Q. Z is a natural number. and sqrt, cube root, 4th root, 5th root, 6th root, or 7th root of Z is also an integer value. If Q = log Z base 32.
What would be the value of Q?
Option
A) 1.4
B) 7
C) 13.37
D) ___ not remember around 15.

• Q = log Z base 32
  32^Q = Z
  32^(Q/7) = Z^(1/7)
  So,Z^(1/7) to be integer
  
  Q/7 = 1/5
  Q = 7/5
  Q = 1.4
  
  answer A
• 10 years agoHelpfull: Yes(15) No(3)
• answer is 84
  
• 10 years agoHelpfull: Yes(8) No(10)
• The answer is 1.4, because 32^1.4 which is equal to 32^7/5 = (2^5)^7/5. so after simplifying it becomes 2^7. Thus 128 is the 7th root (which is integer), 6th root is 64(which is integer), 5th root is 32 (which is integer) .... 1st root2. therefore option A is the correct answer.
• 10 years agoHelpfull: Yes(5) No(1)
• hello nitin ...can u explain how u got q/7 = 1/5
  it should be q/7=1/7 at the utmost case
• 10 years agoHelpfull: Yes(4) No(0)
• if 2^7 is an integer .how cum its 5th root is an integer??
• 10 years agoHelpfull: Yes(3) No(0)
• Ans:-
  Q = log Z base 32
  => 32^Q = Z
  => 32 = Qth root of Z
  So, Q must be {1,2,3,4,5,6,7}
  from options only 7 is possible
  so ans = 7
• 10 years agoHelpfull: Yes(2) No(14)
• how become 32^Q = Z. can explain me
• 10 years agoHelpfull: Yes(0) No(4)
• Q=logZbase32;
  So,(32)^Q=Z=natural no(natural no=1,2,3...ie.whole nos);
  now,(32)^Q/5=(Z)^1/5=integer;((32)^1/5)^Q=Z^1/5;
  so,(2)^Q=(Z)^1/5=INTEGER;
  so,(using the 4 options)Q has to be 7 to make (Z)1/5 an integer
• 10 years agoHelpfull: Yes(0) No(1)
• if q=7
  then z=128
  (128)^1/6 and (128)^1/5 cant be integer..
  i think none of the above will be answer
• 10 years agoHelpfull: Yes(0) No(0)
• here q=logz base 32
  => z=32^q
  now for z ^1/2=32^q/2,z ^1/3=32^q/3,z ^1/4=32^q/4,
  z ^1/5=32^q/5,z ^1/6=32^q/6,z ^1/7=32^q/7,
  so to get z^(1/2,1/3,1/4,1/5,1/6,1/7)=an integer value than q ^ should be divided by aal of these no so, lcm (2,3,4,5,6,7)=420
  so ans is 420.
• 10 years agoHelpfull: Yes(0) No(0)