Latest Elitmus Aptitude Question SOLUTION: 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

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(#M40035330) ELITMUS Numerical Ability question Logarithm Keep an EYE

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.

Asked In Elitmus jitendra kumar (10 years ago)

Unsolved Read Solution (10)

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Read Solution (10) :

Nitin Kumar  (10 years ago)

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

Is this solution Helpfull? Yes (15) | No (3)  

ashes  (10 years ago)

answer is 84

Is this solution Helpfull? Yes (8) | No (10)  

Nandha  (10 years ago)

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.

Is this solution Helpfull? Yes (5) | No (1)  

kodali kranthi  (10 years ago)

hello nitin ...can u explain how u got q/7 = 1/5
it should be q/7=1/7 at the utmost case

Is this solution Helpfull? Yes (4) | No   

Soma Ghosh  (10 years ago)

if 2^7 is an integer .how cum its 5th root is an integer??

Is this solution Helpfull? Yes (3) | No   

jitendra kumar  (10 years ago)

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

Is this solution Helpfull? Yes (2) | No (14)  

Mani kandan  (10 years ago)

how become 32^Q = Z. can explain me

Is this solution Helpfull? Yes | No (4)  

Ayan Gangopadhyay  (9 years ago)

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

Is this solution Helpfull? Yes | No (1)  

DEEPAK SINGHAL  (9 years ago)

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

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MD HASIM  (9 years ago)

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.

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