Question

create and complete a diagram puzzle that only uses factors of 12 as coefficients and constants.

Explanation:

Step1: List factors of 12

Factors of 12: $1, 2, 3, 4, 6, 12$

Step2: Define puzzle rule

We use a 2x2 area model: outer numbers multiply to inner cell values, where row outer × column outer = inner cell.

Step3: Pick row/column outer values

Choose row outer (left): $2, 3$; column outer (top): $4, 6$

Step4: Calculate inner cells

Top-left: $2 \times 4 = 8$ (invalid, re-pick column: $3,4$)
Top-left: $2 \times 3 = 6$, Top-right: $2 \times 4 = 8$ (invalid, re-pick row: $3,4$; column: $2,3$)
Top-left: $3 \times 2 = 6$, Top-right: $3 \times 3 = 9$ (invalid, re-pick column: $2,6$)
Top-left: $3 \times 2 = 6$, Top-right: $3 \times 6 = 18$ (invalid, re-pick row: $2,6$; column: $3,4$)
Top-left: $2 \times 3 = 6$, Top-right: $2 \times 4 = 8$ (invalid, re-pick column: $1,6$)
Top-left: $2 \times 1 = 2$, Top-right: $2 \times 6 = 12$
Bottom-left: $6 \times 1 = 6$, Bottom-right: $6 \times 6 = 36$ (invalid, re-pick bottom row: $3$)
Bottom-left: $3 \times 1 = 3$, Bottom-right: $3 \times 6 = 18$ (invalid, re-pick column: $2,3$)
Top-left: $2 \times 2 = 4$, Top-right: $2 \times 3 = 6$
Bottom-left: $3 \times 2 = 6$, Bottom-right: $3 \times 3 = 9$ (invalid, re-pick row: $4,3$; column: $2,3$)
Top-left: $4 \times 2 = 8$ (invalid, re-pick row: $1,2$; column: $6,3$)
Top-left: $1 \times 6 = 6$, Top-right: $1 \times 3 = 3$
Bottom-left: $2 \times 6 = 12$, Bottom-right: $2 \times 3 = 6$ (all values are factors of 12 or products that are factors? No, rule: coefficients/constants are factors, so inner cells can be products, outer are factors. Final valid set:
Left outer (rows): $3, 4$; Top outer (columns): $2, 3$
Top-left: $3 \times 2 = 6$, Top-right: $3 \times 3 = 9$ (invalid, re-pick column: $2, 6$)
Top-left: $3 \times 2 = 6$, Top-right: $3 \times 6 = 18$ (invalid, re-pick rule: inner cells are sums of row+column outer? No, standard area model: outer multiply to inner, inner must be factors? No, problem says "uses factors of 12 as coefficients and constants" so outer and inner are factors. So we need row × column = factor of 12.
Valid pairs (row, column):
$(1,1)=1$, $(1,2)=2$, $(1,3)=3$, $(1,4)=4$, $(1,6)=6$, $(1,12)=12$
$(2,1)=2$, $(2,2)=4$, $(2,3)=6$, $(2,6)=12$
$(3,1)=3$, $(3,2)=6$, $(3,4)=12$
$(4,1)=4$, $(4,3)=12$
$(6,1)=6$, $(6,2)=12$
$(12,1)=12$

Step5: Fill the puzzle

Left outer (top to bottom): $2, 3$
Top outer (left to right): $3, 4$
Inner top-left: $2 \times 3 = 6$, Inner top-right: $2 \times 4 = 8$ (invalid, re-pick column: $3, 2$)
Inner top-left: $2 \times 3 = 6$, Inner top-right: $2 \times 2 = 4$
Inner bottom-left: $3 \times 3 = 9$ (invalid, re-pick bottom row: $6$)
Inner bottom-left: $6 \times 3 = 18$ (invalid, re-pick row: $1, 6$; column: $2, 3$)
Inner top-left: $1 \times 2 = 2$, Inner top-right: $1 \times 3 = 3$
Inner bottom-left: $6 \times 2 = 12$, Inner bottom-right: $6 \times 3 = 18$ (invalid, re-pick column: $2, 6$)
Inner bottom-right: $6 \times 6 = 36$ (invalid, re-pick bottom row: $4$)
Inner bottom-left: $4 \times 2 = 8$ (invalid, re-pick row: $3, 4$; column: $2, 1$)
Inner top-left: $3 \times 2 = 6$, Inner top-right: $3 \times 1 = 3$
Inner bottom-left: $4 \times 2 = 8$ (invalid, re-pick row: $2, 6$; column: $1, 3$)
Inner top-left: $2 \times 1 = 2$, Inner top-right: $2 \times 3 = 6$
Inner bottom-left: $6 \times 1 = 6$, Inner bottom-right: $6 \times 3 = 18$ (invalid, re-pick column: $1, 2$)
Inner bottom-right: $6 \times 2 = 12$ (valid, all values are factors of 12)

Answer:

Left Outer (Rows)	Top Outer (Columns)	Inner Cells (2x2)
6 (bottom)	2 (right)	Top-right: 4
		Bottom-left: 6
		Bottom-right: 12

Visual layout:
Left side (top to bottom): $\boldsymbol{2}$, $\boldsymbol{6}$
Top side (left to right): $\boldsymbol{1}$, $\boldsymbol{2}$
Inner grid:
Top-left: $\boldsymbol{2}$, Top-right: $\boldsymbol{4}$
Bottom-left: $\boldsymbol{6}$, Bottom-right: $\boldsymbol{12}$