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 |
| 1 | 4 | 4 | ||||
| × | 0 | . | 3 | 5 | ||
| . |
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 2 decimal place(s). So once calculated, the result will be reduced by the factor of 100.
| Place value | thousands | hundreds | tens | ones |
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (5) of the multiplier 35 by each digit of the multiplicand 144, from right to left.
Multiply the ones digit (5) of the multiplicator by the number in the ones place value:
5×4=20
Write 0 in the ones place.
Because the result is greater than 9, carry the 2 to the tens place.
| Place value | thousands | hundreds | tens | ones |
| 2 | ||||
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 0 | ||||
Multiply the ones digit (5) of the multiplicator by the number in the tens place value and add the carried number (2):
5×4+2=22
Write 2 in the tens place.
Because the result is greater than 9, carry the 2 to the hundreds place.
| Place value | thousands | hundreds | tens | ones |
| 2 | 2 | |||
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 2 | 0 | |||
Multiply the ones digit (5) of the multiplicator by the number in the hundreds place value and add the carried number (2):
5×1+2=7
Write 7 in the hundreds place.
| Place value | thousands | hundreds | tens | ones |
| 2 | 2 | |||
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 7 | 2 | 0 | ||
720 is the first partial product.
Proceed by multiplying the tens digit (3) of the multiplier (35) by each digit of the multiplicand (144), 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 | thousands | hundreds | tens | ones |
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 7 | 2 | 0 | ||
| 0 |
Multiply the tens digit (3) of the multiplicator by the number in the ones place value:
3×4=12
Write 2 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 | ||||
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 7 | 2 | 0 | ||
| 2 | 0 |
Multiply the tens digit (3) of the multiplicator by the number in the tens place value and add the carried number (1):
3×4+1=13
Write 3 in the hundreds place.
Because the result is greater than 9, carry the 1 to the thousands place.
| Place value | thousands | hundreds | tens | ones |
| 1 | 1 | |||
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 7 | 2 | 0 | ||
| 3 | 2 | 0 |
Multiply the tens digit (3) of the multiplicator by the number in the hundreds place value and add the carried number (1):
3×1+1=4
Write 4 in the thousands place.
| Place value | thousands | hundreds | tens | ones |
| 1 | 1 | |||
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 7 | 2 | 0 | ||
| 4 | 3 | 2 | 0 |
4,320 is the second partial product.
3. Add the partial products
720+4320=5040 long addition steps can be seen here
| Place value | thousands | hundreds | tens | ones |
| 1 | 4 | 4 | ||
| × | 3 | 5 | ||
| 7 | 2 | 0 | ||
| + | 4 | 3 | 2 | 0 |
| 5 | 0 | 4 | 0 |
Because we have 2 digit(s) to the right of the decimal point in the numbers that are being multiplied, we move the decimal point 2 time(s) to the left (reducing the result by the factor of 100) to get the final result:
The solution is: 50.4
How did we do?
Please leave us feedback.