Lesson What is the last digit of the number a^n ?
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<H2>What is the last digit of the number a^n ?</H2> <H3>Problem 1</H3>What is the last digit of the number {{{7^72}}} ? <B>Solution</B> <pre> To solve this problem, you do not need calculate the numbers {{{7^n}}}, one after another. It is enough to trace the LAST DIGITS of these numbers, only. Last digits of {{{7^n}}} form the sequence n 1 2 3 4 5 . . . 72 last digits of {{{7^n}}} 7 9 3 1 7 . . . ? (1) From this Table, the first four terms of this sequence are different numbers 7, 9, 3, 1, but then the digit 7 arises again as the 5-th term. From this point, it should be clear to you, that this sequence is periodic with the period length of 4. The number of periods of the length 4 in the sequence (1) is exactly {{{72/4}}} = 18, so the last digit in this sequence is the last digit of the period {7, 9, 3, 1}. Thus the last digit of the number {{{7^72}}} is 1. <U>ANSWER</U>. {{{7^72}}} has the last digit 1. </pre> <H3>Problem 2</H3>What is the units digit of the number {{{19^93}}} ? <B>Solution</B> <pre> To solve this problem, you do not need calculate the numbers {{{19^n}}}, one after another. It is enough to trace the LAST DIGITS of these numbers, only. Last digits of {{{19^n}}} form the sequence n 1 2 3 4 5 . . . last digits of {{{19^n}}} 9 1 9 1 9 . . . (2) From this Table, the first two terms of this sequence are different numbers 9, 1, but then the digit 9 arises again as the 3-rd term. From this point, it should be clear to you, that this sequence is periodic with the period length of 2. The number of full periods of the length 2 in the sequence (2) is the integer part of the number {{{93/2}}}, i.e. 46. so the last digit in the sequence (2) is the first digit of the next period {9, 1}, i.e. 9. 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