Erich Friedman: Find the 6 missing digits that make the equation (x^x + xx) / xx = 2005 true.
(5^7 + 70)/39 = 2005.
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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
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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
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