SOLUTION: Each of the integers h,m and n is divisible by 3. which of the following integers is always dividing by 9? I. hm II. h + m III. h + m + n (A) I only (B) II only (C) I

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Question 208720: Each of the integers h,m and n is divisible by 3. which of the following integers is always dividing by 9?
I. hm
II. h + m
III. h + m + n

(A) I only (B) II only (C) III only (D) II and III only (E) I,II and III
Found 2 solutions by Edwin McCravy, RAY100:
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Each of the integers h,m and n is divisible by 3. which of the following integers is always dividing by 9?
I. hm
II. h + m
III. h + m + n
(A) I only (B) II only (C) III only (D) II and III only (E) I,II and III


You can rule out II by 6 + 9 = 15, 6 and 9 
are both divisible by 3, but 15 is not 
divisible by 9,

You can rule out III by 6 + 9 + 15 = 30, 6, 
9 and 15 are all divisible by 3, but 30 is 
not divisible by 9,

However h and m each have a factor of 3, 
therefore hm must contain two factors of 3.  For 
example if h=6 and m = 12 then hm=6*12=72 
which is divisible by 9.

Therefore, the correct choice is (A), I only

Edwin
Answer by RAY100(1637)   (Show Source): You can put this solution on YOUR website!
Let x,y,z be integers
.
let h=3x,,,m=3y,,,n=3z
.
hm = (3x)(3y) = 9xy,,,,,therefore divisible by 9
.
h+m = 3x +3y=3(x+y),,,,,,therefore not necessarily divisible by 9(only3*3,3*6,3*9,etc)
.
h+m+n= 3x+3y+3z =3(x+y+z) ,,,,therefore not necessarily divisible by 9(only3*3,3*6,3*9,etc)
.
Therefore only (a) is correct answer
.
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