x = 13!
a = x^16 - x^8 / x^8 + x ^ 4 =
(x^8+x^4)(x^8-x^4)/(x^8+x^4) =
x^8 - x^4
a/x^4 =
x^4 - 1
digit number of 13! must be 0 because its like (1*2*3...*11*12*13) * 10
so the final answer would be 1
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Difficult Math Question #23 - Algebra
Source: Beat The GMAT — Problem Solving |
Let (13!)^4 = x
(x^4 - x^2)/ (x^2 - x) = a
by solving further we get:
x^3 + x^2 = ax
or a/x = x + 1
a/(13! ^ 4 ) = (13!^ 4) + 1
as 13! ends with 0, so is 13!^ 4; so (13!^ 4) + 1 ends with 1.
Unit's digit is 1.
(x^4 - x^2)/ (x^2 - x) = a
by solving further we get:
x^3 + x^2 = ax
or a/x = x + 1
a/(13! ^ 4 ) = (13!^ 4) + 1
as 13! ends with 0, so is 13!^ 4; so (13!^ 4) + 1 ends with 1.
Unit's digit is 1.
