An increasing average

Let's firstly assume that there are $n$ integers in this sequence excluding 35. This allows us to form 2 equations.

$\frac{a_1+a_2+...+a_n+35}{n+1}=53$

$\frac{a_1+a_2+...+a_n}{n}=54$

Rearranging these 2 equations gives us

$a_1+a_2+...+a_n=53n + 18$ $a_1+a_2+...+a_n=54n$

Solving these equations gives us $n = 18$

So, $a_1+a_2+...+a_{18}=972$

Assuming that $a_{18}$ is the largest possible number, we will have to minimise the $a_1$ to $a_{17}$ to maximise $a_{18}$ .

Clearly, $a_{18}$ is maximised when $a_1=1$ , $a_2=2$ ... and $a_{17}=17$ since all numbers are distinct.

$\therefore a_{18}=972-\frac{17\times18}{2}=819$

53-35=18. After removal of a number the average increased by only 1. Hence total number of numbers in list is 18+1=19 numbers inclusive of 35. Having numbers 1-17 as the first 17 numbers+35 has a total of 188. Total is 19*53=1007. Largest number=1007-188=819.

let x be the sum of the sequence without number 35 and n be the total number of distinct value in the sequence.

so, (35+x)/n = 53

we know that x/(n-1) = 54 so x= 54n - 54

plug in x we get 35 + (54n - 54) = 53

so n = 19 the total sum of the numbers is 53 * 19 = 1007

we need to find the HIGHEST POSSIBLE number in the sequence so we have to make other numbers in the sequence have the lowest possible values. But we must use DISTINCT positive integers only.

so...start from 1 (Remember that there are 19 numbers in the sequence including 35) but we stop at 17 and then go to 35 because 19-2 = 17 (2, 1st is 35, 2nd is the highest possible number)

1+2+3+...+17+35+ H(Highest possible number) = 1007 so using 1+2+3+...+n = n+1(n/2)

so 153 + 35 + H = 1007

so H = 819 YEAHHHH!!!!!!!!!!