Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
| Place value | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (9) of the multiplier 709 by each digit of the multiplicand 503, from right to left.
Multiply the ones digit (9) of the multiplicator by the number in the ones place value:
9×3=27
Write 7 in the ones place.
Because the result is greater than 9, carry the 2 to the tens place.
| Place value | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 2 | ||||||
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 7 | ||||||
Multiply the ones digit (9) of the multiplicator by the number in the tens place value and add the carried number (2):
9×0+2=2
Write 2 in the tens place.
| Place value | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 2 | ||||||
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 2 | 7 | |||||
Multiply the ones digit (9) of the multiplicator by the number in the hundreds place value:
9×5=45
Write 5 in the hundreds place.
Because the result is greater than 9, carry the 4 to the thousands place.
| Place value | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 4 | 2 | |||||
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 4 | 5 | 2 | 7 | |||
4,527 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 (7) of the multiplier (709) by each digit of the multiplicand (503), 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 | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 4 | 5 | 2 | 7 | |||
| 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 | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 2 | ||||||
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 4 | 5 | 2 | 7 | |||
| 1 | 0 | 0 |
Multiply the hundreds digit (7) of the multiplicator by the number in the tens place value and add the carried number (2):
7×0+2=2
Write 2 in the thousands place.
| Place value | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 2 | ||||||
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 4 | 5 | 2 | 7 | |||
| 2 | 1 | 0 | 0 |
Multiply the hundreds digit (7) of the multiplicator by the number in the hundreds place value:
7×5=35
Write 5 in the ten thousands place.
Because the result is greater than 9, carry the 3 to the hundred thousands place.
| Place value | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 3 | 2 | |||||
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 4 | 5 | 2 | 7 | |||
| 3 | 5 | 2 | 1 | 0 | 0 |
352,100 is the second partial product.
3. Add the partial products
4527+352100=356627 long addition steps can be seen here
| Place value | hundred thousands | ten thousands | thousands | hundreds | tens | ones |
| 5 | 0 | 3 | ||||
| × | 7 | 0 | 9 | |||
| 4 | 5 | 2 | 7 | |||
| + | 3 | 5 | 2 | 1 | 0 | 0 |
| 3 | 5 | 6 | 6 | 2 | 7 |
The solution is: 356,627
How did we do?
Please leave us feedback.