Maximum Element Value

Let $M$ denote the arithmetic mean. By its definition $M = \sum_{a \in A} a / |A| .$ We can rewrite this summation as $\sum_{a \in A} (a - M) = 0.$ Thu sum splits into two parts -- elements greater than $M$ contribute positively while those less than $M$ contribute negatively. Because we want $A$ to include an element of maximal possible value it follows that the first sum is over just one element $E$ . Therefore we get $E - M = \sum_{a \in A, a < M} (M - a) .$ Since every term of this sum is positive, it's maximized precisely when we include all of the elements from $1$ up to $20$ . Therefore $E = 1 + \cdots + 21$ so the answer is $\bf 231$ .

Assume there are $n$ other elements than $x$ . We want to minimize $n$ to maximize $x$ , so this minimum would be the integers $1,2,\dots,n$ , whose sum is $\frac{n(n+1)}{2}$ . Then we have

$\frac{x+\frac{n(n+1)}{2}}{1+n}=21\implies x=\frac{42+41x-x^2}{2}\le\boxed{231}$

Let the elements of the set be - $a_1, a_2, \ldots a_n$ and $x$ where $x$ is the maximum possible value of the element of set A

Then $\frac{a_1+a_2+\ldots+a_n+x}{n+1}=21$

To maximize $x$ , we need to minimize $a_1+a_2+\ldots+a_n$

Notice, that if the elements of the set are lower than the mean, then the mean also gets down. Thus this is the only way to minimize $a_1+a_2+\ldots+a_n$

Hence, $a_1 = 1, a_2 = 2, \ldots a_n=20$ . So, $n=20$

Therefore, $a_1+a_2+\ldots+a_{20}=1+2+3+4+\ldots+20=210$

(Also taking 21 also to be an element of the set, would make any change)

Thus, $\frac{210+x}{20+1}=21 \Rightarrow x = \boxed{231}$

Suppose A contains n distinct positive integers. In order to contain the maximum value by which the elements of A have an arithmetic mean of 21, A must contain the elements 1, 2, 3, ..., n – 1 alongside some n th element which is the maximum. Consider the cases in which n is small.

• If | A | = 2, then A = {1, 41} since $\frac{1+41}{2} = \frac{42}{2} = 21$ .

• If | A | = 3, then A = {1, 2, 60} since $\frac{1+2+60}{2} = \frac{63}{3} = 21$ .

• If | A | = 4, then A = {1, 2, 3, 78} since $\frac{1+2+3+78}{4} = \frac{84}{4} = 21$ .

• If | A | = 5, then A = {1, 2, 3, 4, 95} since $\frac{1+2+3+4+95}{5} = \frac{105}{5} = 21$ .

The n th term takes the form $21n-\frac{n(n-1)}{2}$ such that the average of the elements of A is given by the expression $\frac{1+2+3+...+(n-1)+(21n-\frac{n(n-1)}{2})}{n}$ . Let $f(n) = 21n-\frac{n(n-1)}{2} = 21n-\frac{n^{2}-n}{2} = 21n-\frac{n^{2}}{2}+\frac{n}{2} = -\frac{1}{2}n^{2}+\frac{43}{2}n$ . The quadratic function $f(n)$ has a maximum at $n= \frac{-\frac{43}{2}}{2(-\frac{1}{2})} = \frac{43}{2} = 21.5$ . Note that f (21.5) = 231.125. Since n is an integer, take n = 21 or n = 22, noting that f (21) = f (22) = 231. Thus, 231 is the maximum possible value for an element of A .

First we have to find the highest number under optimal circumstances, after that we have to proof that it really is a maximum.

the average is 21.

So every element that is under 21 gets the average down.

So 1,2,...,20 are all in A.

Every integer that is higher than 21 gets the average up. So they shouldnt be in A, except for the maximum possible value, which is 21*21-1-2-3-4-5-6-7-8-9-10-11-12-13-14-15-16-17-18-19-20

Adding any other element (except for 21 itself, which is trivial) to A gets the average up. This means that to maintain the average of 21, there has to be another element that is lower than 21 added to A. But since 1,2,..,20 are all in A, this is not possible.

So there cannot be an element added to A that is higher than 231.

If you remove 231 from A, the average goes down, but if you add a number that is higher than 231 to A, the averae will be higher than 21.

Let n be the total number of elements in the set A. As the AM is 21, the sum of all the elements can be $21 \times n$ Let n=1 then maximum value is 21 Let n=2 then maximum value is $2 \times 21 - 1$ since the set A has distinct positive integers. Let n=3 then maximum value is $3 \times 21 - (1+2)$ since the set A has distinct positive integers. Now let's formulate a function in order to find out the maximum possible value for the set A having number of elements n $max(x) = 21 \times n - \frac{(n-1) \times n}{2}$ Now let's find when the value of this function is zero $\Rightarrow 21 \times n - \frac{(n-1) \times n}{2} = 0$ $\Rightarrow 43 \times n - n^{2} = 0$ $\Rightarrow n(n - 43) = 0$ $\Rightarrow n = 0 | n=43$ $\Rightarrow n = 43$ since, A is not a null set. Now, we know that the nmaximum value increases from n = 1 to n = x and decreases from n = x to n=43 The value of x can be found out as it lies midway between 1 an 43. Thus, $x = \frac{43-1}{2}$ $\Rightarrow x = \frac{42}{2}$ $\Rightarrow x = 21$ The maximum possible value can be found at n=x i.e n=21. Applying the function,we get $21 \times 21 - \frac{20 \times 21}{2}$ $=441-210$ $=231$

Assume that there are n numbers in A. Then the total value is 21n. To find the greatest possible value of an element, assume that the other (n-1) numbers are 1, 2, 3..., (n-2), (n-1), the total of which is n(n-1)/2.

Hence the maximum value is 21n-n(n-1)/2 = \frac{43}{2} n - \frac{n^2}{2}. Differentiating this with respect to n, we have \frac{43}{2} - n. Equating this with 0 would give n= \frac{43}{2}, whose value is halfway between 21 and 22. Hence the maximum possible value is possible whether there are 21 or 22 values in A. Now, the maximum value is \frac{43}{2} x 21 - \frac{21^2}{2} = 231.

Let A consist of all distinct positive integers smaller than 21. $\{1,2,3... 20\} \in A$ Then the sum of these elements is $\sum_{i=1}^{20} = \frac{20*21}{2} = 210$ with 20 elements. With the largest single integer the total number of elements will be 21, where the sum is supposed to be (\21*21 = 441). Therefore, the largest a number can be is $441 - 210 = 231$

If u take 2 numbers, namely 1 and 41, u will get 21 as average If u take 3 numbers, namely 1,2 and (41+40), u will get 21 as average. This means that from 1 to 20, u will need a number (41+40+39...+23+22) to get 21 as average.

The sum of the elements of A divided by the number of elements equals 21. Because all elements must be distinct and positive, the maximum possible value would be in a set with all of the positive integers less than the mean. The sum of the numbers 1 through 20 equals 210, so (210 + maxValue) / 21 = 21. Solving the equation for maxValue yields 231.

Let $x_n$ be the maximum value, then $\begin{aligned} \bar{x}&=21\\ \frac{x_1+x_2+\cdots+x_n}{n}&=21\\ x_n&=21n-(x_1+x_2+\cdots+x_{n-1}). \end{aligned}$ To make $x_n$ is a maximum integer, $x_1, x_2, x_3, \cdots$ must be an arithmetic series $0, 1, 2, 3, \cdots$ . We have $\begin{aligned} x_n&=21n-(0+1+2+\cdots+(n-1))\\ &=21n-\frac{n}{2}(0+(n-1))\\ &=-\frac{1}{2}n^2+21n+\frac{1}{2}. \end{aligned}$ To obtain the maximum value of a function, $f(n)$ , use the standard method: $f'(n)=0$ and $f''(n)<0$ . $\begin{aligned} \frac{dx_n}{dn}&=0\\ -n+21&=0\\ n&=21, \end{aligned}$ and $\frac{d^2x_n}{dn^2}=-1<0.$ Substitute $n=21$ to $x_n$ , we obtain $x_n=-\frac{1}{2}(21)^2+21(21)+\frac{1}{2}=\boxed{231}.\\ \text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

21 n = $\sum_{i=1}^n$ $x_{i}$

Let $x_{n}$ be y, to let y become maximum, sum of the numbers before it must be minimum. But given each number is different positive integer, smallest $\sum_{i=1}^n-1$ $x_{i}$ is arithmetic series starting from 1.

Now we have 21n = $\frac{(n-1)(n)}{2}$ + y

y = (43n- $n^{2}$ ) $\frac{1}{2}$

Using differentiation to get maximum y.

43=2n

After trying few times, y really is maximum when n = 21 and 22

y = (43(21)- $21^{2}$ ) $\frac{1}{2}$ = $\boxed{231}$