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

Home > Just Math
Pretty Potent Power Pair(s) (Posted on 2008-10-22) Difficulty: 3 of 5
Determine all possible pair(s) (2A, 2B), where each of A and B is a nonnegative integer, such that 2B is obtained by deleting the first digit from the left in the base 10 representation of 2A. None of the base 10 representations of 2A and 2B can contain leading zeroes.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer-assisted proof | Comment 1 of 2
   10   P=1
   20   repeat
   30    P=P*2
   40    P10=int(log(P)/log(10))+1
   50    for K=1 to 9
   60      N=K*10^P10+P
   70      if N=2*P or N=4*P or N=8*P or N=16*P or N=32*P or N=64*P or N=128*P then
   80        :print N:print P:print
   90    next
  100   until log(P)/log(10)>2000

The above checks powers of 2 up to 2000-digit numbers (an overkill, as we'll see). It finds the pairs (32,2) and (64,4).

Note that the higher power (2^A) can only be 2x, 4x, 8x, 16x, 32x or 64x the lower power (2^b). The 128x was thrown in for good measure.

Could any higher pair of powers work?

The following program considers all the possibilities mod 1000, to check to see if the last three digits of such power pairs could match, given that one must be 2, 4, 8, 16, 32 or 64 times the other:

   10   for I=2 to 998 step 2
   20   if (2*I)@1000=I then print I,2
   30   if (4*I)@1000=I then print I,4
   40   if (8*I)@1000=I then print I,8
   50   if (16*I)@1000=I then print I,16
   60   if (32*I)@1000=I then print I,32
   70   if (64*I)@1000=I then print I,64
   80   if (128*I)@1000=I then print I,128
   90   next

Zero is not checked as no power of 2 can end in zero as such a number would have 5 as a factor.

The only results were:

 200     16
 400     16
 600     16
 800     16

This indicates that of the endings considered, only numbers ending in 200, 400, 600 or 800 would remain that way after multiplication by a power of 2 under 100. But a number ending in one of these three triplets of digits can't be a power of 2 itself.
So powers of 2 of more than three digits will not work. Therefore, only the two pairs found (32,2) and (64,4) work.

  Posted by Charlie on 2008-10-22 11:44:45
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information