If x is both the cube and the square of an integer and x is between 2 and ... ...
If x is both the cube and the square of an integer and x is between 2 and 800, what is the value of x?
(A) 8
(B) ...
(C) ...
(D) ...
(E) ...
*This question is included in
Nova Math - Problem Set F: Number Theory
Replies to This Thread: 0
|
----
Posted: 05/20/2012 13:20
There is no choice F.
Admin
Replies to This Thread: 0
|
----
Posted: 05/20/2012 17:42
Tiffany, choice F is supposed to say 729. It will be fixed in the next update. Thank you for noticing.
Replies to This Thread: 1
|
----
Posted: 02/12/2013 14:06
Am I the only one who interpret(ed) the question as asking for x being both the cube and the square of an integer; that is the SAME integer?
Thus (c) 64 - 8 squared, and 4 cubed.
Admin
Reply 1 of 1
Replies to This Thread: 0
|
----
Posted: 02/12/2013 21:37
Hi Asa,
The grammar allows the integers to be different. As you point out, x can be 64 (= 8^2 = 4^3). Here, the integers are 8 and 4. The number 729 also works: 729 = 9^3 = 27^2. Here, the integers are 9 and 27.
If we wanted x to be both the cube and the square of single number, we could change the grammar in the problem to
'If x is both the cube and the square of THE SAME integer....'
Now, the answer to the question would be 0 and 1:
0 = 0^3 = 0^2
1 = 1^3 = 1^2
Nova Press
Replies to This Thread: 0
|
----
Posted: 05/15/2013 09:44
Thanks!
Replies to This Thread: 0
|
----
If x is both the cube and the square of an integer and x is between 2 and ... ...
Posted: 06/21/2013 19:29
If I'm not mistaken one of the examples doesn't calculate correctly either. It says something like 14*5 =60. Unless I'm misunderstanding the equation, it would be 70.
