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Finding Patterns in Digits


Date: 04/29/98 at 08:18:33
From: Nathan Spriggs
Subject: Alg II

There are certain numbers that when put in the calculator, are equal. 
For example:

     2^5*9^2 = 2592    and    11^2 * 9 1/3 = 1129 1/3

How does this work? Is there a particular process involved to be able 
to find more examples of this occurrence?


Date: 04/29/98 at 16:08:48
From: Doctor Rob
Subject: Re: Alg II

If you pick a relationship you wish to hold on some number of digits,
you will be able to solve a Diophantine equation (that is, one whose
solutions must be integers) to find all solutions, if any.

Example 1: a^2*b = 100*a + 2*10 + b, where 1 <= a <= 9 and 1 <= b <= 9

This is equivalent to:

     (a^2-1)*b = 20*(5*a+1)

Case 1: a is even, so a^2-1 is odd, and 5*a+1 is odd, so b must be 
a multiple of 4, but not a multiple of 8, so b = 4. Then 
a^2 - 1 = 25*a + 5, a^2 - 25*a - 6 = 0, a = [25 + sqrt(649)]/2, so a 
is not an integer. This case is impossible.

Case 2: a = 4*A+1, with 0 <= A <= 2. Then a^2-1 = 8*A*(2*A+1) and 
5*a+1 = 20*A+6 are both even. Thus, when we substitute a = 4*A+1 into:

     (a+1)*(a-1)*b = 20*(5*a+1)

we find:

     8*A*(2*A+1)*b = 40*(10*A+3)
       A*(2*A+1)*b = 50*A+15

Clearly A = 0 is impossible. A = 1 gives 3*b = 65, b is not an 
integer. A = 2 gives 10*b = 115, so b is not an integer. This case is 
impossible.

Case 3: a = 4*A+3, with 0 <= A <= 1. Then a^2-1 = 16*A^2+24*A+8, 
5*a+1 = 20*A+16, and:

     8*(A+1)*(2*A+1)*b = 80*(5*A+4)
       (A+1)*(2*A+1)*b = 10*(5*A+4)

Now A = 0 yields b = 40, an integer, but too large. A = 1 gives 
6*b = 90, b = 15, an integer, but also too large. This case is 
impossible.

This proves that the desired relationship cannot hold.

Example 2: a*b^3 = 100*a + 10*b + 3 with 1 <= a and 1 <= b <= 9
Here we will allow a to be as big as we want.

Clearly a and b must both be odd. Substituting a = 2*A + 1, 0 <= A, 
and b = 2*B + 1, 0 <= B <= 4, then:

     (2*A+1)*(2*B+1)^3 = 200*A + 20*B + 113
     16*A*B^3+8*B^3+24*A*B^2+12*B^2+12*A*B+6*B+2*A+1 = 200*A+20*B+113
     16*A*B^3+8*B^3+24*A*B^2+12*B^2+12*A*B-14*B-198*A-112 = 0

Trying B = 0, we get A = -56/99, not an integer.
Trying B = 1, we get A = -53/73, not an integer.
Trying B = 2, we get A = 14/25, not an integer.
Trying B = 3, we get A = -85/243, not an integer.
Trying B = 4, we get A = -268/629, not an integer.
Thus there are no solutions.

Example 3: a^5*b^2 = 1000*a + 100*5 + 10*b + 2 with 1 <= a and 
                                                    1 <= b <= 9

Case 1: b  is even, so b = 2*B. Thus a^5*4*B^2 = 1000*a + 502 + 20*B, 
so the left side is divisible by 4, and the right side is not. This is 
impossible.

Case 2: b odd, so b = 2*B+1. Then a is even, so a = 2*A. Dividing out 
a 4 from both sides, we find that B must be a multiple of 4, so 
B = 4*C and b = 8*C+1.

   Subcase 2a: C = 0, then 2*A^2 - 125*A - 32 = 0, which has no      
               integer roots. This is impossible.
   Subcase 2b: C = 1, then 162*A^5 - 125*A - 37 = 0. This has only one
               integer root, A = 1.  Then a = 2, b = 9. 

Thus the only solution is 2^5*9^2 = 2592.

Every form you can dream up can be subjected to this kind of analysis.  
The lucky part is finding one with any solutions at all.

Another problem of a similar kind is 19/95 = 1/5 (erroneous canceling 
of equal digits yields a correct answer). What other two-digit 
numerators and denominators satisfy this property? How about three-
digit ones?

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Number Theory

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