Overview of Cross-Multiplication

Here is a step-by-step description of how to do cross-multiplication, illustrated with a concrete example of a five-digit by five-digit multiplication.

We will use:

12345 × 67890

The same method applies to any pair of numbers.

Why This Works

Cross-multiplication is simply a structured application of the distributive property.

Using the standard school method to solve
123 × 456, we write:

(100 × 456) + (20 × 456) + (3 × 456)

When we look carefully at how digits contribute to each place value in the final answer, we observe the following:

• Ones digit
  The ones digit of the product depends only on the ones digits of the two factors.
  In this case, it is the ones digit of 3 × 6.
  No other products affect the ones digit.
  The tens digit of 3 × 6 is carried forward.

• Tens digit
  The tens digit depends only on:

  • the tens digit of the first number times the ones digit of the second, and

  • the ones digit of the first number times the tens digit of the second,
    plus any carry from the previous step.

• Hundreds digit
  The hundreds digit depends only on:

  • the hundreds digit of the first number × ones digit of the second,

  • the tens digits multiplied together,

  • the ones digit of the first number × the hundreds digit of the second,
    plus any carry.

At each step, only a specific set of digit pairs contributes to that place value.

Cross-multiplication organizes these calculations by grouping together all products that affect the same digit of the final answer. Each step forms a “cross” of digits.

1. Set Up the Numbers

Write the numbers on top of each other, as in long multiplication:

    12345
× 67890

Each digit in the top number will multiply with each digit in the bottom number. Instead of working column-by-column, we will work diagonal-by-diagonal, grouping digits by place value.

2. Understand the Diagonals

Each diagonal corresponds to a specific place value in the final answer.

• Units diagonal
  Rightmost digit × rightmost digit

• Tens diagonal
  Digit pairs whose positions add to 1 digit from the right

• Hundreds diagonal
  Digit pairs whose positions add to 2 digits from the right

• And so on.

3. Units Diagonal

Multiply the rightmost digits:

5 × 0 = 0

This gives the units digit of the answer.

4. Tens Diagonal

The cross-terms are:

• 4 × 0

• 5 × 9

Add them:

0 + 45 = 45

Write down 5 in the tens place and carry 4.

5. Hundreds Diagonal

Now include all digit pairs whose positions add to two digits from the right:

• 3 × 0

• 4 × 9

• 5 × 8

Add them:

0 + 36 + 40 = 76
Add the carry: 76 + 4 = 80

Write down 0, carry 8.

6. Continue Through the Diagonals

Each step works the same way:

• list all digit pairs in the diagonal,

• multiply,

• add,

• include any carry.

Thousands diagonal:

• 2 × 0

• 3 × 9

• 4 × 8

• 5 × 7

Add them:

0 + 27 + 32 + 35 = 94
Add carry 8 → 102

Write down 2, carry 10.

7. Finish the Remaining Diagonals

Once you reach the leftmost digits of the factors, the diagonals begin shrinking:

• first 5×5

• then 4×4

• then 3×3

• then 2×2

• finally 1×1

Continue until all diagonals are complete.

Variant #1: Left-to-Right Cross-Multiplication

The same diagonal idea can be applied left to right.

This approach is slightly slower, but it has two advantages:

• it allows early estimation of the answer, and

• it is often easier to remember intermediate results.

Using the same example:

  12345
× 67890

The initial steps proceed as follows:

• 1 × 6 = 6
  Add a zero → 60

• 1 × 7 + 2 × 6 = 19
  Add to running total → 79
  Add a zero → 790

• 1 × 8 + 2 × 7 + 3 × 6 = 40
  Add to running total → 830
  Add a zero → 8300

This process continues until all diagonals are accounted for.

Variant #2: Two-Digit Blocks

Cross-multiplication can also be applied using multi-digit blocks, such as two-digit chunks.

When using two-digit blocks:

• each step adds two zeros instead of one, and

• the calculation load increases significantly.

If the total number of digits is odd, you may:

• add a leading zero, or

• add a trailing zero and remove it at the end.

This variant is powerful but requires substantial practice to execute reliably under time pressure.