Multiply 99_{10} times 99_{10} :
`9801`

How many decimal places does each operand (99) take:
`2`

.

How many decimal places does the result take:
`4`

.

10110111 B7 183 10100010 A2 162 ———————— —— ——— 00000000 16E 366 10110111. 726. 1098 00000000.. 183 00000000... 00000000.... 10110111..... 00000000...... 10110111....... ———————————————— ———— ————— 111001111001110 73CE 29646

The product of two N-place decimal integers may need 2N places. This is true for numbers expressed in any base. In particular, the product of two integers expressed with N-bit binary may need 2N bits. For example, in the picture, two 8-bit unsigned integers are multiplied using the usual paper-and-pencil multiplication algorithm (but using binary arithmetic).

The two 8-bit operands result in a 15-bit product. Also shown is the same product done with base 16 and base 10 notation.

Is a 32-bit general register always able to hold the result of multiplying two 32-bit integers?