What will be the last three digits of the number $17^{256}$?
This question really intrigued me. First I thought that binomial theorem would help. But miserably failed. The given answer is 681. Please help!
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gt6989b
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2Hint: use repeated squaring and take the result modulo $1000$ at every step. – kccu Feb 03 '20 at 16:02
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1https://math.stackexchange.com/questions/619810/what-is-the-shortest-way-to-compute-the-last-3-digits-of-17256 – lab bhattacharjee Feb 03 '20 at 16:07
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10Does this answer your question? What is the shortest way to compute the last 3 digits of $17^{256}$? – an4s Feb 03 '20 at 16:07
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write the exponent as $3x+1$ or as $400-144$ – Feb 03 '20 at 16:10
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In base $17$, answer is $000$ ... – Déjà vu Feb 03 '20 at 16:18
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It can be easily done by binomial $$17^{256}=(289)^{128}=(290-1)^{128}$$ In the expansion all terms except last three are multiple of 1000. Hence last three digit are by last three digits of last 3 terms i.e Last 3 digit of ${128 \choose 126} (290)^2-{128 \choose 127} (290)+{128 \choose 128}$ $$. $$ =Last three digits of above terms are $(800-120+1)$ $$=681$$
Rajan
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2An answer using this method was linked above 7 minutes prior to your posting. Please check the comments before posting answers to help avoid duplication. – Bill Dubuque Feb 03 '20 at 18:34