In this article, we would learn a simple technique of multiplying two 2-digit numbers and two 3-digit numbers in a much faster way.

Let the 2 numbers be AB and CD.

Their product would be calculated as follows:

AB

__CD__

Step 1: BD (Write only the unit’s digit and carry the rest to the next step)

Step 2: AD + BC + carry over (Cross multiply and add, write a single digit and carry rest to next step)

Step 3: AC + carry over (Write the complete number as this is the final step)

35

__62__

Step 1: 2×5 = 10 (Write 0 and 1 is carried forward to the next step)

Step 2: 2×3 + 6×5 + 1(from carry over) = 37 (Write 7 and 3 is carried over to next step)

Step 3: 6×3 + 3 (from carry over) = 21 (Write 21 as this is the final step)

Thus, the answer is 2170.

Let the 2 numbers be ABC and DEF.

Their product would be calculated as follows:

ABC

__DEF__

Step 1: CF (Write only the unit’s digit and carry the rest to the next step)

Step 2: BF + CE + carried over (Write only the unit’s digit and carry the rest to the next step)

Step 3: AF + CD + BE + carried over (Write only the unit’s digit and carry the rest to the next step)

Step 4: AE + BD + carried over (Write only the unit’s digit and carry the rest to the next step)

Step 5: AD + carried over (Write the complete number as this is the last step)

267

__376__

Step 1: 6×7 = 42 (Write 2 and 4 is carry 4 to next step)

Step 2: 6×6 + 7×7 + 4 = 89 (Write 9 and carry 8 to next step)

Step 3: 6×2 + 3×7 + 6×7 + 8 = 83 (Write 3 and carry 8 to next step)

Step 4: 3×6 + 7×2 + 8 = 40 (Write 0 and carry 4 to next step)

Step 5: 3×2 + 4 = 10 (Write 10 as this is the final step)

Thus the answer is 100392.

This way you can save time in actual exam and become more efficient.