Consider the set $A = \{1,2,3,4,…99,100\}$, i.e. the set of natural numbers less or equal to $100$.
Elements are chosen at random from this set without repetition of elements.
What is the maximum number of elements that can be chosen making sure that their sum does not exceed $1000$?. Clearly justify your answer.
My solution so far is as follows:
Assuming worst case scenario, largest numbers are chosen first so:
$100 + 99 + 98 + … + 91 + 90$, etc.
$100 + 99 + 98 + … + 92 + 91 = (100 + 91) \times 5 = 955$
Adding 90 would make the sum exceed 1000, so the maximum number of elements is 10
However, the answer key says the answer is $11$ elements – is my working out correct or is there simply an error? This is from a Grade 11 maths exam practice.
Best Answer
Your answer is correct. In the worst case scenario, the sum of $11$ numbers will exceed $1000$. So, to be certain that the sum does not exceed $1000$, the maximum number of elements that can be chosen is $10$.