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Number Theory Problem

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 \(67 ?\)

Note by Swapnil Das
2 years ago

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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 \(2 \times 3\times4\times....\times132\times133\) 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 Adarsh Kumar · 2 years ago

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Is the answer 1? Adarsh Kumar · 2 years ago

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@Adarsh Kumar Yes, really thank you Adarsh sir! Swapnil Das · 2 years ago

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@Swapnil Das Hey bud,please don't call me sir. Adarsh Kumar · 2 years ago

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Thank you Swapnil Das · 2 years ago

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Remainder is 0 as m is 133*133! Sayantan Dhar · 2 years ago

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@Sayantan Dhar Delete the post. Swapnil Das · 2 years ago

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