Andrew's parents follow a regular schedule for taking care of their car. They change the oil every 3 000 km, rotate the tyres every 10 000 km and replace the wiper blades every 15 000 km. After how many kilometres will they first have to change the oil, rotate the tyres and replace the wiper blades all at once? (5)
Explanation (if necessary):
To solve this problem, we can define three arephmetic progressions:
1. The first progression А1 is an "oil change" with a step d = 3 000 km;
2. The second progression А2 is "tire rotation" with a step of d = 10 000 km;
3. The third progression А3 is "replacement of wiper blades" with a step of d = 15 000 km;
Since we are interested in the moment (distance traveled, km) when the progressions first intersect, we need to determine this point (distance).
Let's start with the A1 progression, where the biggest step is (15 000 km).
Any (n-th) term of the progression can be calculated using the general term formula:
an= a1 + d*(n-1)
If member 15 000 is included in sequence A2, then:
15 000 = 10 000 + 10 000 (n-1), where n is a natural number.
10 000 (n - 1) = 5 000
n - 1 = 0.5
n = 1.5 (not a natural number).
This means that at the point 15 000 km and all three progressions will not intersect.
We take the second term of the progression A3 = 15 000 + d = 30 000 km and first check whether this number is included in the second progression:
30 000 = 10 000 + 10 000 (n-1), where n is a natural number.
10 000 (n - 1) = 20 000
n - 1 = 2
n = 1 is a natural number (!)
So at the point of 30 000 km the third and second progressions intersect.
Similarly, check this intersection point for the first progression A1 with a step of d = 3 000 km:
30 000 = 3 000 + 3 000 (n-1), where n is a natural number.
3 000 (n - 1) = 27 000
n - 1 = 9
n = 8 is a natural number (!), which means that all three progressions will intersect at the point of 30 000 km.
Comments
Leave a comment