aime26-15

Proposed solution

I need to find all arithmetic sequences with first term 4 that contain both 24 and 34 as terms, then sum their 10th terms.

Let me denote an arithmetic sequence with first term $a_1 = 4$ and common difference $d$. The general term is:
$$a_n = 4 + (n-1)d$$

Since both 24 and 34 are terms in this sequence, there exist positive integers $k$ and $m$ such that:
$$a_k = 4 + (k-1)d = 24$$
$$a_m = 4 + (m-1)d = 34$$

From the first equation: $(k-1)d = 20$, so $d = \frac{20}{k-1}$

From the second equation: $(m-1)d = 30$, so $d = \frac{30}{m-1}$

Since both expressions equal $d$:
$$\frac{20}{k-1} = \frac{30}{m-1}$$

Cross-multiplying:
$$20(m-1) = 30(k-1)$$
$$20m - 20 = 30k - 30$$
$$20m = 30k - 10$$
$$2m = 3k - 1$$
$$2m + 1 = 3k$$

This means $3k = 2m + 1$, so $k = \frac{2m + 1}{3}$.

For $k$ to be a positive integer, $(2m + 1)$ must be divisible by 3.
Since $2m + 1 \equiv 2m + 1 \pmod{3}$, we need $2m \equiv -1 \equiv 2 \pmod{3}$.
This means $m \equiv 1 \pmod{3}$.

So $m = 3t + 1$ for some non-negative integer $t$.

When $m = 3t + 1$, we have:
$$k = \frac{2(3t + 1) + 1}{3} = \frac{6t + 3}{3} = 2t + 1$$

The common difference is:
$$d = \frac{20}{k-1} = \frac{20}{2t + 1 - 1} = \frac{20}{2t} = \frac{10}{t}$$

For $d$ to be an integer, $t$ must be a positive divisor of 10.
The positive divisors of 10 are: 1, 2, 5, 10.

Let me check each case:

Case 1: $t = 1$
- $m = 4$, $k = 3$
- $d = 10$
- Sequence: $4, 14, 24, 34, 44, ...$
- Check: $a_3 = 4 + 2(10) = 24$ ✓, $a_4 = 4 + 3(10) = 34$ ✓

Case 2: $t = 2$  
- $m = 7$, $k = 5$
- $d = 5$
- Sequence: $4, 9, 14, 19, 24, 29, 34, 39, ...$
- Check: $a_5 = 4 + 4(5) = 24$ ✓, $a_7 = 4 + 6(5) = 34$ ✓

Case 3: $t = 5$
- $m = 16$, $k = 11$
- $d = 2$
- Sequence: $4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, ...$
- Check: $a_{11} = 4 + 10(2) = 24$ ✓, $a_{16} = 4 + 15(2) = 34$ ✓

Case 4: $t = 10$
- $m = 31$, $k = 21$
- $d = 1$
- Sequence: $4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, ...$
- Check: $a_{21} = 4 + 20(1) = 24$ ✓, $a_{31} = 4 + 30(1) = 34$ ✓

Now I need to find the 10th term of each sequence:
- For $d = 10$: $a_{10} = 4 + 9(10) = 94$
- For $d = 5$: $a_{10} = 4 + 9(5) = 49$  
- For $d = 2$: $a_{10} = 4 + 9(2) = 22$
- For $d = 1$: $a_{10} = 4 + 9(1) = 13$

The sum of the 10th terms is: $94 + 49 + 22 + 13 = 178$

$\boxed{178}$