In how many ways can 40 be expressed as the sum of different positive odd numbers?
We only consider the addends that make up 40, not the order of the addends, so we will list the terms in ascending order. The representations will be grouped by the number of terms.

Since 40 is an even number, the number of addends will be even and at most 6, because the sum of 8 different positive odd numbers is already not less than $1+3+5+7+9+11+13+15=64$.

a) Since the terms are different, the first term of the two-term representations cannot reach half of 40, that is, it can be at most 19. There are as many odd numbers up to 19 as there are even numbers up to 20, which is $20: 2=10$, this is the number of such representations.

b) When looking for four-term representations, we will select the first two terms in all possible ways, so - as we will see - our question can be reduced to the case in a). Let the first two terms be $1+3=4$. Then the sum of the other two is $x+y=36$, but each of them is greater than 3, and thus greater than the next even number, 4, so $x-4$ and $y-4$ are again positive odd numbers, and $(x-4)+(y-4)=40-3 \cdot 4=28$. Therefore, similarly to the above reasoning, $\frac{28}{2}: 2=7$ representations are obtained. For example, in the case of $x-4=1$, $y-4=27$, and thus $40=1+3+x+y=1+3+5+31$; the last one is from the pair $x-4=13, y-4=15$: $1+3+17+19$.

Taking 5 as the second term instead of 3, similarly $x, y>5+1=6 ;(x-6)+(y-6)=40-3 \cdot 6=22$. Here the first addend can be at most $\frac{22}{2}-1=10$, up to which the number of odd numbers, with the usual notation for the integer part of a number

$$
\left[\frac{\left(\frac{22}{2}-1\right)+1}{2}\right]=\left[\frac{22}{4}\right]=5
$$

Similarly, we get the number of decompositions starting with $1+7$, $1+9$, $1+11$:

$$
\left[\frac{40-3 \cdot 8}{4}\right]=\left[\frac{16}{4}\right], \quad\left[\frac{10}{4}\right], \quad\left[\frac{4}{4}\right]
$$

It is useful to collect our results in a table, the first row of the table consists of the above. A decomposition cannot start with 1 + 13, because $1+13+15+17>40$.

Taking 3 as the first term, we repeat our previous reasoning as follows: every term is greater than 2, so it is enough to look for the part of the terms above 2 - which is again odd - and their sum is $40-4 \cdot 2=32$. The number of decompositions of 40 starting with $3+5$, $3+7$, $3+9$ is the same as the number of decompositions of 32 starting with $1+3$, $1+5$, $1+7$, that is, respectively $\left[\frac{32-3 \cdot 4}{4}\right]=\left[\frac{20}{4}\right], \quad\left[\frac{14}{4}\right], \quad\left[\frac{8}{4}\right]$ (the second row of our table).

Taking 5, then 7 as the first term, 32 is replaced by $40-4 \cdot 4=24$, then 16, and the numbers $\left[\frac{24-12}{4}\right],\left[\frac{24-18}{4}\right]$, and $\left[\frac{16-12}{4}\right]$ are placed in the 3rd, then 4th row of the table. This completes the counting of the four-term representations, because taking 9 as the first term would already give a sum of at least 48.

second term

|  |  | 3 | 5 | 7 | 9 | 11 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| el- |  |  |  |  |  |  |
| só |  |  |  |  |  |  |
| $\operatorname{tag}$ | 1 | 7 | 5 | 4 | 2 | 1 |
|  |  |  |  |  |  |  |
|  | 3 | $\times$ | 5 | 3 | 2 | - |
|  | 5 | $\times$ | $\times$ | 3 | 1 | - |
|  | 7 | $\times$ | $\times$ | $\times$ | 1 | - |

c) The counting of the six-term decompositions starts with the selection of the first 4 terms. Starting with $1+3+5+7$, the sum of the last two terms is $x+y=24$ and $x, y>8$, so $(x-8)+(y-8)=8$. There are 2 such decompositions according to the above pattern: those resulting from $1+7$ and $3+5$: $x+y=9+15$ and $11+13$. The first four terms cannot be chosen otherwise, as $1+3+5+9+11+13>40$.

According to all this, the number of the desired representations is $10+34+2=46$.

Újj Piroska (Nagybátony-Bányaváros, g. II. o. t.)

Gajdács Ibolya (Budapest, Berzsenyi D. g. I. o. t.)