Answer:
Fewest number of envelopes 10 and 1$ , 2$,$4 , 8$ , 16 $ , 32 $,64 $ +128 ,$ 256 $, 489 $
Step-by-step explanation:
Given:
$1000 bills with 1$ bill are start which remains to 999$
To Find:
Fewest number of envelopes you need And how many dollar bills are in each envelope?
Solution:
Consider first 1 $ to get total of 1000$ bills on short way is to find a series which ultimately adds up to 1000$
So using the logical that double the previous bill amount one can easily achieve
So 1st amount is of 1 $ Next will be double of 1 $ i.e 2 $
2$ to 4$ , 4 $ to 8$ then 8 $ to 16 $
16$ to 32 $
32 $ to 64 $ ,
64 $ 128$,
128$ to 256 $
256 $ to 512 $
512 $ to 1024 $ But we started from 1 $ so it will 1023 $
But we exceeded 1000$ by 23 $ i.e.
We need to subtract the 23 $ tofrom last number ,
So last nu,ber 512 $ become 512 -23 $=489$
Hence which Form a series like as
1$ + 2$+ $4 + 8$ + 16 $ + 32 $ +64 $ +128 +$ 256 $+ 489 $=1000$
Hence Total number envelope will 10
And each envelope has double the amount of previous ones with 1 $ at start.