安裝中文字典英文字典辭典工具!
安裝中文字典英文字典辭典工具!
|
- A fast way to find the sum of the sequence $5,5. 5,5. 55,5. 555,5. 5555 . . .
You found a very efficient method For the last step, note that $10^ {-20}$ is utterly negligible in comparison with the other numbers in your expression If you replace it by $0$, the distance from the truth is only $ (50 81)\times 10^ {-20}\:$! In fact, you can then write down the exact answer, since the last digit must be $5$
- How to find a general sum formula for the series: 5+55+555+5555+. . . . . ?
$$5+55+555+5555+\cdots+\overbrace {55\dots5}^ {n\text { fives}}$$ $$=\frac59 (9+99+999+9999+\cdots+\overbrace {99\dots9}^ {n\text { nines}})$$ $$=\frac59 (10^1-1+10^2
- elementary number theory - Mathematics Stack Exchange
Note that $111111$ is divisible by $7$ This follows either from $111111 = \frac {10^6-1} {9}$ where $10^6-1$ is divisible by $7$ by Fermat's little theorem, or from the factorization $111111 = 111 \cdot 1001 = 111 \cdot 7 \cdot 11 \cdot 13$ This implies that $555555$ is also divisible by $7$, hence $$ \underbrace {555 \cdots 5}_ {k \text { times } 5} \mod 7 $$ is periodic with period $6$: if
- Given the numerical succession 5, 55, 555, 5555, 55555. . . Are there . . .
I get the solution (555555555555555555) by using prime factorization But I was wondering if there is a solution using modular arithmetic
- What is the remainder when the number below is divided by $100$?
Hint: $55555555^ {n} = 55^ {n} \mod 100$ In fact for all of those terms, you can simply cross out all digits to the left of the hundreds place in the base
- Find the sum of $5. 5+55. 55+555. 555. . $ up till n terms?
Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges,
- arithmetic - How do you add and subtract repeating decimals . . .
You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
- Longest chain of digits in $\pi$. - Mathematics Stack Exchange
Playing on this site, I reproduced below the longest string of repeated numbers, the position and number of times they appear in the first 200 million of digits of $\pi$ 00000000 172330850 2 11111111 159090113 3 22222222 175820910 1 33333333 36488176 1 44444444 22931745 2 55555555 168743355 1 666666666 45681781 1 777777777 24658601 1 888888888 46663520 1 99999999 66780105 1 The problem is that
|
|
|