Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
| Place value | thousands | hundreds | tens | ones |
| 3 | ||||
| × | 7 | 5 | 0 | |
2. Multiply the numbers using long multiplication method
Because the ones digit of the multiplicator equals 0, skip to the next digit.
Proceed by multiplying the tens digit (5) of the multiplier (750) by each digit of the multiplicand (3), from right to left.
Because digit (5) is in tens place, we shift partial result by 1 place(s) by placing 1 zero(s).
| Place value | thousands | hundreds | tens | ones |
| 3 | ||||
| × | 7 | 5 | 0 | |
| 0 | ||||
Multiply the tens digit (5) of the multiplicator by the number in the ones place value:
5×3=15
Write 5 in the tens place.
Because the result is greater than 9, carry the 1 to the hundreds place.
| Place value | thousands | hundreds | tens | ones |
| 1 | ||||
| 3 | ||||
| × | 7 | 5 | 0 | |
| 1 | 5 | 0 | ||
150 is the first partial product.
Proceed by multiplying the hundreds digit (7) of the multiplier (750) by each digit of the multiplicand (3), from right to left.
Because digit (7) is in hundreds place, we shift partial result by 2 place(s) by placing 2 zero(s).
| Place value | thousands | hundreds | tens | ones |
| 3 | ||||
| × | 7 | 5 | 0 | |
| 1 | 5 | 0 | ||
| 0 | 0 |
Multiply the hundreds digit (7) of the multiplicator by the number in the ones place value:
7×3=21
Write 1 in the hundreds place.
Because the result is greater than 9, carry the 2 to the thousands place.
| Place value | thousands | hundreds | tens | ones |
| 2 | ||||
| 3 | ||||
| × | 7 | 5 | 0 | |
| 1 | 5 | 0 | ||
| 2 | 1 | 0 | 0 |
2,100 is the second partial product.
3. Add the partial products
150+2100=2250 long addition steps can be seen here
| Place value | thousands | hundreds | tens | ones |
| 3 | ||||
| × | 7 | 5 | 0 | |
| 1 | 5 | 0 | ||
| + | 2 | 1 | 0 | 0 |
| 2 | 2 | 5 | 0 |
The solution is: 2,250
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