f(n) = |3n - 20| + |11 - n | + |2n - 17|
f(n+1) = |3n + 3 - 20| + |11 - n -1 | + |2n + 2 - 17| = |3n - 17| + |10 - n | + |2n - 15|

Now since both are expressions in modulus, we know by property of modulus that it means absolute value implying that anything inside a modulus can be positive or negative so we can say that:

f(n) {on simplification using all modulus as themselves} = 4n - 26 or f(n) {all modulus negative} = -4n + 26
f(n+1) = 4n - 22 or f(n+1) = -4n + 22

Question asks us for how many integers is f(n) = f(n+1):

Now here we can have 4 different situations:

1. Both positive: 4n - 26 = 4n - 22; both variables get cancelled so not possible
2. Both negative: Same as above both variables will get cancelled so not possible
3. f(n) positive and f(n+1) negative: 4n - 26 = -4n + 22; we can solve for n
4. f(n) negative and f(n+1) positive: Similar expression and same value for n

So, there is just 1 integer that satisfied f(n) = f(n+1) hence answer B

B)One for n = 7

Solve for all values of x -> 20/3,17/2,11
Then solve for all values of x+1->20/3,15/2,17/3,10

You will see that it collides for only one integer value, i.e. 7

It wasn't tough but I had to check manually for multiple values of x. Therefore, it took me longer to solve the problem. Does anyone have a better and quick approach?

How did you choose the values for x?

Random - started from zero and tried the different values of x. I noticed how the value of the whole expression changes with one unit change in the value of x.

anyone have a quick approach to this without trial and error?

Hi, Can you please explain how did you come up with no.7?

Putting values from zero seems like a lengthy option
there must be a more time-saving option

If f(x)=|3x−20|+|11−x|+|2x−17|f(x)=|3x−20|+|11−x|+|2x−17|, for how many integers n is f(n)=f(n+1f(n)=f(n+1) ?


(A) None
(B) One
(C) Two
(D) Three
(E) More than three

Here is a way to solve this problem..
Hope it assists..

Asked: If \(f(x) = |3x - 20| + |11 - x | + |2x - 17|\), for how many integers n is \(f(n) = f(n + 1\)) ?

Values of x for which the modulus changes signs: -
x = 20/3 = 6 2/3 = 6.66
x = 11
x = 17/2 = 8.5

f(n) = f(n+1)
|3n-20| + |11-n| + |2n-17| = |2n-17|+|10-n|+ |2n-15|

f(6) = |18-20|+|11-6|+|12-17| = 2 + 5 + 5 = 12
f(7) = |21-20| + |11-7| + |14-17| = 1 + 4 + 3 = 8
f(8) = |24-20| + |11-8| + |16-17| = 4 + 3 + 1 = 8
f(9) = |27-20| + |11-9| + |18-17| = 7 + 2 + 1 = 10
f(10) = |30-20| + |11-10| + |20-17| = 10 + 1 + 3 = 14
f(11) = |33-20| + |11-11| + |22-17| = 13+0+5 = 18
f(12) = |36-20| + |11-12| + |24-17| = 16 + 1 + 7 = 24

Only f(7) = f(8) = 8

IMO B
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