The average (arithmetic mean) of three different positive integers is $12$ . If the first of these integers is $9$ times the second integer, what is the least possible value of the third integer?

6

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4

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3

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Solution

The correct option is A $6$
Let the three integers be $a, b, c$ . arithmetic mean of those three integers is $\frac{(a + b + c)}{3} = 12$ which implies $a + b + c = 36$ .
Given that $a = 9 b$ .
Therefore the equation becomes $9 b + b + c = 10 b + c = 36$ .
Now here the least possible positive value of $c$ is $6$ , which occurs at $b = 3$ and $a = 27$ .

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The average (arithmetic mean) of three different positive integers is 12. If the first of these integers is 9 times the second integer, what is the least possible value of the third integer?

The correct option is A 6Let the three integers be a,b,c. arithmetic mean of those three integers is (a+b+c)3=12 which implies a+b+c=36.Given that a=9b. Therefore the equation becomes 9b+b+c=10b+c=36.Now here the least possible positive value of c is 6, which occurs at b=3 and a=27.

The average (arithmetic mean) of three different positive integers is $12$ . If the first of these integers is $9$ times the second integer, what is the least possible value of the third integer?