Let n be a positive integer such that 2n1 is a digit anagram of n (in its decimal representation). An example of such a pair would be n=37, 2n1=73.
Show that if n's least significant digit is 3, n must also contain an 8.
Suppose n's last digit is 3. Then, 2n1 ends in 5. Since n is an anagram of 2n1, then n has a 5. This 5 will turn into either a 0 or 1 in 2n1. If the 5 turns into a 1, then n also has a 1. If the 5 turns into a 0, then n also has a 0. This 0 will turn into either a 0 or 1 in 2n1. If the 0 turns into a 1, then n has a 1. If the 0 turns into a 0, then n has another 0. This 0 will turn into either a 0 or 1. If it turns into a 1, then n has a 1. If it turns into a 0, then n has another 0. Eventually, the 0 will turn into a 1. Therefore, n contains a 1.
Let small digits be 0, 1, 2, 3, 4, and let big digits be 5, 6, 7, 8, 9. Then, an even digit in 2n1 is created from a digit before a small digit in n, and an odd digit in 2n1 other than the last digit is created from a digit before a big digit in n. Therefore, there is 1 more odd digit in 2n1 than big digits in n. Since n and 2n1 have the same digits, n has 1 more odd digit than big digits. Then, n has 1 more small odd digit than big even digits.
Suppose n has no 8. Then, the only big even digit that can be in n is 6. Since n has no 8, every 6 in 2n1 is created from a 3 in n. Let x be the number of 6's in n. Then, 2n1 has x 6's. Then, n has x 3's that create the 6's. It also has a 1 and the 3 at the end of n. Since 3 is a small odd digit, n has at least x+2 small odd digits. Since 6 is the only possible big even digit in n, then n has exactly x big even digits. Then, n has at least 2 more small odd digits than big even digits. That is a contradiction. Therefore, n must contain an 8.

Posted by Math Man
on 20190124 15:23:50 