math problem

[shahar]

In a question in math, I ask to show that there no solution to next problem:
There are 10 given pockets, How you can to insert 44 coins to them (=10 pockets) so in every pocket that will be a different number of coins?

1. How can I do it?
2. Have we have systematic way to show it? Which?

 

[pka]

In a question in math, I ask to show that there no solution to next problem:
There are 10 given pockets, How you can to insert 44 coins to them (=10 pockets) so in every pocket that will be a different number of coins?
1. How can I do it?
2. Have we have systematic way to show it? Which?

This sort of problem is in the study known as integer partitions.
Note that you can leave at most one pocket empty. Otherwise you have to pockets with zero coins.
If we begin with nine in the first, eight in the second, \(\displaystyle \large\cdots~,0\) in the tenth pocket, how many coins have we used?
If you need help click here.
With this sort of question recall that the sum of the first \(\displaystyle n\) positive integers is \(\displaystyle \frac{n(n+1)}{2}\)

 

[JeffM]

You must consider that you have a limited number of coins. Label the pockets as # 1 for the pocket with the smallest number of coins, # 2 for the pocket with the next smallest number of coins, and so on. Let n_p be the the number of coins in pocket # p. (Notice that they must all differ.)

[MATH]\sum_{k=1}^{10}n_k \le 44.[/MATH]
With me to here?

[MATH]n_{p+1} \ge n_p + 1 \implies n_2 \ge n_1 + 1 \implies n_3 \ge n_2 + 1 \ge (n_1 + 1) + 1 = n_1 + 2 \implies[/MATH]
[MATH]n_{p+1} \ge n_1 + p - 1.[/MATH]
[MATH]n_1 \ge 0.[/MATH]
[MATH]\therefore \sum_{k=1}^{10}n_k \ge \sum_{p=1}^{10} (0 + p - 1) = \sum_{p=2}^{10} (p - 1) = \sum_{k=1}^9 k = \dfrac{9 * 10}{2} = 45.[/MATH]
[MATH]\sum_{k=1}^{10}n_k \le 44 \text { and } \sum_{k=1}^{10} n_k \ge 45 \text { is impossible.}[/MATH]

 

[Steven G]

I would start off by trying to put 0 coins in the 1st pocket, 1 coins in the 2nd pocket, ..., 9 coins in the 10th pocket. This uses 45 coins, but you have on 44 coins. You must remove 1 coin from a pocket but then two pockets will have the same number of coins. So no solution.