5(a-2)(b-2)(c-2)=abc

If a,b, and c are integers then what are all the possible values for abc?

forker Dec 5, 2023

#1**-2 **

Since the original expression is 5 times an integer, abc must be an integer, as well. We can factorize the original expression to get 5(a+2)(b+2)(c+2), so abc=5(a+2)(b+2)(c+2), and we need to find the possible values for 5(a+2)(b+2)(c+2).

If any of a+2, b+2, c+2 is equal to 1, then abc is equal to 5. Since a, b, and c are integers, we know that 5=5(a+2)(b+2)(c+2) is the only case where abc is equal to 5.

If any of a+2, b+2, c+2 is equal to 2, then abc is equal to 10. Since a, b, and c are integers, we know that 10=5(a+2)(b+2)(c+2) is the only case where abc is equal to 10.

If any of a+2, b+2, c+2 is equal to 5, then abc is equal to 50. Since a, b, and c are integers, we know that 50=5(a+2)(b+2)(c+2) is the only case where abc is equal to 50.

If all of a+2, b+2, c+2 are equal to 3, then abc is equal to 125. Since a, b, and c are integers, we know that 125=5(a+2)(b+2)(c+2) is the only case where abc is equal to 125.

Therefore, the possible values for abc are 0,5,10,50,125.

BuiIderBoi Dec 5, 2023

#2**0 **

That is completely wrong as you made a ton of mistakes in your proof and that you somehow got 5(a+2)(b+2)(c+2)=abc when it is actually a-2, b-2, and c-2.

ALLEMPROBS
Dec 7, 2023