Problem 3

3. Consider the number $2^{9876543}$. Determine (a) the number of digits it has; (b) its last three (rightmost) decimal digits; and (c) its first three (leftmost) decimal digits.

Commentary: This problem's part (c) really scared me because I hadn't ever heard of how to calculate digits starting from the left in any simple fashion. I put it off for a later time and maybe some research, but before then, the thought had already passed through my mind of simply raising $10$ to the decimal part to get the digits, scientific-notation-style. But it seemed to good to be true, so I put off actually thinking about it until it was nearly the last of the remaining problems.

Proof: (a) The number $10^x$ has $\lfloor x \rfloor + 1$ digits to the left of the decimal for nonnegative $x$. So we solve $10^x=2^9876543$ by $x=9876543ln(2)/ln(10) \approx 2,973,135.696$. Hence there are $2,973,136$ digits in full.

(b) A basic exercise to calculate modulo $1000$. The most efficient way to do this by hand is to calculate the smallest $n$ such that $2^n ≤ 9876543$ (in this case $n=23$), then successively calculate the reductions $2^i$ modulo $1000$ for $0 ≤ i ≤ n$ and line them up left to right, and according to the binary representation of $9876543$ multiply and reduce modulo $1000$ each of these terms in line. The result of this effort is $208$.

(c) We have $10^x=2^{9876543}$ so in scientific notation this number is $10^d · 10^{\lfloor x \rfloor}$ where $d=x-\lfloor x \rfloor$ is the decimal part of $x$. In this case we see $10^d \approx 4.97$ so the first three digits are $497$.