text: Let\nh = hundreds digit\nt = tens digit\nu = units digit\n\nwe know that\nt = u + 5 (since the tens digit is 5 more than the ones digit)\nh = t - 8 (since the hundreds digit is 8 less than the tens digit)\n\nLet us take a look at the second equation:\nh = t - 8\n\nSince h and t are digits of a number, then we know that h and t are single digits.\n\nNow, h and t cannot be negative, so the only possible value for t is 9 or 8 (since 10 is no longer a digit and any value less than 8 will make h negative).\n\nThus, if t = 9, h = 9 - 8 = 1, and\nif t = 8, h = 8 - 8 = 0.\n\nBut since we know that h is the hundreds digit, h cannot be equal to zero. Therefore, h = 1.  And also t = 9.  Now, using the first equation, \nt = u + 5\n\nWe can solve for u\nu = t - 5\n\nSince t = 9, then\nu = 9 - 5\n\nTherefore,\nu = 4\n\nNow, we know the values\nu = 4,\nt = 9, and\nh = 1. \n\nTherefore, the 3-digit number is \n194.\n\n^_^