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(#M40029603) TCS question Keep an EYE

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

Asked In TCS Abhishek Kumar (11 years ago)

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chinku  (11 years ago)

When dividend is of the form a^n+b^n or a^n-b^n:
we have 3 rules
Thrm 1:a^n+b^n is divisible by a+b when n is odd
Thrm2: a^n-b^n is divisible by a+b when n is even
Thrm3:a^n-b^n is always divisible by a-b
Solution: 20^2004 + 16^2004-3^2004-1=(20^2004-3^2004)+(16^2004-1^2004)
Now 20^2004-3^2004 is divisible by 17(Theorem 3)and 16^2004-1^2004 is divisible by 17 (Theorem 2)
Hence the complete expression is divisible by 17
20^2004+16^2004-3^2004-1=(20^2004-1^2004)+(16^2004-3^2004)
Now 20^2004-1^2004 is divisible by 19 (Theorem 3) and 16^2004-3^2004 is divisible by 19 (Theorem 2).
Hence the complete expression is also divisible by 19
Hence the complete expression is divisible by 17 × 19 = 323

Is this solution Helpfull? Yes (34) | No (5)  

ramya  (11 years ago)

divisible by 2..

Is this solution Helpfull? Yes (4) | No (1)  

Prerak Jain  (11 years ago)

Ans:- 17

(17+3)^2004 + (17-1)^2004 - 3^2004 - 1
Use Binomial

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

harish budhwani  (11 years ago)

20^2004=even no.
16^2004=even no
3^2004=odd no
3^2004+1=even no.
thus 20^2004+16^2004-(3^2004 +1)=even no.
and every even no is divisible by 2,thus whole expression is divisible by 2.

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

soham majumder  (11 years ago)

in 20^2004 unit digit is o
in 16^2004 unit digit is 6
in 3^2004 unit digit is 1
so the unit digit in the expression in (0+6-1-1)=4
so it is divisible by 2 or 4.

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

Abhishek Kumar  (11 years ago)

please post options for these type of questions

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nikhil kumar  (11 years ago)

divisible by 4....

Is this solution Helpfull? Yes | No (2)  

Ajay Poshak  (11 years ago)

20^2004 + 16^2004 - 3^2004 - 1^2004
then (20+16-3-1)^2004 so 32^2004
Thus it is divisible by 2 and its multiples.

Is this solution Helpfull? Yes | No (2)  

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