Solution
- Before we begin solving, let us first know what the result of the original product is
[tex]253\times31=7843[/tex]
- Now, let us analyze what was done on both diagrams.
Diagram A:
- In this diagram, 253 was divided into 3 numbers and 31 was divided into two numbers.
- We can visualize this better if we wrote them in the product form as shown above:
[tex]\begin{gathered} 253\times31 \\ 253=200+50+3 \\ 31=30+1 \\ \\ \therefore253\times31=(200+50+3)\times(30+1) \\ \text{ Expanding the brackets, we have:} \\ \\ 253\times31=200(30)+200(1)+50(30)+50(1)+3(30)+3(1) \\ \\ \text{Now, notice that} \\ 200(30)\text{ is associated with the first box at the top left of the entire diagram,} \\ \text{where 200 and 30 me}et \\ \\ 200(1)\text{ depicts where 200 and 1 me}et\text{ on the diagram} \\ \\ 50(30)\text{ depicts where 50 and 30 me}et\text{ on the diagram} \\ \\ \text{And so on}\ldots \end{gathered}[/tex]
- We can therefore see that the empty boxes simply represent the product of these numbers stated above.
- Thus, we can easily populate the boxes if we just perform the products associated with them.
- That is:
- Adding up all the numbers in all 6 squares, we have
[tex]6000+1500+90+200+50+3=7843[/tex]
- This is the same result as the original product
Diagram B:
- For this diagram, we take a similar approach
[tex]\begin{gathered} 253\times31=253\times(30+1) \\ =253(30)+253(1) \end{gathered}[/tex]
- Placing these products in their respective boxes, we have:
- Again, Adding up these numbers in the boxes we have:
[tex]7590+253=7843[/tex]
