There are n petrol pumps in a circle of circumference L, distributed in an arbitrary manner. Each petrol pump gives $a_i$ quantity of petrol s.t. $\sum_i a_i = L$. You have a car which consumes $x$ quantity for covering distance $x$. You start at an petrol pump with empty tank. Show that there exists an petrol pump starting at which you would be able to go round the circle. (Asked by Rustomji)
Showing posts with label average_trick. Show all posts
Showing posts with label average_trick. Show all posts
Friday, 6 May 2016
Sunday, 24 August 2014
Maths Puzzle #1
Source: CSE blog
Question: A random permutation of integers from 1 to N is arranged in a circle. Prove that there exist k consecutive integer with sum >= k(n+1)/2
Observation: In several puzzles where a comment has to be made about the sum, it is worthwhile to look at the average. Here we ask the question what is the average of each group of k-integers? We know the sum of 1 to n is n(n+1)/2. But in our case every number is part of multiple groups, for ex: number at kth position from start is part of group number 1,2,....k. So every number appears k times, making total sum of all the k-integer groups to be k*n*(n+1)/2. As there are total of n such groups (again count!), average sum of these n groups is k*(n+1)/2, implying there exist a group with value greater than equal to k*(n+1)/2
A stronger statement can also be made if n/k is an integer. There are (n/k) disjoint groups of size at least k. These (n/k) groups can be identified by mentioning the start position. Example, if we start at element 1, the n/k groups will start from 1, k+1, 2k+1,... ((n/k)-1)*k+1. Total sum of these (n/k) groups is n(n+1)/2, thus average sum of each group is n(n+1)/(2*(n/k)) = k(n+1)/2, thus there is at least one group among these n/k groups with sum >= n(k+1)/2
If instead of starting at 1, we start at two, we will get a entirely different set of n/k groups of size k each. In this way we can continue till the start point is k. Thus there will be k different groups of size k each such that there is sum is >=k(n+1)/2
Average trick works in several other puzzles as well. Kaizad rustomji gave me some of those which I'll post later.
Question: A random permutation of integers from 1 to N is arranged in a circle. Prove that there exist k consecutive integer with sum >= k(n+1)/2
Observation: In several puzzles where a comment has to be made about the sum, it is worthwhile to look at the average. Here we ask the question what is the average of each group of k-integers? We know the sum of 1 to n is n(n+1)/2. But in our case every number is part of multiple groups, for ex: number at kth position from start is part of group number 1,2,....k. So every number appears k times, making total sum of all the k-integer groups to be k*n*(n+1)/2. As there are total of n such groups (again count!), average sum of these n groups is k*(n+1)/2, implying there exist a group with value greater than equal to k*(n+1)/2
A stronger statement can also be made if n/k is an integer. There are (n/k) disjoint groups of size at least k. These (n/k) groups can be identified by mentioning the start position. Example, if we start at element 1, the n/k groups will start from 1, k+1, 2k+1,... ((n/k)-1)*k+1. Total sum of these (n/k) groups is n(n+1)/2, thus average sum of each group is n(n+1)/(2*(n/k)) = k(n+1)/2, thus there is at least one group among these n/k groups with sum >= n(k+1)/2
If instead of starting at 1, we start at two, we will get a entirely different set of n/k groups of size k each. In this way we can continue till the start point is k. Thus there will be k different groups of size k each such that there is sum is >=k(n+1)/2
Average trick works in several other puzzles as well. Kaizad rustomji gave me some of those which I'll post later.
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