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# Find integer D. (1) |D|!=|D|!^2 (2) |D|!=|D|!^3

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Joined: 03 Feb 2003
Posts: 1608
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Kudos [?]: 76 [0], given: 0

Find integer D. (1) |D|!=|D|!^2 (2) |D|!=|D|!^3 [#permalink]  18 Jun 2003, 22:52
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Find integer D.

(1) |D|!=|D|!^2
(2) |D|!=|D|!^3
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

[#permalink]  19 Jun 2003, 05:21
It is E, let us post the solution:

(1) let |D|!=A; thus A=A^2, A=0 or A=1
|D|!=0 has no sense
|D|!=1 means that D={0, 1, -1} not enough

(2) employ the same approach
A=A^3; thus, A=0, A=1, and A=-1
|D|!=0 has no sense
|D|!=-1 has no sense
|D|!=1 means that D={0, 1, -1} not enough

Combine D={0, 1, -1} not enough. Therefore, E.
[#permalink] 19 Jun 2003, 05:21
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# Find integer D. (1) |D|!=|D|!^2 (2) |D|!=|D|!^3

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