Question

Sally is tracking her spending. Every day for n days, Sally tracks the exact amount of dollars he spends that day, which is a nonnegative real number but not necessarily an integer. For positive integers i ≤ n, he spends xi dollars on day i. For example, on day 1, he spends x1 dollars. At the end of the n days, he calculates the average daily amount he spends on his food to be averagex = (x1 + x2 + · · · + xn)/n.

a) Prove that on at least one of the n days, Jarett paid at least averagex dollars.
(b) Prove that on fewer than half of the n days, Jarett paid strictly more than 2averagex dollars.

Answer 1

a) Assume that Jarett paid less than avgx in all the days.

x1 + x2 + x3 + ... = n.avgx

Let xi = (avgx - ki) be the amount spend on ith day. Where ki is a positive number

(avgx-k1) + (avgx-k2) + (avgx-k3) + (avgx-k4) + ..... = n.avgx

n.avgx - (k1 + k2 + k3 + k4 +...) = n.avgx

(k1 + k2 + k3 + k4 +...) = 0

since ki is a positive number their sum cannot be 0. So this is a contraditction. Which means that Jarett cannot spend less than avgx in all the days.

The only ase where (k1 + k2 + k3 + k4 +...) = 0 is when k1 = k2 = k3 =... = ki = 0.

So in at least one day he has to pay avgx

b) Let us assume that jarett paid more than 2avgx for m number of days where m = n/2 or more days.

m = n/2+a; a>0

n/2+a+b = n; a+b = n/2

Total spend for these days is m.(2avgx+ji)

= (n/2+a).(2avgx) +(j1+j2+j3+..jm)

= n.avgx + 2a.avgx +  (j1+j2+j3+..jb) > n.avgx

Which is wrong since ji>0. So our assumption that m > n/2 is wrong.

Therefore m < n/2