pinky7219 Posted August 16, 2010 Share Posted August 16, 2010 X = 2^97 + 3^97 + 6^97 + 7^97 + 9^97 What is the remainder when X is divided by 9? OPTIONS 1) 0 2) 1 3) 2 4) 3 5) 4 Quote Link to comment Share on other sites More sharing options...
MBAchase Posted August 16, 2010 Share Posted August 16, 2010 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. Quote Link to comment Share on other sites More sharing options...
Abhishek009 Posted August 17, 2010 Share Posted August 17, 2010 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. Quote Link to comment Share on other sites More sharing options...
sjmit4 Posted August 17, 2010 Share Posted August 17, 2010 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? Quote Link to comment Share on other sites More sharing options...
Abhishek009 Posted August 17, 2010 Share Posted August 17, 2010 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. Quote Link to comment Share on other sites More sharing options...
sjmit4 Posted August 23, 2010 Share Posted August 23, 2010 Check the following link : Prove (a^n - b^n) is divisible by (a - b) help!? - Yahoo! Answers Even I checked it by trail and accepted it as a rule , but did not understood the proof at all. Quote Link to comment Share on other sites More sharing options...
GetItDone Posted August 27, 2010 Share Posted August 27, 2010 Valid question. Please refer to the formulae for a^2-b^2; a^3-b^3, a^3+b^3. These three will clarify the theorem. So if the sign is subtraction, the power being even or odd does not matter. However, if the sign is addition, then the power has to be odd. Quote Link to comment Share on other sites More sharing options...
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