# 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
5 years, 8 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Thank you

- 5 years, 8 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

- 5 years, 8 months ago

Remainder is 0 as m is 133*133!

- 5 years, 8 months ago

Delete the post.

- 5 years, 8 months ago

Is the answer 1?

- 5 years, 8 months ago

- 5 years, 8 months ago

Yes, really thank you Adarsh sir!

- 5 years, 8 months ago

Hey bud,please don't call me sir.

- 5 years, 8 months ago