Find the lowest positive integer that has its digits reversed after dividing it by 2.

(In reply to

Answer by K Sengupta)

Let p = 2q, where both p and q are positive integers, so that q = p/2, and each of p and q are reversals of each other. Then, we must have the respective forms of p and q as n.........m and m.......n, for some m and n.

Accordingly,

2*(m.......n) = n.........m -------- (*)

Comparing the respective last digits in each of the two numbers, we have:

m (mod 10) = 2n -------(i)

The valid values of m and n satisfying (i) are:

(m, n) = (2,1), (2,6), (4,2), (4,7), (6,3), (6,8), (8,4), (8,9) --------- (ii)

Since the number of digits of p and q must be the same, so that there are no leading zeroes in either of p and q, a comparison of the respective first digits of (*) yield:

n = 2m, 2m+1, or 2m+2

If n=2m+2, then in terms of (ii), only (m,n) = (2,6) satisfy the given conditions. But in this situation, we must have a carryover of 2 from the second digit of p to the first, from the third digit to the second digit, and so on. This is only possible when all the digits following 2 in q are 9. But, if the rightmost digit is 9, then we must have n=9, which directly contradicts n=6.

Again, we observe that none of the pairs in (ii) satisfy eiter of the relation n = 2m, or 2m+1. this leads to a contradiction.

Consequently, it follows that no such positive integer can exist that satisfy the given conditions.