Nick's favourites

Since the four integers have an average of $10$ , the sum of the integers would be $10 \times 4 = 40$ . To maximise one of the integers, we should minimise the other three. This can be done by letting the other three integers be $1$ , $2$ and $3$ respectively, since these are the three smallest distinct positive integers. Hence, we can calculate the largest possible value of his favourite integer to be $40 - (1+2+3) = \fbox{34}$ .

$(a+b+c+d)/4=10$

The least possible values of the three other integers are 1, 2, and 3.

$(a+1+2+3)/4=10$

$(a+6)=40$ a=34 is the greatest possible value for Nick's favorite integer.

We know that the average of $n$ numbers is the sum of those numbers divided by $n$ . Therefore, let us assume $a$ , $b$ , $c$ and $d$ to be the four distinct positive numbers that are Nick's favorite.

So, according to question, we have,

$\frac {a + b + c + d}{4} = 10$

Multiplying $40$ on both sides of the equation,

$a + b + c + d = 4 \times 10$

$a + b + c + d = 40$

Now, let $a$ be the largest number. So, let us assume $b$ to be the smallest number. As the smallest value that $b$ can have is $1$ . Also since the numbers are distinct, and the smallest increment that is possible is $(+1)$ then we can have the following assumptions,

$a = a$

$b = 1$

$c = b + 1 = 1 + 1 = 2$

$d = c + 1 = 2 + 1 = 3$

So now we have,

$a + 1 + 2 + 3 = 40$

$a + 6 = 40$

Subtracting $6$ from both the sides,

$a = 34$

Therefore the maximum possible value of his favorite integer must be $34$ .

to get the largest possible value of his favorite integer put the other three to the least possible positive integer which is 1,2 and 3 so, (1+2+3+x)/4=10 1+2+3+x=40 x=40-6 x=34 :)

let the four distinct positive integers be a, b, c, d

as given, average of these four number is 10

so a+b+c+d/4=10

a+b+c+d=10*4

a+b+c+d=40

now we know that the sum of the four distinct positive integers is 40

now we have to take those numbers whose sum is 40

if we take 34, 1, 2, 3

then the sum of 34, 1, 2, 3 is 40

but we want largest number from them

so the answer is 34

The sum of the four numbers must be $40$ . To maximize the large integer, we want to minimize the smaller ones. The three smallest positive integers are $1$ , $2$ , and $3$ , which have a sum of $6$ , so the last one must be $40 - 6 = \boxed {34}$ .

To have the biggest fourth number, u need to have the smallest three numbers or else the arithmetic mean would be big. As its positive, the smallest three distinct numbers are obviously 1,2,3. Therefore we had (1+2+3+x)/4= 10. Answer 34

Label the four distinct integers as A,B, C, D if A+B+C+D/4=10 then all of them added together equals 40.. So make A, B, C the lowest distinct integers possible (1,2,3) so 40-1-2-3= 34

n+1+2+3=10*4; so, n=34

(a+b+c+d)/4 = 10 a+b+c+d = 40 Let d be the value of his favorite integer, then a b and c must add up to 6 (1+2+3), which is the smallest sum of three positive integers. Thus, d = 40 - 6 =34 ...

the solution is: nick has 4 favorite distinct integer whose average is 10 it means that those product must be 40 the most possibilities is 1+2+3+34 = 40 so the possibilities large integer is 34

Let the numbers be a,b,c,d

According to question:

Average of a,b,c & d = 10.

$\frac{a+b+c+d}{4}$ = 10

Implies:

a+b+c+d = 40

Let 'd' be the maximum number.

If d is max, we should minimize a+b+c accordingly

a,b,c,d are distinct numbers

Hence min value of (a+b+c) = 6;

Hence

d = 40 - 6 = 34

we can say that the sum of all positive integer must be equal to 40 to to get the average of 10 since there's only four positive integer must be add with the highest positive integer that can be satisfy the problem .. in trial and error method in further you will get 34+1+2+3=40 those integer 1,2,3,34 are satisfy to the problem since the average of the sum of the positive integer are equal to 10,,, therefore the answer is "34" since it is the largest positive integer that can be the highest possible integer..

It is clear from the question that the sum of the four numbers= $4 \times 10$ = 40 to have the largest 4th integer we have to take the other three as minimum.they must be distinct and positive so we take them as 1,2 and 3. so the 4th one is 40 - (1+2+3) = 34

Let x be the largest possible value of his favorite integer. In order to maximize x , we need to minimize the other integers that is ,

(1+2+3+x)/4 = 10

6+x=40 x=40-6 = 34