Numbers Identical in Base 10 and Base 3

Date: 02/14/2001 at 07:51:49
From: Gwendolyn Scott
Subject: Base ten word problem

Please help me with this word problem!

Find all base-10 numbers that, when multiplied by eight, have exactly 
the same digits as their base-3 counterparts.

I really need an explanation of how you got the answer. I don't know 
where to start. Thanks.

Date: 02/14/2001 at 11:25:32
From: Doctor Rob
Subject: Re: Base ten word problem

Thanks for writing to Ask Dr. Math, Gwendolyn.

I think we can assume that N >= 0. If N were negative, then -N would 
have the same property.

Let N be the number in question. If N has d digits, then 

     8*N < 8*10^d

On the other hand, the smallest base-3 number corresponding to N is 

     10^(d-1) (base-10) = 101^(d-1) (base-3) >
              100^(d-1) (base-3) = 10^(2*d-2) (base-3)

Now

     8*10^d < 10^(2*d-2) iff
     8 < 10^(d-2)        iff
     log(8) < d - 2      iff
     0.9031 < d - 2      iff
     2.9031 < d

That means when d >= 3, you get

     N < 8*10^d < 10^(2*d-2) < N

a contradiction, so no such N can exist. That means we can restrict 
attention to N < 100.

For 1-digit numbers, you would have

     8*N = 10*a + b   (0 <= a <= 2, 0 <= b <= 2),
       N = 3*a + b
     7*N = 7*a
       N = a
       a = 3*a + b
    -2*a = b

This implies that a = b = 0, so N = 0. That is one solution.

For 2-digit numbers, you would have

     8*N = 100*a + 10*b + c
       N =   9*a +  3*b + c
     7*N =  91*a +  7*b
       N =  13*a +    b

Here a, b, and c are all in the set {0,1,2}.  The above imply
that

     0 = -4*a + 2*b + c
     c = 4*a - 2*b

Try all nine possibilities for a and b. Throw out those for which c 
isn't 0, 1, or 2. Those will give you the other positive solutions.  
Take their negatives to get all the negative solutions, and you will 
have a complete list.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/