Posted 14 Mar 2013 228^251 * 126^23416 * 37^217 * 213^19. what will be the last digit of the resulting Answer? 0 Share this post Link to post Share on other sites

0 Posted 14 Mar 2013 (edited) 228^251 * 126^23416 * 37^217 * 213^19. what will be the last digit of the resulting Answer? The short answer: 8 The long answer: The last digit of any number raised to a power will be equal to the last digit of a similarly ending number raised to the same power. (the last digit of 4^5 will equal the last digit of 14^5 will equal the last digit of 5374^5 and so on) Since we are only interested in the last digit of the resulting answer, the equation can be condensed to: 8^251 * 6^23416 * 7^217 * 3^19 Additionally, the same concept can be applied to multiplying different numbers together (in the context of finding the last digit). For example, the last digit of 7 * 3 will equal the last digit of 67 * 123 will equal the last digit of 887 * 76543 and so on, so only the last digit of each segment of the equation is needed to be multiplied. Last digit of 8^251 = 2 Last digit of 6^23416 = 6 Last digit of 7^217 = 7 Last digit of 3^19 = 7 Last digit of 2 * 6 * 7 * 7 = 8 Edited 14 Mar 2013 by BobbyGo 0 Share this post Link to post Share on other sites

0 Posted 14 Mar 2013 The last digit of a number raised to a power cycles with a given periodicity. 8=> 8 4 2 6 8 4 ... every 4 powers it repeats 6 => 6 6 6 6 6 ... always ends in 6 7 => 7 9 3 1 7 9 ... repeats every 4 powers 3 => 3 9 7 1 3 9 ... repeats every 4 powers. 251 mod 4 = 3 so first term ends in 223416 mod 1 = 1 so second term ends in 6 217 mod 4 = 1 so third term ends in 7 19 mod 4 = 3 so fourth term ends in 7 Product of 2 6 7 7 is 588 so last digit of the expression is 8. 0 Share this post Link to post Share on other sites

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228^251 * 126^23416 * 37^217 * 213^19. what will be the last digit of the resulting Answer?

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