In the problem "Mirror, mirror on the wall" it was proved that no number in the decimal system doubles on reversing its digits, and answered for bases 3, 5 and 8 (base 2 has leading zero, so itīs not valid).

Generalise the answer for positive integer bases.

First, consider the base B (an integer > 2). A "mirror" number pair cannot exist when B = 1 mod 3.

Proof: suppose there were a mirror number whose first digit was a and whose last digit was b (no relation to B). then since doubling the mirror number reverses the digits, 2b = a mod B (multiplying the two rightmost places) and either 2a = b or 2a+1=b (multiplying the two leftmost places. The +1 may or may not carry over from earlier digits. No number greater than 1 can carry over becase all digits are < B and so 2*(any digit) < 2B.

2b = a mod B implies 2b = a + kB for some integer k. but 2b < 2B so a + kB  < 2B. Since a >= 0, k <2. But b > a by the second equation so k > 0. That means k = 1, and either:

2b = a + B
2a = b

OR

2b = a + B
2a+1=b

Substitute for b in each of the first equations yields either:

4a = a + B -> 3a = B

or

4a + 2 = a + B  -> 3a+2 = B

In the first case, B must be divisible by 3; in the second case B = 2 mod 3. In no case can B = 1 mod 3.



For each B != 1 mod 3 there is at least one mirror number.

For B = 2 mod 3, the two-digit number whose digits are [(B-2)/3][(2B-1)/3] is a solution. (Note that both expressions are whole numbers less than B given the initial congruence.)

To see why, recall that a two digit number ab base B = aB + b. For the number above, that value is B^2/3 - 2B/3 + 2B/3 - 1/3 = (B^2 - 1)/3. Twice that number is 2(B^2-1)/3.

The mirror number [(2B-1)/3][(B-2)/3] is equal to 2B^2/3 - B/3 + B/3 - 2/3 = 2(B^2-1)/3 -- the same value.

Similarly, when B = 0 mod 3, the four digit number [B/3][B/3 - 1][2B/3 - 1][2B/3] is a mirror number. Expanding this number as a polynomial in B, doubling, and comparing to the expansion in B of the number with digits reversed yields the same value, 2(B^4 + B^3 - B^2 - B)/3

As the computer exploration has shown, there are other solutions; I'd conjecture (but will not try to prove) that there are an infinite number of "mirror" numbers in all bases which allow them in the first place.

Posted by Paul on 2008-09-02 23:12:54