All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers
2^(10n) (Posted on 2015-04-15) Difficulty: 2 of 5
210 = 1024
220 = 1048576
Note that raising 2 to each of the first two multiples of 10 results in a number whose first digit is 1.

Find the smallest multiple of 10 where 2 raising to that power results in a number that does not begin with 1.

See The Solution Submitted by Jer    
Rating: 2.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution: Easy Way Comment 9 of 9 |
The supposed required number = 10*n (given)
Now, 2^(10*n)
= (1024)^n
= {(10)^(3n)} * (1.024)^n

Obviously, the smallest n occurs whenever:
(1.024)^n >= 2, since the other factor is only  an exponent of 10.
or, n >= (log 2)/(log 1.024) ~= 29.226....(by calculator)
Since n > 29, it follows that the least positive integer value of n is 30.
This checks out by calculator, as:
(1.024)^30 ~= 2.037....
(1.024)^29 ~= 1.989 ....

Consequently,  the required smallest multiple of 10 is 10*30 = 300

CHECK
-----------
Calculator verification gives:
2^300  ~= 2.037036 * 10^90
As 2^300 begins with 2, it is verified that 300 is INDEED the sought for 10-multiple exponent of 2.

Edited on June 16, 2022, 12:45 am
  Posted by K Sengupta on 2022-06-16 00:39:24

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information