20.800838230519041145300568243579 * 20.800838230519041145300568243579 * 20.800838230519041145300568243579 = 9000
As a product of its prime factors in exponents: 23*32*53 = 9000
(2) x (3) x (1,500.1) will do it.
3 x 3 x 1000
The number 9000 can be expressed as the product of its prime factors, which are 2^3 * 3^2 * 5^3. In terms of mathematical operations, 9000 can be reached by multiplying 9 by 1000, or by adding 8000 and 1000. Additionally, 9000 can be represented as the sum of consecutive numbers, such as 4499 + 4501.
The product of 5 x 6 x 7 is 210.
As a product of its prime factors in exponents: 23*32*53 = 9000
(2) x (3) x (1,500.1) will do it.
9000x1 1x9000 900x1
3 x 3 x 1000
Prime factorisation of 9000 = 2*2*2*3*3*5*5*5 Well 9000 can be expressed as the product of (2*2*2)*(3*3)*(5*5*5) or 8*9*125 So, 9000 can be expressed as the product of 8,9 and 125. It can be expressed as 2*(2*2*5*5)*(3*3*5) i.e. 9000 = 2*100*45 By using the method of Prime factorisation, we can find many cases in which product of 3 numbers is 9000. We can also express it as 900*(1/10)*100, which cannot be done using P.F. of 9000. We can find a large number of cases in which product of three numbers is 9000. We can find numbers by this method: Assume any three variables and put their product equal to 9000 i.e. x*y*z = 9000 where x,y and z are real numbers. Put any value of x, say 34 and any value of y, say 23, then on putting the values we can find the value of z i.e. 34*23*z = 9000 782*z = 9000 z = 11.5089 So, this method works better than Prime Factorisation.
The number 9000 can be expressed as the product of its prime factors, which are 2^3 * 3^2 * 5^3. In terms of mathematical operations, 9000 can be reached by multiplying 9 by 1000, or by adding 8000 and 1000. Additionally, 9000 can be represented as the sum of consecutive numbers, such as 4499 + 4501.
9000 to the nearest thousand, 10000 to the nearest ten thousands and 0 to the nearest million.
There are many possible answers to this question ... for example, 9000, 2000, and 111. Can you be more specific?
The product of 5 x 6 x 7 is 210.
As a product of its prime factors: 3*3*3*11 = 297
There is no set of 3 consecutive odd numbers whose product is 6,873. However, there is a set of 3 consecutive odd numbers whose sum is 6,873: 2289, 2291, and 2293.
To find the fraction of four-digit natural numbers with an even product of their digits, we first note that a four-digit number ranges from 1000 to 9999, giving us a total of 9000 four-digit numbers. The product of the digits is even if at least one digit is even. The only case where the product is odd is if all four digits are odd. The odd digits are 1, 3, 5, 7, and 9, offering 5 choices for each digit. Thus, the total odd-digit combinations for four-digit numbers is (5^4 = 625). Therefore, the number of four-digit numbers with an even product is (9000 - 625 = 8375). The fraction is then ( \frac{8375}{9000} = \frac{335}{360} ), which simplifies to approximately ( \frac{67}{72} ).