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5(a-2)(b-2)(c-2)=abc
If a,b, and c are integers then what are all the possible values for abc?

 Dec 5, 2023
 #1
avatar+222 
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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​.

 Dec 5, 2023
 #2
avatar+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

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