Q: Give three numbers whose product is about 9000?

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300 x 300

9000x1 1x9000 900x1

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.

30,300,and 1

The product is: 81,000,000

Related questions

300 x 300

3 x 3 x 1000

As a product of its prime factors in exponents: 23*32*53 = 9000

9000x1 1x9000 900x1

20.800838230519041145300568243579 * 20.800838230519041145300568243579 * 20.800838230519041145300568243579 = 9000

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.

(2) x (3) x (1,500.1) will do it.

30,300,and 1

The LCM of the given three numbers is 9000

There are 2,966 numbers between 100 and 9,000 that are divisible by three.

The product is: 81,000,000

9000 = 9000. No other number equals 9000.