Question

the average(arithmetic mean）of five different integers is 30. if the least of these integers is 7 what is the greatest possible value of any of the numbers

Answer 1

Ok basically you think of this problem as an optimisation problem where you are to maximise one of the two remaining integers. You know the arithmetic mean is 30 and 1 of the numbers is 7 and the other 4 are unknown hence:\[\bf \frac{ 7+w+x+y+z }{ 5 }=30 \implies 7+w + x + y +z= 150 \implies w+x + y+z= 143\]We know that each of the numbers w,x,y,z is greater than 7. If we want to maximise, i.e. find the maximum value one of these 4 remaining integers can hold, we will first need to find the SMALLEST sum that the remaining 3 can hold. So 1 integer, the smallest one is 7. If let's say 'w' is the largest, then the remaining 3 x,y,z are all different integers such that each is greater than 7 and the sum x + y + z is the smallest we can get such that x,y,z are different integers greater than 7. Well obviously, if know these are all greater than 7 and x + y + z is the smallest possible sum than \(\bf x+y+z=8+9+10=27\). We are able to conclude this because 27 is the smallest sum we can make with 3 integers, each greater than 7 and we do this by adding up the 3 smallest different integers we get that are all smaller than 7. Now we can solve for 'w', which we assumed to be the largest possible number. If x + y + z = 27 then:\[\bf w+x+y+z=143 \rightarrow w+(27)=143 \implies w = 116\]
@lalaioio

Answer 2

thx for helping

Answer 3

@lalaioio Do you understand? To lay out the steps briefly for you:

First use the info that the smallest integer is 7 and all integers are different. Then realise that if 1 integer takes on the largest possible such that the sum of the 5 integers is 150, then the remaining 3 must have the smallest possible sum which would be 8 + 9 + 10 since these are the smallest 3 integers you can find after 7. Hence the largest possible integer in this sum would 150 - (7 + 8 + 9 + 10) = 116