There are only 4 positive integers N such that
multiplying sod(N) by inverted(sod(N)) equals N.
The trivial number 1 qualifies !
The magic Ramanujan number 1729 qualifies as well ,
Since 19*91=1729
Now, find the missing two and prove that no other exist.
Mod 9, N = sod(N) = inverted(sod(N))
Mod 9,
0*0 = 0 Works
1*1 = 1 Works
2*2 = 4
3*3 = 0
4*4 = 7
5*5 = 7
6*6 = 0
7*7 = 4
8*8 = 1
Thus, Mod 9, sod(N) can only be 0 or 1
Charlie has proved that N is at most a four digit number.
Therefore, sod(N) can only be 1,9,10,18,19,27,28,and 36
We already know that 1 and 19 work.
Testing the others,
9 * 9 = 81 (sod 9) Works
10*01 = 10 (sod 1)
18*81 = 1458 (sod 18) Works
27*72 = 1944 (sod 18)
28*82 = 2296 (sod 19)
36*63 = 2268 (sod 18)
so, the only other two N are 81 and 1458
Edited on December 31, 2021, 2:21 pm
D2 approach (spoiler)
