Binary to BCD question

[bobby19]

This question has really got me stumped.

Consider a ROM eith 7 address(input) lines and 8 data (output) lines. How many bits fro the ROM need to be used to implement a binary to two-digit BCD converter?

The answer is 800bits.


All I can think of is the capcity of the ROM is 2^7 x 8 = 1024 bits, so I know the answer must be less than that. Other than that, I am clueless here. Any insight?

Thanks

 

[Project_maker]

A binary to two-digit BCD converter
Input = 7 bit,
Binary count = 00000000-1100011
Output Required = 4Bit + 4Bit (=8 bits for two digit)
Output count = 00,01,02,03,.....99
Bits need = 100*8 = 800 bits
Use ROM is 2^7 x 8 = 1024 bits,
Only implement in 800 bits and 240 bits is unusable (28 Address)

 

[bobby19]

Makes sense! Thanks.

So for this question

Consider a ROM with 14 address lines and 16 data lines. How many bytes from the ROM need to
be used to implement a binary to four-digit BCD converter?

Would the answer be 10000 bytes?

 

[umup]

9999 bytes (decimal 4 digits) needed, 14 address bits