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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 26 May 2020, 02:42
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If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5


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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 26 May 2020, 08:41
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Please excuse as I could not present the number line screenshot.

Representing the numbers on number line will provide a visual and easily comprehensible solution.
The terminal points are 0 and -21.
Let's consider the extreme terms only. Any value of x that lies between these two values (inclusive) will provide us the minimum result of the modulus operands.

Since the sequence has even number of terms, therefore, the minimum value will occur for all values of x between the two mid terms (inclusive). So min value will occur at -9, -10, -11, -12.

Note:
1) First arrange the terms of a modulus function in ascending order.
2) If the number of terms in the function is odd, then the min value will occur at the mid term. If the number of terms is even, then the min value will occur for all the values between the two mid terms.
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + Updated on: 28 May 2020, 22:34
Originally posted by sambitspm on 26 May 2020, 03:13.
Last edited by sambitspm on 28 May 2020, 22:34, edited 1 time in total.
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f(x)=|x|+|x+3|+|x+6|+|x+9|+|x+12|+|x+15|+|x+18|+|x+21|
For the equation to be minimum, individual items should be minimized.
As all are absolute values, They can occur only on the inflection points.
As there are 8 terms in AP, it should be at mean point = average of 4th and 5th term
Avg(9,12) = 10.5
But as x is an integer, x should be near the average point
x = -11, -10
So 2 values, B

Correction - I should have checked the expression value before jumping to answer, Apparently expression value on x =-9, -10,-11,-12 remains the same.
Hence 4 values
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 26 May 2020, 03:30
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x= 0 -> 0 + 3 + 6 + 9 + 12 + 15+ 18 + 21
x = - 3 -> 3 + 0 + 3 + 6 + 9 + 12 + 15 + 18
x = - 6 -> 6 + 3 + 0 + 3 + 6 + 9 + 12 + 15
x = - 9 -> 9 + 6 + 3 + 0 + 3 + 6 + 9 + 12
x= - 12 -> 12 + 9 + 6 + 3 + 0 + 3 + 6 + 9
x= - 15 -> 15 + 12 + 9 + 6 + 3 + 0 + 3 + 6
x= - 18 -> 18 + 15 + 12 + 9 + 6 + 3 + 0 + 3
x= = -21 -> 21 + 18 + 15 + 12 + 9 + 6 + 3 + 0

-> 2 values when x = - 9 and x = -12 so that the value of f(x) is minimum
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 26 May 2020, 03:34
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Bunuel
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions
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Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum

there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 28 May 2020, 00:27
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Bunuel
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions

Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum

there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D
Show more

In actual exam there would be enough time to do math for all values. My question is why answer is 4 values and not 2, If 10.5 is midterm, can't we only consider -10 and -11 as values. why we included -9 and -12 . Appreciate reply. Thanks
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 28 May 2020, 09:41
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Bunuel
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions

Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum

there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D

In actual exam there would be enough time to do math for all values. My question is why answer is 4 values and not 2, If 10.5 is midterm, can't we only consider -10 and -11 as values. why we included -9 and -12 . Appreciate reply. Thanks
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Hi Rishab,

The question deals with modulus operands. That basically is the distance (absolute value) of a point with reference to another point. It will always be positive.
Now coming to your question, consider any point between -9 and -12. The sum of the distances (absolute values) from -9 to x and x to -12 will always remain the same.

To simplify it further, consider x=-10. Distance from -9 to -10, d1=1. Distance from -10 to -12, d2=2. Sum of d1+d2=3.
If x=-9, d1=0 and d2=3. Again, sum=3. You can see that the absolute value will always be 3 for any point between -9 and -12 inclusive. Any value of x beyond these points will increase the absolute value and thus, we won't get the minimum value.

Num of integer values from -9 to -12 is 4. Ans: 4

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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 28 May 2020, 13:26
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Bunuel
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5[/textarea]


Are You Up For the Challenge: 700 Level Questions

Point#1: everything is in modulus therefore nothing can be NEGATIVE
Point#2: i.e. Minimum value will be obtained when absolute values are kept to their minimum

there should be only one value of x and that would be the midpoint of equally spaced separations i.e. x = -10.5

But since x is an Integer, therefore, x may be either -10 or -11 or -9 and -12 rendering the same minimum result of the function (f(x)=48)

Hence, 4 values {-9, -10. -11, -12}

Answer: Option D

In actual exam there would be enough time to do math for all values. My question is why answer is 4 values and not 2, If 10.5 is midterm, can't we only consider -10 and -11 as values. why we included -9 and -12 . Appreciate reply. Thanks

Hi Rishab,

The question deals with modulus operands. That basically is the distance (absolute value) of a point with reference to another point. It will always be positive.
Now coming to your question, consider any point between -9 and -12. The sum of the distances (absolute values) from -9 to x and x to -12 will always remain the same.

To simplify it further, consider x=-10. Distance from -9 to -10, d1=1. Distance from -10 to -12, d2=2. Sum of d1+d2=3.
If x=-9, d1=0 and d2=3. Again, sum=3. You can see that the absolute value will always be 3 for any point between -9 and -12 inclusive. Any value of x beyond these points will increase the absolute value and thus, we won't get the minimum value.

Num of integer values from -9 to -12 is 4. Ans: 4

Kudos if it helps!
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Kudos to u. I am finding hard to understand modulus question, It seems your concepts are rock solid in this. Can you suggest me from where to study concept of modulus, mostly 700 level ques of gmat come from this section and i am struggling, Have you found any good material on this forum or anywhere you can suggest.
thanks
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 09 Aug 2021, 06:41
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Hi chetan2u
I am not able to grasp why we are taking 4 points, once I got the midpoint as -10.5, I moved on after marking -10 and -11 as we need integers. Please guide me on this aspect
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 09 Aug 2021, 08:25
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Bunuel
If x is an integer and \(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\), then for how many values of x, f(x) will be minimum?

A. 1
B. 2
C. 3
D. 4
E. 5
Show more


|x| means distance of x from 0
|x+3| means distance of x from -3
|x+21| means distance of x from -21.

\(f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 12| + |x + 15| + |x + 18| + |x + 21|\)

So, we have points
1) -21….-18….-15….-12….-9….-6….-3….0(x)
Say x=0, so distance from each point is 3,6,9,12…,18,21 and the sum will be 3+6+9+….+18+21

2) -21….-18….-15….-12..(x)..-9….-6….-3….0
Say x is right at the middle, that is (0+21)/2 or 10.5, so distance from pair of equidistant points will be:
a) (-21,0): -21 to x to 0 =|-21-0|= 21
b) (-18,-3): -18 to x to -3=|-18-(-3)|=15
c) (-15,-6) : -15 to x to -6=|-15-(-6)|=9
d) (-12,-9) : -12 to x to -9=|-12-(-9)|=3
and the sum will be 21+15+9+3=48

3) Whenever x is in between -9 and -12, both inclusive, all the above cases a, b, c and d will remain same and we will get the sum as 48, the minimum possible.
So integer values are -9, -10, -11 and -12.

4) What happens when x shifts away from middle range 9-12, say x=
-21….-18….-15..(x)..-12....-9….-6….-3….0
a) (-21,0): -21 to x to 0 =|-21-0|= 21….SAME
b) (-18,-3): -18 to x to -3=|-18-(-3)|=15….SAME
c) (-15,-6) : -15 to x to -6=|-15-(-6)|=9…..SAME
d) (-12,-9) : (x or -13 to -12) + ( x or -13 to -9)=1+4=5….Different and more by 2(x-(-12)) or 2*1 here.
and the sum will be 21+15+9+5=50

5) As we move further, the sum will keep going up.

Total 4 values from -9 to -12.

RamseyGooner
Therefore the reason is that when x is from -9 to -12, sum of |x+9| and |x+12| will be constant 3.
However, the moment we move away, say make x as -14, |x+9|+|x+12| becomes 5+3 or 8, while it’s minimum value is 3.
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If x is an integer and f(x) = |x| + |x + 3| + |x + 6| + |x + 9| + |x + 14 Aug 2021, 19:58
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How about this for a solution:

The terms are in ascending order, so take the average of the middle 2 terms: -9 + -12 = -10.5, so you can say that x would be -10 or -11.

Keeping in mind that the terms are in ascending order, you also see that all of them contain CONSECUTIVE MULTIPLES of one number (here that number is 3), so then you include the numbers of the middle two terms, and those numbers are -9 and -12.

Then you have -9, -10, -11, -12. That gives you all four numbers with a shortcut! You just have to remember this rule when you have consecutive multiples that are in ascending (or descending -- but someone check me on that) order!
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