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If \(a, b,\) and \(c\) are all integers divisible by \(3\) and \(a > b > c > 0,\) then which of the following must be

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If \(a, b,\) and \(c\) are all integers divisible by \(3\) and \(a > b > c > 0,\) then which of the following must be

by Gmat_mission » Thu Oct 08, 2020 4:01 am

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If \(a, b,\) and \(c\) are all integers divisible by \(3\) and \(a > b > c > 0,\) then which of the following must be divisible by \(3?\)

I. \(a+b+c\)

II. \(a-b+c\)

III. \(\dfrac{abc}9\)


A) I only

B) III only

C) I and II

D) I and III

E) I, II, and III

Answer: E

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Re: If \(a, b,\) and \(c\) are all integers divisible by \(3\) and \(a > b > c > 0,\) then which of the following must b

by Scott@TargetTestPrep » Mon Oct 12, 2020 4:20 am
Gmat_mission wrote: ↑
Thu Oct 08, 2020 4:01 am
 If \(a, b,\) and \(c\) are all integers divisible by \(3\) and \(a > b > c > 0,\) then which of the following must be divisible by \(3?\)

I. \(a+b+c\)

II. \(a-b+c\)

III. \(\dfrac{abc}9\)


A) I only

B) III only

C) I and II

D) I and III

E) I, II, and III

Answer: E

Solution:

Since a, b, and c are all divisible by 3, any sum and difference combo of the three integers will be also divisible by 3. So I and II are true. Furthermore, since each of a, b, and c is divisible by 3, their product is divisible by 3^3 = 27. That is, abc is divisible by 27. Since abc is divisible by 27, abc/9 is divisible by 3 (notice that (abc/9)/3 = abc/27 is an integer). So III is also true.

Answer: E

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