Latest Elitmus Aptitude Question SOLUTION: If mn=1024 and k=(1025!)/m!^n and k has 128 and more zeroes.what is the minimum value of n such dat k is +ve natural no. Dnt knw the exact questn.

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Number System

Elitmus Question Papers:

(#M40030648) ELITMUS Numerical Ability question quant Keep an EYE

If mn=1024 and k=(1025!)/m!^n and k has 128 and more zeroes.what is the minimum value of n such dat k is +ve natural no.

Dnt knw the exact questn.

Asked In Elitmus abc (10 years ago)

Unsolved Read Solution (5)

Is this Puzzle helpful?   (4)   (0) Submit Your Solution Number System

Read Solution (5) :

Akshay Kanthed  (10 years ago)

m=1024
n=1
then k =1025

Is this solution Helpfull? Yes (14) | No (8)  

rishabh singh bisht  (10 years ago)

k=(1025!)/m!^n where m*n=2^10

1025! has 255 zeros, k can have 128 zeros so m!^n can have a maximum of 128 zeros.

Now let m=2^3 and n=2^7
m! has 1 zero
m!^n will have 2^7 zeros ie 128 zeros that is not possible..

Let m=2^2 and n=2^8
m! has no zeros so m!^n will have no zeros..
so least possible value for n is 2^8.

Is this solution Helpfull? Yes (13) | No (1)  

m.sindhuri  (10 years ago)

can u explain d process akshay kanthed.....

Is this solution Helpfull? Yes | No   

praveen  (10 years ago)

visit this for solution http://www.onlineadhyayana.com/ elitmus section...:)

Is this solution Helpfull? Yes | No (11)  

sourav kumar  (8 years ago)

can u explain? rishabh singh bisht

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.