Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
| Place value | thousands | hundreds | tens | ones | . | tenths | hundredths | thousandths | ten thousandths | hundred thousandths | millionths | ten millionths | hundred millionths | billionths | ten billionths | TABLE_COL_DECIMAL_DIGIT_PLACE11 |
| 0 | . | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | ||||||||||||
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 11 decimal place(s). So once calculated, the result will be reduced by the factor of 100,000,000,000.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
2. Multiply the numbers using long multiplication method
Because the tens digit of the multiplicator equals 0, skip to the next digit.
Proceed by multiplying the hundreds digit (5) of the multiplier (1,500) by each digit of the multiplicand (27), from right to left.
Because digit (5) 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 |
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
| 0 | 0 | ||||
Multiply the hundreds digit (5) of the multiplicator by the number in the ones place value:
5×7=35
Write 5 in the hundreds place.
Because the result is greater than 9, carry the 3 to the thousands place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 3 | |||||
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
| 5 | 0 | 0 | |||
Multiply the hundreds digit (5) of the multiplicator by the number in the tens place value and add the carried number (3):
5×2+3=13
Write 3 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 | 3 | ||||
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
| 1 | 3 | 5 | 0 | 0 | |
13,500 is the first partial product.
Proceed by multiplying the thousands digit (1) of the multiplier (1,500) by each digit of the multiplicand (27), from right to left.
Because digit (1) is in thousands place, we shift partial result by 3 place(s) by placing 3 zero(s).
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
| 1 | 3 | 5 | 0 | 0 | |
| 0 | 0 | 0 |
Multiply the thousands digit (1) of the multiplicator by the number in the ones place value:
1×7=7
Write 7 in the thousands place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
| 1 | 3 | 5 | 0 | 0 | |
| 7 | 0 | 0 | 0 |
Multiply the thousands digit (1) of the multiplicator by the number in the tens place value:
1×2=2
Write 2 in the ten thousands place.
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
| 1 | 3 | 5 | 0 | 0 | |
| 2 | 7 | 0 | 0 | 0 |
27,000 is the second partial product.
3. Add the partial products
13500+27000=40500 long addition steps can be seen here
| Place value | ten thousands | thousands | hundreds | tens | ones |
| 2 | 7 | ||||
| × | 1 | 5 | 0 | 0 | |
| 1 | 3 | 5 | 0 | 0 | |
| + | 2 | 7 | 0 | 0 | 0 |
| 4 | 0 | 5 | 0 | 0 |
Because we have 11 digit(s) to the right of the decimal point in the numbers that are being multiplied, we move the decimal point 11 time(s) to the left (reducing the result by the factor of 100,000,000,000) to get the final result:
The solution is: 0.000000405
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