# 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
3 years, 10 months ago

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Thank you

- 3 years, 10 months ago

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

- 3 years, 10 months ago

Remainder is 0 as m is 133*133!

- 3 years, 10 months ago

Delete the post.

- 3 years, 10 months ago

- 3 years, 10 months ago

- 3 years, 10 months ago

Yes, really thank you Adarsh sir!

- 3 years, 10 months ago

Hey bud,please don't call me sir.

- 3 years, 10 months ago