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Difficult Math Question #23 - Algebra

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Source: — Problem Solving |
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by gamemaster » Thu Oct 05, 2006 12:29 am
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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by sanjeevs50 » Thu Oct 05, 2006 2:26 pm
(E) is the anwer

0 of unit digit - 1 = 9
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by shail_8206 » Thu Oct 05, 2006 9:35 pm
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.
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by gamemaster » Fri Oct 06, 2006 8:29 am
the answer is 9

i had a typo...
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OA

by 800guy » Mon Oct 09, 2006 4:38 pm
here's the oa, look at the attachment
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23-answer.gif
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