Can Oracle store a decimal number with trailing zeros like 19.00?
create table xyz_flattest (col_one number, col_two float);insert into xyz_flattest values (1,1.344);insert into xyz_flattest values (2,19.00);insert into xyz_flattest values (2,19.001);commit;But if you run select to_char(col_two,'9999.99999') from xyz_flattestSo if you give me the exact need i can suggest.
Does Pi have an infinite number of integers after the decimal point?
To be exact, all numbers have an infinite number of digits after the decimal point, as, for instance, the number 1 can be written as [math]1.00000...[/math]. You were probably asking whether [math]\pi[/math] has a repeating decimal representation starting from some point in its decimal representation or not.It so happens that any number having a repeating decimal representation is a fraction (technically a rational number), and every fraction has a repeating decimal representation. However, it can be proved that [math]\pi[/math] is not a rational number, i.e. is an irrational number. This was first proved by Lambert in 1761 (see Proof that π is irrational). Thus, its decimal representation does not ever repeat itself, and this means that it cannot be, for example, equal to 3.14159265359000000.... = 3.14159265359 exactly.
Since there are infinitely many primes, then every number ending in 1,3,7, and 9 should have a non-zero probability of being prime. Would this imply a solution to the twin primes conjecture?
If there are infinitely many primes,You mean, since there are infinitely many primes. There’s no “if” here.then every number ending in 1,3,7, and 9 should have a non-zero probability of being prime.How so?We can rigorously define a magnitude which can be interpreted as the probability of a random natural number ending in [math]1[/math] to be prime (in an appropriate sense), but it comes out as [math]0[/math]. On the other hand, the likelihood that a random prime number will end on [math]1,3,7[/math] or [math]9[/math] is [math]1/4[/math].(Those magnitudes are limits of simple proportions, meaning we look at the relevant ratios among the first [math]N[/math] integers and then push [math]N[/math] to infinity.)Would this imply a solution to the twin primes conjecture?Absolutely not. Why would it imply such a thing?Consider numbers ending in [math]21[/math], [math]43[/math], [math]87[/math] or [math]09[/math]. There are infinitely many such numbers, and it is true that a random integer ending in [math]1,3,7[/math] or [math]9[/math] has a well-defined positive probability of being one. But no two such numbers are “twins” in the sense that they are 2 apart.Numbers aren’t random events, and primes aren’t either. If they had been, we could have attempted to estimate the probability of adjacent numbers to be prime, and if it came out positive, we could have shown that infinitely many such pairs must exist. In some ways, prime numbers do exhibit random-like properties, but the emphasis is on “like”. They aren’t actually random, and there’s no way to draw conclusions about twin primes from such probabilities.
How many zeroes are there in 100 million?
100 million = 100, 000,000 ;So, (As Place Holders, the figure has 8 (Eight) Zeroes.But, as Values, if you ask how many Zeroes are there in a Million ?How many Zeroes do you have to add (or How many times do you have to add Zeroes) to make 100 Million ?It should be INFINITY or UNDEFINED or Something like that.Someone with more Math Expertise than me, is Certainly Welcome to add.
How do I round off a float to 2 decimal points in C, like 3.01943 to 3.02?
If you have to round input floating point number then for that the format is“%wf” where w = Integer number for total width of data.(including the digit before and after decimal and decimal itself)Ex- scanf(“%7f”,&x) For x = 3.01943 Result = 7.8If you have to round output floating point number then for that the format is“%w.nf” where w = Integer number for total width of data.(including the digit before and after decimal and decimal itself) and n is the number of digit printed after decimal point.Ex- printf(“x=%7.2f”,x) For x = 3.01943 Result = 3.02You can see some programsC Tutorial - printf, Format Specifiers, Format Conversions and Formatted OutputInput/Output Formatting in CUsing some math functions
Factorial (function): How many zeroes are there in 100!?
-A2A-Well, this is really a tough one to count manually! However, if your question is modified a little to "How many trailing zeros are there in 100!?" then we can definitely work it out. The only trick is to remember that a zero is created by the factor pair (2,5) as 2*5 = 5*2 =10. Therefore, it is necessary to count the number of 5s and 2s. One could possibly argue that the count of 5s alone or 2s alone might give us the result. But that would actually cause a problem by giving us an exaggerated or underestimated count of 0s. As it is possible that the count of 2s is greater or lesser than the count of 5s. So, the count of 0s in the end result would rather depends on the minimum count between the count of 2s and the count of 5s.100! represents the product of all the natural numbers up to and including 100.Clearly, the count of 2s is more than the count of 5s. (Let me know if you could visualize it!)To count the number of 5s, we must note that there are 20 multiples of 5 among the natural numbers up to and including 100. Only these multiples can contribute 5!!It would be worth to note that 16 multiples of 5 would contribute one 5 each except 25, 50, 75 and 100. The reason to expel out these 4 multiples of 5 from the first group is that each of these multiples is a multiple of 25 too. That means each of these multiples contribute two 5s each.So, the total count of 5s = 16*1 + 4*2 = 16 +8 = 24Therefore, the total number of trailing 0s in 100! is = 24Otherwise the total count of 0s in 100! is 30 if all the 0s , not just the trailing 0s, are to be counted.100! = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000I hope it helps!
How can I evaluate √3 up to two places of decimals?
Using the division method, we may find the value of √3; In this method,Step 1: we place a dot and zeros (number of zero depends upon places upto which square root is to be found)Step 2: make pairs and it is done in two steps (start from decimal and go right making pairs and then come back to decimal go right to left of decimal as shown).Step 3: choose a number and square it such that it is less than or equal to 3 And then subtract it.Step 4: add with itself(we add the number with the number was divided, in second part we can see we take 27.Here we multiplied 27*7 hence we add 7 to 27 and get 34). The divisor which in this case was 1 and hence we place 2 on the left as shown.Step 5: As you see the dividend is 200(we bring the next pair after subtracting 3-1).Step 6: As we can see we had added our divisor to 2 and we have 200 to divide. So we choose a number say "X" such that 2X*X is less just less than or equal to dividend.Step 7: continue this process until desired (next time we double 27 and do 54X*X).Therefore, √3 = 1.7324 ⇒ √3 = 1.73
How can we find the longest continuous subsequence of 0's in binary representation of an integer in O(log n) time complexity?
To solve this problem, notice that if we have the binary representation of the number, we can simply iterate through the bits and find the longest substring of zeros in [math]O(L)[/math] where [math]L[/math] is the length of the binary string. Also, note that [math]L = \lceil \log{(n+1)} \rceil[/math]A simple approach to solve this problem is to keep a count of the longest substring of zeros so far and update if we find a longer one. I'll simply find the length of this substring, but finding the start and end position is also trivial.Here's a pseudocode of the algorithm:maxLength = 0currentLength = 0while n >0 if( n mod 2 = 1) currentLength = 0 else currentLength = currentLength + 1 if (currentLength > maxLength) maxLength = currentLength n = n/2 A simple Ruby (programming language) implementation is given belown = 313131532332342423585349487428 #sample numbermaxLength = 0currentLength = 0while n >0 if( n%2 == 1) currentLength = 0 else currentLength = currentLength + 1 end if (currentLength > maxLength) maxLength = currentLength end n = n/2endputs maxLengthThe algorithm takes [math]O(\log{n})[/math].