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 mm satisfying 11+12+13++1133=m133!\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{133}=\frac{m}{133!} upon division by 67?67 ?

Note by Swapnil Das
4 years, 5 months ago

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Thank you

Swapnil Das - 4 years, 5 months ago

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To get the 133!133! in the denominator's place of the sum of fraction,you will have to multiply 11 with 2×3×4×....×132×1332 \times 3\times4\times....\times132\times133 and 22 with 1×3×4×5...×132×1331\times3\times4\times5...\times132\times133 and so on till 133133.Then observe that mm 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,167\dfrac{1}{67} that is,1×2×3×...66×68×69×....132×1331\times2\times3\times...66\times68\times69\times....132\times133.You basically have to find that mod 6767.That will be,123...66123...66=(66!)2=(1)2=1((mod67)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 - 4 years, 5 months ago

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Remainder is 0 as m is 133*133!

Sayantan Dhar - 4 years, 5 months ago

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Delete the post.

Swapnil Das - 4 years, 5 months ago

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Is the answer 1?

Adarsh Kumar - 4 years, 5 months ago

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@Swapnil Das

Adarsh Kumar - 4 years, 5 months ago

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Yes, really thank you Adarsh sir!

Swapnil Das - 4 years, 5 months ago

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@Swapnil Das Hey bud,please don't call me sir.

Adarsh Kumar - 4 years, 5 months ago

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