Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
| Place value | hundreds | tens | ones | . | tenths | hundredths | thousandths |
| 0 | . | 0 | 6 | 1 | |||
| × | 3 | 9 | 7 | ||||
Ignore the decimal points and multiply as if these are whole numbers (as if each most right digit is the ones digit):
In this case we removed 3 decimal place(s). So once calculated, the result will be reduced by the factor of 1,000.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (7) of the multiplier 397 by each digit of the multiplicand 61, from right to left.
Multiply the ones digit (7) of the multiplicator by the number in the ones place value:
7×1=7
Write 7 in the ones place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 7 | |||||
Multiply the ones digit (7) of the multiplicator by the number in the tens place value:
7×6=42
Write 2 in the tens place.
Because the result is greater than 9, carry the 4 to the hundreds place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 4 | |||||
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
427 is the first partial product.
Proceed by multiplying the tens digit (9) of the multiplier (397) by each digit of the multiplicand (61), from right to left.
Because digit (9) 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 |
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
| 0 | |||||
Multiply the tens digit (9) of the multiplicator by the number in the ones place value:
9×1=9
Write 9 in the tens place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
| 9 | 0 | ||||
Multiply the tens digit (9) of the multiplicator by the number in the tens place value:
9×6=54
Write 4 in the hundreds place.
Because the result is greater than 9, carry the 5 to the thousands place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 5 | |||||
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
| 5 | 4 | 9 | 0 | ||
5,490 is the second partial product.
Proceed by multiplying the hundreds digit (3) of the multiplier (397) by each digit of the multiplicand (61), from right to left.
Because digit (3) is in hundreds place, we shift partial result by 2 place(s) by placing 2 zero(s).
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
| 5 | 4 | 9 | 0 | ||
| 0 | 0 |
Multiply the hundreds digit (3) of the multiplicator by the number in the ones place value:
3×1=3
Write 3 in the hundreds place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
| 5 | 4 | 9 | 0 | ||
| 3 | 0 | 0 |
Multiply the hundreds digit (3) of the multiplicator by the number in the tens place value:
3×6=18
Write 8 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 | |||||
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
| 5 | 4 | 9 | 0 | ||
| 1 | 8 | 3 | 0 | 0 |
18,300 is the third partial product.
3. Add the partial products
427+5490+18300=24217 long addition steps can be seen here
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 6 | 1 | ||||
| × | 3 | 9 | 7 | ||
| 4 | 2 | 7 | |||
| 5 | 4 | 9 | 0 | ||
| + | 1 | 8 | 3 | 0 | 0 |
| 2 | 4 | 2 | 1 | 7 |
Because we have 3 digit(s) to the right of the decimal point in the numbers that are being multiplied, we move the decimal point 3 time(s) to the left (reducing the result by the factor of 1,000) to get the final result:
The solution is: 24.217
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