# Number Theory Problem

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
6 years, 1 month 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

- 6 years, 1 month ago

- 6 years, 1 month ago

- 6 years, 1 month ago

Yes, really thank you Adarsh sir!

- 6 years, 1 month ago

Hey bud,please don't call me sir.

- 6 years, 1 month ago

Thank you

- 6 years, 1 month ago

Remainder is 0 as m is 133*133!

- 6 years, 1 month ago

Delete the post.

- 6 years, 1 month ago