Consider a number system in which the k-th digit can have values 0, . . ., k; so

Question

Consider a number system in which the k-th digit can have values 0, . . ., k; so the
rightmost digit is 0 or 1, the next 0, 1, or 2, the next 0, 1, 2, or 3 etc. Counting in
this system from one to twenty is 1, 10, 11, 20, 21, 100, 101, 110, 111, 120, 121, 200,
201, 210, 211, 220, 221, 300, 301, 310. The largest numbers that can be expressed
with one to four digits are 1 (value 1), 21 (value 5), 321 (value 23), 4321 (value 119).
Show that the value of the k-digit number represented by 100 · · · 0 (with k 1 zeros)
is k!. Hint: the number of steps from 1 to 1000 is the same as the number of steps
from 1001 to 2000, or from 2001 to 3000 etc; but there is no number 5000, instead
we have 10000.

Explanation / Answer

For any number with   k digits,

1st digit can represent 2 values 0 & 1
2nd digit can have 3 values - 0,1 & 2.

So, 1st & 2nd place (2 digits) can represent 2*3 = 6 value (0 to 5 in decimal)

3rd place can have 4 values - 0,1,2 and 3.
So, 1st, 2nd and 3rd place (3 digits) together can represent 2*3*4 = 24 value (0 to 23 in decimal)

Continuing in the same manner, k-1 digits will represent 2*3*4......*(k-1) value (0 to k!-1)
The highest value represented by k-1 bits in decimal is k!-1.

The next possible value in this number system is obtained by adding 1 more bit to k-1.
This number is 100....00 (with k-1 0's)
Since the highest value of k-1 bits is k!-1, the next value in this number system will be k!-1 + 1 = k!

Hence, 1000..00 (k-1 0's) represents k! in decimal

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