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HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y|

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HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y| [#permalink] New post   02 Sep 2020, 20:00
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69% (02:02) correct 31% (02:10) wrong based on 291 sessions
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x, y and z are integers such that |x|, |y| and |z| are distinct numbers. If xyz = 36, what is the least possible value of the average (arithmetic mean) of x, y and z?

A. -12
B. -7
C. -19/3
D. 7
E. 38/3


 

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C
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Re: HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y| [#permalink] New post   Updated on: 03 Sep 2020, 20:05
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Given: x, y and z are integers such that |x|, |y| and |z| are distinct numbers.

Asked:If xyz = 36, what is the least possible value of the average (arithmetic mean) of x, y and z?

Since product xyz is positive and he have to find least value of average, 2 large numbers need to be negative and 1 number positive.

Let us take x=-36; y=-1; z=1; but since |y| & |z| are distinct numbers; not feasible.
Let us take x = - 18; y = -2; z = 1; |x|, |y| and |z| are distinct numbers; Average =( - 18 - 2+ 1)/3 = - 19/3

IMO C

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Originally posted by Kinshook on 02 Sep 2020, 20:10.
Last edited by Kinshook on 03 Sep 2020, 20:05, edited 2 times in total.
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Re: HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y| [#permalink] New post   02 Sep 2020, 20:17
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Answer - C

Given - |x|, |y| and |z| are distinct numbers. If xyz = 36

In order to minimize the value of the average, we need to take 2 values as negative , so that xyz remains positive

36 = 2²*3²= -18*-2*1

Average = (x+y+z)/3
= (-18-2+1)/3
= -19/3

Answer Option C

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Re: HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y| [#permalink] New post   03 Sep 2020, 03:05
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For us to have a positive number after multiplying 3 numbers, x, y and z have to either be all positive, or two have to be negative and one needs to be positive for.

As we want the smallest possible mean, we are looking two negatives and one positive.

In this case we want 1 to be our largest number to maintain the positive sign when x, y and z are multiplied together and for the other two numbers to yield 36.

As we cannot use -1, the best pair of will be -2 and -18.

(1)(-2)(-18)= 36

(1)+(-2)+(-18)= -19

Answer is C.

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Re: HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y| [#permalink] New post   03 Sep 2020, 05:49
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Constraints:

1. x, y, z = Integers

2. |x|, |y|, and |z| are not equal (i.e. distinct integers)

Given: xyz = 36

What can we infer from xyz = 36? Since the product of 3 integers is positive, then only two scenarios are possible:

Scenario 1: All three integers = POSITIVE

Scenario 2: Two integers (out of 3) = NEGATIVE

Since we have to find the least possible value of average (x,y,z), scenario 1 can be rejected since it will always give a positive value.

Now, under scenario 2, multiple cases are possible (shown below)

Case 1: x = 1, y = -2, z =-18 ---> Average = (1-2-18)/3 = -19/3

Case 2: x = 1, y = -3, z =-12 --->Average = (1-3-12)/3 = -14/3

Case 3: x = 1, y = -4, z =-9 ---> Average = (1-4-9)/3 = -12/3

Case 4: x = 2, y = -3, z =-6 ---> Average = (1-3-6)/3 = -7/3

Among four cases above, case 1 gives the least value of average.

Option C (correct)

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Re: HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y| [#permalink] New post   04 Sep 2020, 05:58
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Given

    • x, y, and z are integers such that |x|, |y| and |z| are distinct numbers.
    • x y z = 36


To Find

    • The least possible value of the average (arithmetic mean) of x, y, and z.



Approach and Working Out

    • As we need to minimize the number and need to take the different absolute values, we can take it as,
      o x = - 18,
      o y = - 2,
      o z = 1

    • Average = \(\frac{-19}{3}\)

Correct Answer: Option C
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Re: HOT Competition 2 Sep/8PM: x, y and z are integers such that |x|, |y| [#permalink]
04 Sep 2020, 05:58
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