Question

Let x be the number of digits in begin math size 16px style 2 to the power of 2001 end style when calculated. Let y be the number of digits in begin math size 16px style 5 to the power of 2001 end style when calculated. Find the sum of x and y

Answer 1

Answer:

2002

Step-by-step explanation:

The number of digits in a number is the ceil value of the base-ten logarithm of the number (ceil value means rounding up).

$\text{Number of digits in }N=\left \lceil{\log N}\right \rceil$

When $N = 2^{2001}$

$x=\left \lceil{\log 2^{2001}}\right \rceil=\left \lceil{2001\log2} \right \rceil= 603$

When $N = 5^{2001}$

$y=\left \lceil{\log 5^{2001}}\right \rceil=\left \lceil{2001\log5} \right \rceil= 1399$

$x+y=603+1399 = 2002$

Note that the sum x + y could be derived by finding the number of digits in $2^{2001}\times5^{2001}=10^{2001}$

The number of zeros in $10^{2001}$ is 2001. Including the 1 in leftmost digit position gives 2001 + 1 = 2002