Please post solution to the given problem. I am not able to figure out the solution, so please help. Thank You.
What is the remainder of \(m\) satisfying \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{133}=\frac{m}{133!}\] upon division by \(67 ?\)
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To get the 133! in the denominator's place of the sum of fraction,you will have to multiply 1 with 2×3×4×....×132×133 and 2 with 1×3×4×5...×132×133 and so on till 133.Then observe that m will be the sum of these products,now observe that every product,except for one,will have 67 in it.The one which will not have it will be,671 that is,1×2×3×...66×68×69×....132×133.You basically have to find that mod 67.That will be,1∗2∗3∗...∗66∗1∗2∗3∗...66=(66!)2=(−1)2=1((mod67),this we have from Wilson's theorem!Swapnil Das
Is the answer 1?
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@Swapnil Das
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Yes, really thank you Adarsh sir!
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Thank you
Remainder is 0 as m is 133*133!
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