Show that from a group of seven people whose (integer) ages add up to 332 one can select three people with the total age at least 142. [closed]

I need help with this problem, and I was thinking in this way: $$ x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} = 332 $$ and I need to find three of these which sum is at least 142.

But I don't know what next. Any help would be appreciated

Solution 1:

Suppose that any $3$ chosen people have at most a total age of $141$. Then, $x_1 + x_2 + x_3 \le 141$ and so is $x_4 + x_5 + x_6$. This indicates that $x_7 \ge 50$.

Repeating the same process exchanging the roles of $x_i$'s yields $x_i \ge 50$ for any $i$. Which leads to a contradiction.

Solution 2:

You can form $\binom{7}{3}$ triplets, i.e. 35 triplets. Every men participates in $\binom{6}{2}$ triplets, i.e 15. So the sum of all 35 triplets is $15 \times 332 = 4980$, which means the average is $\frac{4980}{35} = 142.427...$ so there has to be at least one triplet with total age of at least 142. Actually since all years are integers this proves that there has to be a triplet with total sum of at least 143.

Solution 3:

Assume $x_1\le x_2\le\ldots\le x_7$. Here's a proof that $x_5+x_6+x_7\ge144$ (which answers an observation made by TonyK in comments).

Suppose $x_5+x_6+x_7\le143$. Then $x_1+x_2+x_3+x_4\ge332-143=189$, which implies $x_4\ge189/4=47.25$. Since the ages are assumed to be integers, we have $48\le x_4\le x_5\le x_6\le x_7$, which contradicts the supposition.

Solution 4:

I find the complexity of the methods shown here to be strange. There is a much simpler solution. $332$ years divided among $7$ people works out to be a little more than $47$ years per person. Specifically, because $7 \cdot 47 = 329$, there are three extra years. So the most even distribution of ages is:

$$47,47,47,47,48,48,48.$$

Obviously, there is a group of three whose ages add up to at least $142$ (in fact, it is $144$). Further note that any time you take a year from someone, you must give it to someone else to keep the sum constant. Taking a year from an older person just redistributes the years and taking a year from a younger person does not help in trying to get a lower sum.

To elaborate on this, you can do one of four redistributions at the beginning: from a $47$ to a $47$, from a $47$ to a $48$, from a $48$ to a $47$, and from a $48$ to a $48$. After these, the outcomes (when sorted again) will be:

$$46,47,47,48,48,48,48$$ $$46,47,47,47,48,48,49$$ $$47,47,47,47,48,48,48$$ $$47,47,47,47,47,48,49$$

As you can see, in every case, the three greatest ages have a sum that is at least $3 \cdot 48 = 144$. Now note that the next time you take a year from someone and give it to someone else, if the first person is older, then you will get one of the five configurations seen already. The only way to get a configuration not seen before is to take a year from a younger person, and this will never lower the sum of the ages of the three oldest people.

To look at it another way, if every person was $47$ years old, then every triple of persons has an age sum of $141$. The instant you give anyone an extra year, there is a triple with an age sum of $142$. Adding two more extra years doesn't help.