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Bunuel
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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
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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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