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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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TopNewestTo get the \(133!\) in the denominator's place of the sum of fraction,you will have to multiply \(1\) with \(2 \times 3\times4\times....\times132\times133\) and \(2\) with \(1\times3\times4\times5...\times132\times133\) 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,\(\dfrac{1}{67}\) that is,\(1\times2\times3\times...66\times68\times69\times....132\times133\).You basically have to find that mod \(67\).That will be,\(1*2*3*...*66*1*2*3*...66=(66!)^2=(-1)^2=1(\pmod{67}\),this we have from Wilson's theorem!Swapnil Das

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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

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Remainder is 0 as m is 133*133!

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Delete the post.

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