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What is the remainder


pinky7219

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Good question,

 

X = 2^97 + 3^97 + 6^97 + 7^97 + 9^97

 

X = (2^97 + 7^97) + (3^97 + 6^97) + 9^97

 

(a^n + b^n) is always divisible by (a + b) for integers a, b and odd values of n.

 

Thus, note that (2^97 + 7^97) is divisible by 2 + 7 i.e. 9.

 

Similarly, (3^97 + 6^97) is divisible by 3 + 6 i.e. 9

 

The last term (9^97) is obviously divisible by 9.

 

Since each term of X is divisible by 9, X is also divisible by 9.

 

Hence, X leaves a remainder of 0 on division by 9.

 

Hence, option 1.

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Good question,

 

X = 2^97 + 3^97 + 6^97 + 7^97 + 9^97

 

X = (2^97 + 7^97) + (3^97 + 6^97) + 9^97

 

(a^n + b^n) is always divisible by (a + b) for integers a, b and odd values of n.

 

Thus, note that (2^97 + 7^97) is divisible by 2 + 7 i.e. 9.

 

Similarly, (3^97 + 6^97) is divisible by 3 + 6 i.e. 9

 

The last term (9^97) is obviously divisible by 9.

 

Since each term of X is divisible by 9, X is also divisible by 9.

 

Hence, X leaves a remainder of 0 on division by 9.

 

Hence, option 1.

 

 

Yes this is the best and quickest approach. Otherwise you may go by finding out individual remainder , but that is really time consuming.

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Good question,

 

X = 2^97 + 3^97 + 6^97 + 7^97 + 9^97

 

X = (2^97 + 7^97) + (3^97 + 6^97) + 9^97

 

(a^n + b^n) is always divisible by (a + b) for integers a, b and odd values of n.

 

 

Can anyone tell me why value of n has to be an odd value?

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Can anyone tell me why value of n has to be an odd value?

 

Well dude this is a rule . Check the following link : Prove (a^n - b^n) is divisible by (a - b) help!? - Yahoo! Answers

 

This is a proof if you don't get it this way just plug in some value to remember this concept.

 

Say a = 4 ; b = 2 and n = 3

 

Hence we get 4^3 - 2^3 = 64 - 8 = 56 which is definitely divisible by 2 (4 - 2).

 

 

Now take a = 5 and b = 2 and n = 3

 

so ; 5^3 - 2^3 = 125 - 8 = 117 which is definitely divisible by 3 (5 - 2 ).

 

 

Remember : a^n - b^n is divisible by a - b only when n is ODD.

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