Choosing Numbers That Sum to 50

suppose 5 numbers will be like this (1+2+3+4+X) and their average is (1+2+3+4+X)/5=10 after solving the equation - x will be 40

Let's call these five distinct positive integers as X, Y, W, Z and K, if the average is 10 we know that (X+Y+W+Z+K)/5 equals 10, then X+Y+W+Z+K = 50. Let's imagine that the largest possible is K, then X, Y, W, Z must be the smallest possible and, remember, distinct and different of zero. Therefore: x = 1, y = 2, w = 3, z = 4. => 1+2+3+4+K = 50 => K = 40

For example we have five num ber : a, b, c, d, e Everage : $=> \frac{a + b + c + d + e}{5} = 10$ We have : a, b, c ,d, e are distinct => we choose a, b, c ,d are smallest numbers : 1, 2, 3, 4 $=> 1 + 2 + 3 + 4 + e = 50$ $e = 40 (is largest positive number)$

(x1+x2+x3+x4+x5)/5 = 10

so, x1+x2+x3+x4+x5= 50

And we want that x5 be the largest so

1+2+3+4+x5= 50

so x5= 40

1 + 2 + 3 + 4 + 40 = 50

50/5 = 10

AVG = 10 Hence ; --> (a1 + a2 + a3 + a4 + a5 ) / 5 = 10 --> (a1 + a2 + a3 + a4 + a5 ) = 50 Now try to minimize the value of FIRST FOUR TERMS , SO THAT THE 5TH TERM BECOMES MAXIMUM..!! Also Note that they must be DISTINCT..!! So, 1 + 2 + 3 + 4 + 49 = 50 ..... So, Ans = 49

Let: A,B,C,D,E be 5 arbitrary positive integers; (A+B+C+D+E)/5=10 --> (A+B+C+D+E)=50 --> say, A be the largest possible integer value, A= 50-(B+C+D+E), where B+C+D+E are counting from the smallest positive integers, thus (1+2+3+4); result A=50-(1+2+3+4)= 40 .

Let the five numbers be $a_{1}$ , $a_{2}$ , $a_{3}$ , $a_{4}$ and $a_{5}$ .

Given, $\frac{a_{1} + a_{2} + a_{3} + a_{4} + a_{5}}{5} = 10$

$a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 50$ .......... (i)

Let us say $a_{5}$ is the largest possible integer. Since, we want $a_{5}$ as largest, we have to consider remaining numbers as smallest as their sum is constant.

But, the numbers are all distinct and positive, hence we choose $a_{1} = 1$ , the smallest positive integer,

$a_{2} = 2$ , $a_{3} = 3$ , $a_{4} = 4$ .

When we substitute these values in equation (i), we get

$a_{5} = 40$

That's the answer!

a+b+c+d+e=50

since a,b,c,d,e aren't the same then

one of them must be the smallest positive integer(read=1)

the rest of them aren't 1 right,then, choose other small positive integers(read=2,3,4) and then e is the largest integer

e=50-(1+2+3+4)=40

sorry for my grammar

As average is 10(of 5 nos.) the sum is 50.

If you select 4 smallest nos. u will get 5th the largest possible.

Therefore, 1,2,3,4 and x are the nos.

Here, x turns out to be 40 which is the largest possible!

Average of distinct 5 positive integers is 10 i.e. Total of them is 5x 10 = 50 sum of the least distinct 4 positive integers = 1+2+3+4=10. so, the remaining 5th positive integer can be 50-10=40 so, largest possible integer can be 40

sum of no.s /5=10 sum of no.s =50 four mini no. are 1,2,3,4 whose sum is 10 remaining largest no.=50-10=40

As there are five distinct number and their average is 10 then we can write, ( a + b + c + d + e ) / 5 = 10 or, a + b + c + d + e = 50 or, 40 + 4 + 3 + 1 = 50 this is the combination where 40 is largest

First let's think about some small integers. For one to be the largest, others need to be small. Sum is 1+2+3+4+40

If (a+b+c+d+e)/5=10 when (a,b,c,d,) >= 1 & where no two of the numbers are same , and a<b<c<d<e then (a+b+c+d+e)=5*10, (a+b+c+d+e)=50, If a=1,b=2,c=3,d=4 as the lowest positive integers possible in (a+b+c+d+e)=50 then, (1+2+3+4+e)=50, 10+e=50, e=50-10 e=40

(a+b+c+d+e)/5=10 <=> 5 (a+b+c+d+e)/5=10 5 a+b+c+d+e=50 four smallest distinct positive integers 1, 2, 3 and 4 1+2+3+4+e=50 10+e=50 <=> -10+10+e=50-10 e=50-10 e=40

5 intergers can be represented as a, b, c, d and e consequently the problem could be re-written as (a+b+c+d+e)=50, another words e=50-(a+b+c+d), smallest postive intergers so the maximum is left over, is 1, 2, 3, 4, they add up to 10. the Largest e is 40

Since the no. required is greatest the other no. should be the least possible. but they shuold be distinct too. Average required is 10 =>for five nos. the sum of five no.= 50.(i.e. 5 * 10) since lowest distinct nos can be 1 2 3 and 4 add up them we get sum 1+2+3+4 = 10. Subtract from the total of five nos. we get greatest no. => 50-10 = 40

average is 10 of 5 number so sum is =50 , so we need to find out the sum five distinct numbers which would be 50. & the condition is they must be distinct . so we have to choose one maximum number & other 4 numbers. as we know that sum of 1:10 numbers is 55. so we need to find out the least sum of 4 numbers , which leads to (1+2+3+4=10). subtract this number from total sum 50 which is =50-10=40 . ie our largest possible integer.

multiply 10 by 5 to get 50 (the sum of the five numbers). 1, the smallest positive number, is one of the numbers. I forgot: choose four least distinct positive numbers to get largest number. then 2,the next to 1. 3, next to 2. and 4, next to 3. 1, 2, 3 and 4 are the least positive distinct numbers. add them the result is 10. subtract 10 from 50. then you get 40, largest number of the five numbers that has an average of 10. paragraph 1

equation:

(a+b+c+d+e)/5=10

[(a+b+c+d+e)/5] 5=10 5 multiply both sides by 5 to cancel "divided by 5"

(a+b+c+d+e)=50

1+2+3+4+e=50 think of four least positive distinct problems they are 1,2,3,4

10+e=50 combine like terms

10+e-10=50-10 subtract 10 to cancel positive 10

e=40 viola!!!!

50/5 = 10 (Given Average Number) 50 = 40+4+3+2+1 (5 Distinct numbers = 5 Different numbers)