4 ms·

How many digits does the number 125^100 have? Im breaking my head here :(

• mjfl 6y ago
  100*log10(125) ~ 210?
  • naniwaduni 6y ago
    We've reduced a tedious elementary problem to a slightly harder problem! (We can observe that 125 = 1000/8, so 100log10(125) = 300 - 300log10(2) = 300 - 30log(1024). From here we merely need the bound 90 = 30log(1000) < 30log(1024) < 91, or equivalently 1 < 1.024^30 < 10. There are probably better approaches but I'm partial to 1.024^30 = (1.024^15)^2 < (1.024^16)^2 < 3^2 < 10, since 1.024^16 is easy to upper bound by repeated squaring and rounding up: 1.024^16 < 1.03^16 < 1.1^8 < 1.3^4 < 1.7^2 < 2.9.)
    • mjfl 6y ago
      observe that log10(125) = log10(100) + log10(1.25) log10(100) = 2 log10(1.25) = log(1 + 0.25) / log(10) ~ 0.25 / log(10) < 0.25 / 2 = 0.125 Thus log10(125) ~ 2.12 but probably a little less.
      • naniwaduni 6y ago
        The extra contortions are to get a bound tight enough to admit only one integer.