Erich Friedman: Find the 6 missing digits that make the equation (x^x + xx) / xx = 2005 true. (5^7 + 70)/39 = 2005. ------------------------------------------------------------------------ DENIS BORRIS Find the 6 missing digits that make the equation (x^x + xx) / xx = 2005 true digits = a,b,c,d,e,f a^b + 10c + d = 2005(10e + f) right side's last digit = 0 or 5 (multiplication by 2005); then d = 0 or 5 10c ends with 0; since a^b cannot end with 0, then must end with 5; so a = 5 so: a=5, b=7, c=7, d=0, e=3, f=9; (5^7 + 70) / 39 = 2005 ------------------------------------------------------------------------ I cheated and used two mathematica lines digit6 = Table[IntegerDigits[n, 10, 6], {n, 0, 10^6 - 1}]; Select[digit6, (#[[1]]^#[[2]] + 10 #[[3]] + #[[4]])/(10 #[[5]] + #[[6]]) == 2005 &] after warnings, answer is {{5, 7, 7, 0, 3, 9}}, checking, (5^7 + 70)/39 outputs 2005. Chris Lomont ------------------------------------------------------------------------- Doug Gwyn Alan Lemm Mark Michell Cynthia Disenfeld Mario Roederer Justin Smith Joseph DeVincentis Evgeni Lukin Konstantin Knop Rafael Almeria Paulo Sachs Zbyszek Zarzycki Amol Lele Ritch McBride