# Puzzle 29 | (Car Wheel Puzzle)

A car has 4 tyres and 1 spare tyre. Each tyre can travel a maximum distance of 20000 miles before wearing off. What is the maximum distance the car can travel before you are forced to buy a new tyre? You are allowed to change tyres (using the spare tyre) unlimited number of times.

Answer: 25000

Divide the lifetime of spare tire into 4 equal part i.e., 5000 and swap it at each completion of 5000 miles distance.

Let four tyres be A, B, C and D and spare tyre be S.

5000 KMs: Replace A with S. Remaining distances (A, B, C, D, S) : 15000, 15000, 15000, 15000, 20000

10000 KMs: Put A back to its original position and replace B with S. Remaining distances (A, B, C, D, S) : 15000, 10000, 10000, 10000, 15000

15000 KMs: Put B back to its original position and replace C with S. Remaining distances (A, B, C, D, S) : 10000, 10000, 5000, 5000, 10000

20000 KMs: Put C back to its original position and replace D with S. Remaining distances (A, B, C, D, S) : 5000, 5000, 5000, 0, 5000

25000 KMs : Every tyre is now worn out completely.

All tyres are used upto their full strength.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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**Related Amazon interview question **

There are n pencils, each having l length. Each can write 4 kilometers. After writing 4 kilometers it has l/4 length. One can join 4 pencils which are having l/4 length and can make 1 pencil. One can’t make pencil of pieces if remaining pieces are 3 or 2 or 1 in number but one can include these remaining pieces whenever needed.

Write a relation independent of l, length of given pencil, for how much one can write from n pencils.

**Examples:**

Input :4Output :20

**Approach: **

Suppose We use 3 pencils that will 12 and generate 3 used pencils, now if the remaining pencils are greater than zero at least 1 unused pencil can be used with those 3 unused to write 4 and that will generate 1 more unused pencil. This will keep repeating.

if(n-3 >= 1){ f(n) = f(n-3) + 12 + 4 } else{ // Used pencil that cannot be used f(n) = 4 + 4 } Above relation can be optimized in O(1) // x is max no of time we can subtract 3 without n-3 <= 3

n - 3 + x <= 3 x > (n-3)/3 i.e. n/3 - 1 if it divides exactly else n/3

`// C++ program to find relation independent of l ` `// length of given pencil, for how much one ` `// can write from n pencils. ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function to find no of pencils ` `int` `count(` `int` `n) ` `{ ` ` ` `int` `x = (n / 3) - 1; ` ` ` `if` `(n % 3) { ` ` ` `x++; ` ` ` `} ` ` ` `return` `(4 * x + 4 * n); ` `} ` ` ` `// Driver function ` `int` `main() ` `{ ` ` ` `int` `n = 5; ` ` ` `cout << count(n) << endl; ` `} ` |

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**Output:**

24

Recursively it can be written as:

int count(int n){ if(n<4){ return (n*4); } else{ return (16+count(n-3)); } }

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