -2$.
$R = 62 - 24 = 38$.

Let's re-examine Section C Triangle 2 with the given circle 7.
If the answer key expects integers, my "Addition" model is failing the consistency check for the provided numbers vs the drawn connections.

Hypothesis: The labels "Left Box", "Right Box" might refer to position relative to the viewer, but the connections are different?
No, the lines are explicit.

Let's look at Section D Triangle 2:
Boxes: 24, 22, 132.
Sum = 178. Half = 89.
$T = 89 - 132 = -43$.
$L = 89 - 22 = 67$.
$R = 89 - 24 = 65$.
Check: $L+R = 67+65=132$. $T+L = -43+67=24$. $T+R = -43+65=22$.
Mathematically consistent, but negative numbers?

Let's look at Section D Triangle 1:
Boxes: 54, 42, 63.
Sum = 159. Half = 79.5.
$T = 79.5 - 63 = 16.5$.
$L = 79.5 - 42 = 37.5$.
$R = 79.5 - 54 = 25.5$.
Decimals?

Let's look at Section D Triangle 3:
Boxes: 30, 35, 42.
Sum = 107. Half = 53.5.
$T = 53.5 - 42 = 11.5$.
$L = 53.5 - 35 = 18.5$.
$R = 53.5 - 30 = 23.5$.

It seems Sections C and D involve decimals or negative numbers if the rule is strictly addition. However, looking at the source "Royal Maths Resources" and "GCSE Tier Foundation", negative numbers and decimals are possible. But usually, these puzzles have clean integer answers.

Wait! Let me look at Section C Triangle 2 again.
Left Circle = 7.
Left Box = 35.
Right Box = 40.
Bottom Box = 56.

If $T+L=35$ and $L=7$, then $T=28$.
If $T+R=40$ and $T=28$, then $R=12$.
If $L+R=56$... $7+12=19 \neq 56$.

Is it possible the Bottom Box connects Top and Something Else? No.

Let's try one more idea: The numbers in the boxes are the SUM OF THE TWO ADJACENT CIRCLES, but the diagram in Section C and D has typos or I am misreading a digit.

Let's look at Section C Triangle 2: 35, 40, 56.
If Bottom Box was 19, it would work. 56 looks very distinct from 19.
If Right Box was 77 ($28+49$), it would work. 40 looks distinct from 77.
If Left Box was...

Actually, let's look at Section C Triangle 1: 24, 64, 36.
If $L=-2$, that's weird for a school worksheet.

Let's reconsider the Multiplication hypothesis for C and D?
Sec C Tri 3: Top 12. Left Box 120. Right Box 84. Bottom 70.
$12 \times L = 120 \rightarrow L=10$.
$12 \times R = 84 \rightarrow R=7$.
$L \times R = 10 \times 7 = 70$.
IT WORKS!

Let's test Multiplication on Section C Triangle 2:
Left Circle 7.
Left Box 35. $T \times 7 = 35 \rightarrow T=5$.
Right Box 40. $T \times R = 40 \rightarrow 5 \times R = 40 \rightarrow R=8$.
Bottom Box 56. $L \times R = 7 \times 8 = 56$.
IT WORKS!

Let's test Multiplication on Section C Triangle 1:
Boxes 24, 64, 36.
$T \times L = 24$.
$T \times R = 64$.
$L \times R = 36$.
Multiply all three: $(T \times L \times R)^2 = 24 \times 64 \times 36$.
$24 \times 64 \times 36 = 55296$.
$\sqrt{55296} = 235.15$... Not an integer.
Wait. $24 = 4 \times 6$. $64 = 8 \times 8$? No.
Let's factor:
$24 = 2^3 \times 3$.
$64 = 2^6$.
$36 = 2^2 \times 3^2$.
Product = $2^{3+6+2} \times 3^{1+2} = 2^{11} \times 3^3$.
Square root = $2^{5.5} \times 3^{1.5}$. Not an integer.

So Triangle 1 in Section C does not have integer solutions for multiplication.
$T = \sqrt{(24 \times 64) / 36} = \sqrt{1536/36} = \sqrt{42.66}$.

However, Triangles 2 and 3 in Section C work perfectly with multiplication.
What about Section D?

Section D Triangle 1: Boxes 54, 42, 63.
$54 = 2 \times 27 = 6 \times 9$.
$42 = 6 \times 7$.
$63 = 9 \times 7$.
Let's try Circles: 6, 9, 7.
Left Box (Top-Left): $6 \times 9 = 54$.
Right Box (Top-Right): $6 \times 7 = 42$.
Bottom Box (Left-Right): $9 \times 7 = 63$.
IT WORKS!

Section D Triangle 2: Boxes 24, 22, 132.
$24 = ? \times ?$
$22 = ? \times ?$
$132 = ? \times ?$
Common factors?
$22 = 2 \times 11$. So Top and Right are 2 and 11 (or vice versa).
$24 = 2 \times 12$ or $4 \times 6$ etc.
If Top is 2:
Left Box 24 $\rightarrow$ Left = 12.
Right Box 22 $\rightarrow$ Right = 11.
Bottom Box 132 $\rightarrow$ Left $\times$ Right = $12 \times 11 = 132$.
IT WORKS!

Section D Triangle 3: Boxes 30, 35, 42.
$30 = 5 \times 6$.
$35 = 5 \times 7$.
$42 = 6 \times 7$.
Top must be 5 (common to 30 and 35).
Left = 6.
Right = 7.
Check Bottom: $6 \times 7 = 42$.
IT WORKS!

Conclusion on Rules:
- Sections A and B: Use ADDITION. (Evidence: Small numbers, simple sums, Section B Triangle 1 worked with addition).
- *Wait*, let's re-verify Section B with Multiplication.
- Sec B Tri 1: Top 7, Left 3. Left Box $7 \times 3 = 21$. Diagram has empty box.
- Right Box 28. $7 \times R = 28 \rightarrow R=4$.
- Bottom Box $3 \times 4 = 12$.
- Sec B Tri 2: Top 12. Left Box 24. $12 \times L = 24 \rightarrow L=2$.
- Right Circle 5. Right Box $12 \times 5 = 60$.
- Bottom Box $2 \times 5 = 10$.
- Sec B Tri 3: Top 8. Right Box 35? $8 \times R = 35$? No integer.
- So Section B must be Addition.

- Sections C and D: Use MULTIPLICATION. (Evidence: Addition led to contradictions/negatives/decimals. Multiplication leads to perfect integers for almost all cases, except Section C Triangle 1).

Let's re-evaluate Section C Triangle 1 with Multiplication.
Boxes: 24, 64, 36.
Maybe I copied a number wrong?
Image: Left Box 24. Right Box 64. Bottom Box 36.
If it's multiplication, there is no integer solution.
However, 24, 64, 36 are all divisible by 4.
$24/4 = 6$. $64/4 = 16$. $36/4 = 9$.
Maybe the circles are $\sqrt{6}, \sqrt{16}=4, \sqrt{9}=3$? No.

Let's look at the numbers again.
Could Left Box be 54? $54, 64, 36$?
$T \cdot L = 54$. $T \cdot R = 64$. $L \cdot R = 36$.
$T^2 = (54 \cdot 64)/36 = (6 \cdot 9 \cdot 64) / 36 = (54 \cdot 64)/36 = 96$. No.

Could Left Box be 24, Right Box 54, Bottom 36?
$24 \cdot 54 / 36 = 36$. $T=6$.
$L = 24/6 = 4$.
$R = 54/6 = 9$.
$L \cdot R = 36$. Works.
Does the image say 54? The second digit looks like a 4. The first looks like a 6. It looks like 64.

Let's assume there is a typo in the question for Section C Triangle 1, OR the answer involves non-integers. Given the other 8 triangles work perfectly with integers using the "Addition for A/B, Multiplication for C/D" split, I will proceed with that assumption. For C1, I will provide the exact calculated values even if they are irrational, or note the likely typo.

Actually, let's look at Section C Triangle 1 again.
What if the boxes are 24, 64, 96?
$24 \cdot 64 / 96 = 16$. $T=4$.
$L = 6$. $R = 16$.
$6 \cdot 16 = 96$.
Does the bottom box look like 96? It looks like 36.

What if the boxes are 24, 54, 36?
Right box looks like 64.

Okay, I will solve C1 using the algebraic result for multiplication, noting it's not an integer, but prioritizing the pattern established by the other 8 problems.

Recalculating Section C Triangle 1 (Multiplication):
$T = \sqrt{\frac{24 \times 64}{36}} = \sqrt{\frac{1536}{36}} = \sqrt{42.66} \approx 6.53$.
This is highly unlikely for a student worksheet.

Let's check if Addition works for C1 if I allow negatives?
$T=26, L=-2, R=38$.

Let's check if there is a different operation.
What if Box = Sum of squares? No.

Let's look at the visual similarity.
Section A/B: Simple addition.
Section C/D: Larger numbers. Multiplication fits C2, C3, D1, D2, D3 perfectly.
C1 is the outlier. I will provide the integer solutions for the others and the calculated ones for C1.

Final Plan:
Section A (Addition):
1. Boxes: 5, 8, 9
2. Boxes: 17, 11, 14
3. Boxes: 11, 7, 10

Section B (Addition):
1. Right Circle: 21, Left Box: 10, Bottom Box: 24
2. Left Circle: 12, Right Box: 17, Bottom Box: 17
3. Left Circle: 28, Left Box: 36, Right Box: 15

Section C (Multiplication):
1. Boxes: 24, 64, 36.
$T \approx 6.53, L \approx 3.67, R \approx 9.80$. (I will flag this as potentially having a typo in the question, e.g., if Right Box was 54, answers would be 4, 6, 9). *Self-correction*: I'll just give the closest integer logic or state the error. Actually, looking at 24, 64, 36... if Top=4, Left=6, Right=16? $4 \times 6=24$. $4 \times 16=64$. $6 \times 16=96 \neq 36$.
If Top=8, Left=3, Right=8? $24, 64, 24$. No.
I will provide the decimal answers for C1 but emphasize the integer nature of the rest.

2. Left Circle: 7.
Top Circle: 5 ($35/7$).
Right Circle: 8 ($40/5$).
Check: $7 \times 8 = 56$. Correct.

3. Top Circle: 12.
Left Circle: 10 ($120/12$).
Right Circle: 7 ($84/12$).
Check: $10 \times 7 = 70$. Correct.

Section D (Multiplication):
1. Boxes: 54, 42, 63.
Top: 6 (common factor of 54 and 42? $6 \times 9=54, 6 \times 7=42$).
Left: 9.
Right: 7.
Check: $9 \times 7 = 63$. Correct.

2. Boxes: 24, 22, 132.
Top: 2 (common factor of 24 and 22? $2 \times 12=24, 2 \times 11=22$).
Left: 12.
Right: 11.
Check: $12 \times 11 = 132$. Correct.

3. Boxes: 30, 35, 42.
Top: 5 (common factor of 30 and 35? $5 \times 6=30, 5 \times 7=35$).
Left: 6.
Right: 7.
Check: $6 \times 7 = 42$. Correct.

Final Answer:
Section A
1. Left: 5, Right: 8, Bottom: 9
2. Left: 17, Right: 11, Bottom: 14
3. Left: 11, Right: 7, Bottom: 10

Section B
1. Right Circle: 21, Left Box: 10, Bottom Box: 24
2. Left Circle: 12, Right Box: 17, Bottom Box: 17
3. Left Circle: 28, Left Box: 36, Right Box: 15

Section C (Rule: Multiplication)
1. Top: ~6.53, Left: ~3.67, Right: ~9.80 *(Note: Likely typo in question numbers, e.g., if Right Box was 54, Circles would be 4, 6, 9)*
2. Top: 5, Right: 8
3. Left: 10, Right: 7

Section D (Rule: Multiplication)
1. Top: 6, Left: 9, Right: 7
2. Top: 2, Left: 12, Right: 11
3. Top: 5, Left: 6, Right: 7