Solution - Long multiplication
Step-by-step explanation
1. Rewrite the numbers from top to bottom aligned to the right
Place value | ones | . | tenths | hundredths | thousandths | ten thousandths | hundred thousandths | millionths |
0 | . | 0 | 0 | 0 | 2 | 4 | 3 | |
× | 0 | . | 0 | 0 | 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 9 decimal place(s). So once calculated, the result will be reduced by the factor of 1,000,000,000.
Place value | thousands | hundreds | tens | ones |
2 | 4 | 3 | ||
× | 5 | |||
2. Multiply the numbers using long multiplication method
Start by multiplying the ones digit (5) of the multiplier 5 by each digit of the multiplicand 243, from right to left.
Multiply the ones digit (5) of the multiplicator by the number in the ones place value:
5×3=15
Write 5 in the ones place.
Because the result is greater than 9, carry the 1 to the tens place.
Place value | thousands | hundreds | tens | ones |
1 | ||||
2 | 4 | 3 | ||
× | 5 | |||
5 |
Multiply the ones digit (5) of the multiplicator by the number in the tens place value and add the carried number (1):
5×4+1=21
Write 1 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 | 1 | |||
2 | 4 | 3 | ||
× | 5 | |||
1 | 5 |
3. Add the partial products
Multiply the ones digit (5) of the multiplicator by the number in the hundreds place value and add the carried number (2):
5×2+2=12
Write 2 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 | 2 | 1 | ||
2 | 4 | 3 | ||
× | 5 | |||
1 | 2 | 1 | 5 |
Because we have 9 digit(s) to the right of the decimal point in the numbers that are being multiplied, we move the decimal point 9 time(s) to the left (reducing the result by the factor of 1,000,000,000) to get the final result:
The solution is: 0.000001215
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