3. Final Assembly

Since the condition is met, the numbers are assembled left-to-right:

Final Product: 2,376

Comparison: Handling the “Carry” Scenario

As you noted, if the sum of the internal digits (d1+d2) had been ≥10, a 1 would be carried over to the Left Segment.

• If Sum < 10: Result = [Left][Sum][Right]
• If Sum > 10: Result = [Left+1][Sum mod 10][Right]

In the “Carrying Scenario,” the mental load increases slightly because a carry-over digit must be added to the leftmost segment of your product. This happens when the internal sum of the partial products equals or exceeds 10.

The “Carrying” Scenario

Handling Positional Overlap in Mental Addition

When the sum of the internal digits—specifically the tens-value of the Units product and the units-value of the Tens product—is ≥10, a value of 1 must be carried over to the hundreds/thousands segment.

Case Study: 487×7

1. Identify Positional Products

• Units: 7×7=49 (d1=4)
• Tens: 8×7=56 (d2=6)
• Hundreds: 4×7=28.

2. The Carry Verification

• Sum Check (d1 + d2): 4+6=10.
• Condition: Since 10≥10, this is a Carrying Scenario.

3. Final Assembly with Carry-Over

In this scenario, we must increment the Left Segment by 1 and use only the remainder of the internal sum for the middle position.

Final Product: 3,409

Comparison of Logic Flow

The difference between the two scenarios can be summarized by how you handle the “Middle Bin”:

• No Carry (Sum<10): [(Hundreds×n)+Tens(Tens×n)][Sum][Units(Units×n)]
• With Carry (Sum≥10):
   [(Hundreds×n)+Tens(Tens×n)+1][Sum−10][Units(Units×n)]

Technical Analysis of the Algorithm

The algorithm works because it treats a 3-digit number ABC as (100A+10B+C). When multiplied by n, the product is:

n(100A+10B+C)=100(nA)+10(nB)+(nC)

1. Strengths

• Left-to-Right Processing: Unlike the standard school algorithm (right-to-left), yours allows the "Human Calculator" to begin speaking the answer while still calculating the end.
• Chunking: You break the 12-digit potential product into three distinct mental "bins":
  1. Bin 1 (Thousands/Hundreds): nA+tens digit of nB
  2. Bin 2 (Tens): units digit of nB+tens digit of nC
  3. Bin 3 (Units): units digit of nC

2. The Critical "Carry" Logic

Your "No Carrying" rule (d1+d2<10) is a verification of Bin 2.

• If the sum is <10, Bin 1 is "locked" and won't change.
• If the sum is ≥10, a value of 1 must ripple back to Bin 1.

3. Example Verification

Problem: 297×8

• Partial Products: 16 (Hundreds), 72 (Tens), 56 (Units).
  1. Left Segment: 16+7=23
  2. Internal Sum: 2+5=7 (Condition: 7<10, so "No Carry")
  3. Right Digit: 6
• Result: 2376. (Correct)

Summary of the algorithm’s logic flow:

Where PH is the Hundreds product, PT1/PT2 are the Tens/Units of the Tens product, and PU1/PU2 are the Tens/Units of the Units product.