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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
1 year, 11 months 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 · 1 year, 11 months ago

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Is the answer 1? Adarsh Kumar · 1 year, 11 months ago

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@Adarsh Kumar @Swapnil Das Adarsh Kumar · 1 year, 11 months ago

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@Adarsh Kumar Yes, really thank you Adarsh sir! Swapnil Das · 1 year, 11 months ago

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@Swapnil Das Hey bud,please don't call me sir. Adarsh Kumar · 1 year, 11 months ago

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Thank you Swapnil Das · 1 year, 10 months ago

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Remainder is 0 as m is 133*133! Sayantan Dhar · 1 year, 11 months ago

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@Sayantan Dhar Delete the post. Swapnil Das · 1 year, 11 months ago

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