Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
Place value | ten thousands | thousands | hundreds | tens | ones |
4 | 7 | 2 | |||
× | 3 | 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 (3) of the multiplier (30) by each digit of the multiplicand (472), from right to left.
Because digit (3) is in tens place, we shift partial result by 1 place(s) by placing 1 zero(s).
Place value | ten thousands | thousands | hundreds | tens | ones |
4 | 7 | 2 | |||
× | 3 | 0 | |||
0 |
Multiply the tens digit (3) of the multiplicator by the number in the ones place value:
3×2=6
Write 6 in the tens place.
Place value | ten thousands | thousands | hundreds | tens | ones |
4 | 7 | 2 | |||
× | 3 | 0 | |||
6 | 0 |
Multiply the tens digit (3) of the multiplicator by the number in the tens place value:
3×7=21
Write 1 in the hundreds place.
Because the result is greater than 9, carry the 2 to the thousands place.
Place value | ten thousands | thousands | hundreds | tens | ones |
2 | |||||
4 | 7 | 2 | |||
× | 3 | 0 | |||
1 | 6 | 0 |
Multiply the tens digit (3) of the multiplicator by the number in the hundreds place value and add the carried number (2):
3×4+2=14
Write 4 in the thousands place.
Because the result is greater than 9, carry the 1 to the ten thousands place.
Place value | ten thousands | thousands | hundreds | tens | ones |
1 | 2 | ||||
4 | 7 | 2 | |||
× | 3 | 0 | |||
1 | 4 | 1 | 6 | 0 |
14,160 is the first partial product.
3. Add the partial products
14160=14160 long addition steps can be seen here
Place value | ten thousands | thousands | hundreds | tens | ones |
4 | 7 | 2 | |||
× | 3 | 0 | |||
+ | 1 | 4 | 1 | 6 | 0 |
1 | 4 | 1 | 6 | 0 |
The solution is: 14,160
How did we do?
Please leave us feedback.