Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
| Place value | hundreds | tens | ones |
| 3 | 3 | ||
| × | 2 | 3 | |
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (3) of the multiplier 23 by each digit of the multiplicand 33, from right to left.
Multiply the ones digit (3) of the multiplicator by the number in the ones place value:
3×3=9
Write 9 in the ones place.
| Place value | hundreds | tens | ones |
| 3 | 3 | ||
| × | 2 | 3 | |
| 9 | |||
Multiply the ones digit (3) of the multiplicator by the number in the tens place value:
3×3=9
Write 9 in the tens place.
| Place value | hundreds | tens | ones |
| 3 | 3 | ||
| × | 2 | 3 | |
| 9 | 9 | ||
99 is the first partial product.
Proceed by multiplying the tens digit (2) of the multiplier (23) by each digit of the multiplicand (33), from right to left.
Because digit (2) is in tens place, we shift partial result by 1 place(s) by placing 1 zero(s).
| Place value | hundreds | tens | ones |
| 3 | 3 | ||
| × | 2 | 3 | |
| 9 | 9 | ||
| 0 |
Multiply the tens digit (2) of the multiplicator by the number in the ones place value:
2×3=6
Write 6 in the tens place.
| Place value | hundreds | tens | ones |
| 3 | 3 | ||
| × | 2 | 3 | |
| 9 | 9 | ||
| 6 | 0 |
Multiply the tens digit (2) of the multiplicator by the number in the tens place value:
2×3=6
Write 6 in the hundreds place.
| Place value | hundreds | tens | ones |
| 3 | 3 | ||
| × | 2 | 3 | |
| 9 | 9 | ||
| 6 | 6 | 0 |
660 is the second partial product.
3. Add the partial products
99+660=759 long addition steps can be seen here
| Place value | hundreds | tens | ones |
| 3 | 3 | ||
| × | 2 | 3 | |
| 9 | 9 | ||
| + | 6 | 6 | 0 |
| 7 | 5 | 9 |
The solution is: 759
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