Latest TCS Aptitude Question SOLUTION: 20^2004+16^2004-3^2004-1 is divisible by

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

TCS Question Papers: TCS Interview Experience(91) TCS Company Profile

(#M40030091) TCS Numerical Ability question Keep an EYE

20^2004+16^2004-3^2004-1 is divisible by

Asked In TCS shubham kumar (11 years ago)

Unsolved Read Solution (4)

Is this Puzzle helpful?   (2)   (2) Submit Your Solution Number System

Read Solution (4) :

RAKESH  (11 years ago)

(x^n-y^n) is divisible by (x-y) for all values of n
(x^n-y^n) is divisible by (x+y) for all even values of n

20^2004+16^2004-3^2004-1
(20^2004-3^2004)+(16^2004-1^2004)
(20^2004-3^2004) is divisible by 20-3=17 for all n
(16^2004-1^2004) is divisible by 16+1=17 for even values of n
(20^2004-1^2004) is divisible by 20-1=19 for all n
(16^2004-3^2004) is divisible by 16+3=19 for even values of n
hence (20^2004+16^2004-3^2004-1) is divisible by 17*19=323
ANS 323

Is this solution Helpfull? Yes (38) | No (2)  

akshita agarwal  (11 years ago)

2 as (even+even)-(odd+odd)=even-even=even so no is divisible by 2

Is this solution Helpfull? Yes (15) | No (1)  

AVINASH KUMAR  (11 years ago)

20+16-3=33

Is this solution Helpfull? Yes (1) | No (2)  

sanjeet  (10 years ago)

(20^2004+3^2004) divisible by (20-3)=17 {by rule (a^n-b^n) is divisible by a-b}
(16^2004-1^2004) divisible by (16+1_=17 {by rule (a^n-b^n)is divisible by a+b}
again,
(20^2004-1^2004) divisible by (20-1)=19 {rule used above}
(16^2004-3^2004) divisible by (16+3)=19 {rule used above)
so, 17*19=323 ans.....

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.