Ever notice how there are practically no nines in pi? it's true. what's the longest continuous sequence not containing a nine in
the first million digits? (i want the sequence itself.)
1 answer:
Not sure if it is a school problem, but it is of interest to find the answer.
WithIn the first million digits of Pi, there are
5118-9 's
794-99 sequences
79-999 sequences
6-9999 sequences
2-99999 sequences
1-999999 sequence
Feel free to check the accuracy of above counts.
The 6-digit sequence starts at the 762th digit after the decimal point.
Digits of Pi including the first 800 digits after the decimal are included below:
3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914127372458700660631558817488152092096282925409171536436789259036001133053054882046652138414695194151160943305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912983367336244065664308602139494639522473719070217986094370277053921717629317675238467481846766940513200056812714526356082778577134275778960917363717872146844090122495343014654958537105079227968925892354201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859
You might be interested in
Answer:
There are 232 students at the school.
Step-by-step explanation:
Let s represent the number of students at the school. We are told ...
174 = 75% × s
174/0.75 = s = 232 . . . . . divide by the coefficient of s
There are 232 students at the school.
2a/5b = 6
(2a-5b)/5b could be written as: (2a/5b) -(5b/5b)
Repace by its equivalent value (6) -(1) = 5
What is the ribbon thing for?
Answer:
The variable is z
Step-by-step explanation:
Answer = (7p) + (7q) + 63