Lesson Find the last three digits of these numbers
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<H2>Find the last three digits of these numbers</H2> <H3>Problem 1</H3>Using binomial theorem, find the Last three digits of the number {{{27^27}}}. <B>Solution</B> <pre> 1. {{{27^27}}} = {{{((27^3)^3)^3}}}. Let us go up on this upstairs from the bottom to the top step by step. 2. First consider the number {{{27^3}}}. It is 19683: {{{27^3}}} = 19683. Write it in the form {{{27^3}}} = {{{19*10^3 + 683}}}. It is clear that for {{{(27^3)^3}}} and for {{{((27^3)^3)^3}}} the last three digits are determined by last three digits "683" of the number {{{19*10^3 + 683}}}. The part {{{19*10^3}}} does not affect the last three digits of the number {{{(27^3)^3}}}. It is exactly what the binomial theorem says and provides in this situation. Therefore, in finding the three last digits of the number {{{(27^3)^3}}} we can track only for {{{683^3}}} and do not concern about other terms. It implies that the last three digits of the number {{{(27^3)^3}}} are exactly the same as the last three digits of the number {{{683^3}}}. The number {{{683^3}}} = 318611987, as easy to calculate (I used Excel in my computer), so its last three digits are 987. 3. Now we can make the next (and the last) step up on this upstairs in the same way. The last three digits of the number {{{((27^3)^3)^3}}} are the same as the last three digits of the number {{{987^3}}}, and it is easy to calculate. {{{987^3}}} = 961504803, and its last three digits are 803. Therefore, the last three digits of the number {{{27^27}}} are 803. </pre> <H3>Problem 2</H3>Find the last three digits of the number {{{126^2018}}}. <B>Solution</B> <U>Intermediate statement</U>. The last three digits of the number {{{126^n}}} are 376, for any n >= 3. <pre> First consider the number {{{126^2}}}. It is 15876. Next {{{126^3}}}. It is 2000376. Next {{{126^4}}}. It is 252047376. Notice that {{{126^3}}} and {{{126^4}}} have the last 3 digits 376. Now it is clear that for {{{126^(n+1)}}} the last three digits are determined by last three digits of the number {{{126^n}}}. To prove it, write {{{126^n}}} in the form {{{126^n}}} = 1000*N + 376, and notice that 376*126 = 47376 has the three last digits 376. It implies, by the method of Mathematical induction, that ALL the numbers {{{126^n}}} have the three lest numbers 376, starting from n = 3. </pre> Thus the intermediate statement is proved, and the solution to the problem is completed. 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