x of x 2 of x 3 of an unknown integer perfect square integer unknown integer will be x is

10) none of these

Infinite numbers are possible.

If x=1, and unknown number is y , then
1 % of 1^2 % of 1^3 % of y = perfect square integer
y = ( perfect square integer)*10^3

Put any perfect square integer and get new value of y.

so infinite values.
12 years agoHelpfull: Yes(3) No(3)
10) none of these

Infinite possible integers.

x % of x^2 % of x^3 % of an unknown integer = perfect square integer
(x^6)*(10^(-6)) of unknown number = perfect square integer
Hence unknown number can be 10^6,10^8,10^10 and so on..........
If unknown number is 10^6 , RHS is x^6 which is perfect square of x^3
If unknown number is 10^8 , RHS is 100*x^6 which is perfect square of 10*x^3

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12 years agoHelpfull: Yes(2) No(2)
if u take unknown integer to be "y" ,
at the end.....
perfect square interger = y((x/10)^3)^2 ,,
that means y also hs to be a number which is square of sm number , in options it s "25" n "49" n "9" ,,

bt again , here to make whole number a square of sm integer , (x/10) should be integer , which we are nt sure of !

so ans : 10) none of these !!
12 years agoHelpfull: Yes(2) No(1)
Let the unknown integer is y
Then the equation is y*((x/10)^3)^2=perfect square.
The solution must satisfy for all x, an integer.
for x=1 and y=5 RHS is not perfect square.
Hence option 1 is not a solution.
for y=9,12,19,25,32,49 and x=1 again RHS is not perf square.
Hence all of the 7 options are wrong and hence
ans is option 10
12 years agoHelpfull: Yes(2) No(1)

x % of x^2 % of x^3 % of an unknown integer = perfect square integer.
unknown integer will be
(x is any integer)

Options 1) 5
2) 9
3) 12
4) 19
5) 25
6) 32
7) 49
8) 2 options are possible
9) 3 options are possible
10) none of these