YISTA

john ormond says:

Question 17, about f(n)=n. Are there more solutions? Yes, after 199,981 the next 9 numbers are all valid. Then, 199,991 adds two 1′s, so that f(n) is now ahead of n, for the first time. So they remain, until n=200,000 (no 1′s) and so catches up with f(n) which is 200,000 and 200,001 is also a valid f(n)=n. After that, n progresses, but f(n) only goes up about 20 per hundred. My ASM program got up to 655,360 and then crashed with a divide overflow error. My guess is that the difference between n, and f(n) increases until n = 1,000,000 after that, f(n) slowly catches up, until 1,100,000 when there are at least two 1′s for every increase in n. Until n=2,000,000 when n would again go ahead. Using 32-bit ASM would find more solutions, but DEBUG doesn’t do 32-bit, and I don’t know the relevant hex values. And, lets face it… does it matter? I wonder what Google really achieved by this question. As a programming exercise, it is quite simple, even in ASM.