There is no formula that will generate all the prime numbers less than or equal to 500.
Perhaps the "next best thing" is that there are some formulas that will generate prime numbers for certain values that are plugged in to the formula, but not necessarily all the prime numbers.
For example, the formula n2 - n + 41 will generate prime numbers for all values of n from 0 to 40, but not for all values greater than or equal to 41. But even for values of n that are less than or equal to 40, while the formula will result in a Prime number, it doesn't generate all the prime numbers. The first few prime numbers generated by this formula (for n = 0, 1, 2, 3, 4, and 5) are 41, 41, 43, 47, 53, and 61. But many prime numbers get "skipped over" by using this, or any other, formula.
The cube root of 5000 is approx 17.1 So the numbers 1 to 17 have cubes which are smaller than 5000 that is, there are 17 such numbers.
Here's a start umbers=[True]*5001 index=2 primes=[] while index<5000: multiplier=2 while index*multiplier <= 5000: Numbers[index*multiplier]=False multiplier+=1 index+=1 while Numbers[index]==False and index < 5000: index+=1 for x in range(0,5000): if Numbers[x]==True: primes.append(x) x+=1 print primes
5000 is an even numberEven numbers end with 0, 2, 4, 6, and 8.Odd numbers end with 1, 3, 5, 7, and 9.
On converting between Arabic and Roman Numbers ,we get : 5000 IN ROMAN NUMERALS is written as : _ V
40. Excluding numbers starting with 0.
The two numbers that multiply to get 5000 are 50 and 100. This is because 50 multiplied by 100 equals 5000. These two numbers are factors of 5000, which means they divide evenly into 5000 without leaving a remainder.
The cube root of 5000 is approx 17.1 So the numbers 1 to 17 have cubes which are smaller than 5000 that is, there are 17 such numbers.
5081
To count only numbers that exceed a particular value use the COUNTIF function. In your example, where the numbers to be counted are in the range A1:G1 use the following formula: =COUNTIF(A1:G1,">5000")
5000
Here's a start umbers=[True]*5001 index=2 primes=[] while index<5000: multiplier=2 while index*multiplier <= 5000: Numbers[index*multiplier]=False multiplier+=1 index+=1 while Numbers[index]==False and index < 5000: index+=1 for x in range(0,5000): if Numbers[x]==True: primes.append(x) x+=1 print primes
It is a composite number.
Half of 1000 is 500, so numbers in the range 5000 ± 500 round to 5000 to the nearest 1000 EXCEPT that numbers exactly half way round up. ie numbers greater than or equal to 5000 - 500 = 4500 and less than 5000 + 500 = 5500 will round to 5000 to the nearest 1000. So the largest whole number that rounds to 5000 to the nearest 1000 is 5500 - 1 = 5499.
5000
3
As a product of its prime factors: 2*2*2*5*5*5*5 = 5000
I think that you write it out :in words : five thousand percentin numbers : 5000 %