If x and y are positive integers 1800x = y^3, what is the minimum poss

\(2^3 * 3^2 * 5^2 * x = y^3\)

Hence, the minimum value of y will be 3*5 = 15

Thus, answer will be (C) 15

1800= (2^3)(3^2)(5^2)
1800x= y^3 ==> x= 3*5 =15
C

We are given that 1800x = y^3, which means that 1800x is a perfect cube. We must remember that all perfect cubes break down to unique prime factors, each of which has an exponent that is a multiple of 3. So let’s break down 1800 into primes to help determine the minimum value of x.

1800 = 18 x 100 = 2 x 9 x 10 x 10 = 2 x 3 x 3 x 2 x 5 x 2 x 5 = 2^3 x 3^2 x 5^2

In order to make 1800x a perfect cube, the smallest value of x is 3^1 x 5^1, so that 1800x = (2^3 x 3^2 x 5^2) x (3^1 x 5^1) = 2^3 x 3^3 x 5^3, which is a perfect cube.

Thus, x = 3^1 x 5^1 = 15.

Answer: C

x y + int

1800 = 18*2^2*5^2
1800 = 2^2*3*5^2*6
1800 = 2^3*3^2*5^2
2^3*3^2*5^2 * x = Y^3

missing 3 and 5

3x5=15
so c

Hi, sry I could not solve after 1step. Pls explain.
Thank you

Since y is an integer, then y^3 is a perfect cube. Therefore, 1800x must also be a perfect cube. For \(1800x = 2^3*3^2*5^2*x\) to be a prefect cube x must complete the powers of 3 and 5 to the closest multiple of 3, thus the least value of x is 3*5.

Hope it's clear.

If x and y are positive integers 1800x = y^3, what is the minimum possible value of x?

This question tests the rules of exponents. This asks for the minimum value of x so that 1800*x becomes a perfect cube.

The quickest way to solve this is to prime factories 1800. The result is 3^2*5^2*2^3. The exponents of prime factors of perfect cube number will always be closest multiple to 3. Here prime factors 3 & 5 do not have exponents which are multiples of 3.

Since we are asked the minimum number for x, we have to use the closest multiple in the equation, i.e. will be 3. Hence by multiplying another 3 & 5. The prime factors of 3&5 will have exponents which are multiple of 3 and thus 3*5 is 15 the minimum value possible for X is 15.

Option C 15