Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
| Place value | hundreds | tens | ones |
| 9 | |||
| × | 1 | 0 | 7 |
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (7) of the multiplier 107 by each digit of the multiplicand 9, from right to left.
Multiply the ones digit (7) of the multiplicator by the number in the ones place value:
7×9=63
Write 3 in the ones place.
Because the result is greater than 9, carry the 6 to the tens place.
| Place value | hundreds | tens | ones |
| 6 | |||
| 9 | |||
| × | 1 | 0 | 7 |
| 6 | 3 | ||
63 is the first partial product.
Because the tens digit of the multiplicator equals 0, skip to the next digit.
Proceed by multiplying the hundreds digit (1) of the multiplier (107) by each digit of the multiplicand (9), from right to left.
Because digit (1) is in hundreds place, we shift partial result by 2 place(s) by placing 2 zero(s).
| Place value | hundreds | tens | ones |
| 9 | |||
| × | 1 | 0 | 7 |
| 6 | 3 | ||
| 0 | 0 |
Multiply the hundreds digit (1) of the multiplicator by the number in the ones place value:
1×9=9
Write 9 in the hundreds place.
| Place value | hundreds | tens | ones |
| 9 | |||
| × | 1 | 0 | 7 |
| 6 | 3 | ||
| 9 | 0 | 0 |
900 is the second partial product.
3. Add the partial products
63+900=963 long addition steps can be seen here
| Place value | hundreds | tens | ones |
| 9 | |||
| × | 1 | 0 | 7 |
| 6 | 3 | ||
| + | 9 | 0 | 0 |
| 9 | 6 | 3 |
The solution is: 963
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