aime26-15

Proposed solution

To find the sum of the 10th terms of all arithmetic sequences of integers that have first term equal to 4 and include both 24 and 34 as terms, we start by defining the general form of an arithmetic sequence. An arithmetic sequence can be expressed as:

\[ a_n = a + (n-1)d \]

where \( a \) is the first term and \( d \) is the common difference. Given that the first term \( a = 4 \), we can write the 10th term as:

\[ a_{10} = 4 + 9d \]

We need to find the common difference \( d \) such that both 24 and 34 are terms in the sequence. This means there exist integers \( m \) and \( n \) satisfying:

\[ 4 + (m-1)d = 24 \]
\[ 4 + (n-1)d = 34 \]

From the first equation:

\[ 4 + (m-1)d = 24 \]
\[ (m-1)d = 20 \]
\[ d = \frac{20}{m-1} \]

From the second equation:

\[ 4 + (n-1)d = 34 \]
\[ (n-1)d = 30 \]
\[ d = \frac{30}{n-1} \]

For \( d \) to be the same in both equations, we must have:

\[ \frac{20}{m-1} = \frac{30}{n-1} \]

Cross-multiplying gives:

\[ 20(n-1) = 30(m-1) \]
\[ 20n - 20 = 30m - 30 \]
\[ 20n - 30m = -10 \]
\[ 2n - 3m = -1 \]

We need to find integer solutions \( (m, n) \) to this equation. Rearranging, we get:

\[ 2n = 3m - 1 \]
\[ n = \frac{3m - 1}{2} \]

For \( n \) to be an integer, \( 3m - 1 \) must be even, implying \( m \) must be odd. Let's find the values of \( m \) and corresponding \( n \):

1. \( m = 1 \):
   \[ n = \frac{3(1) - 1}{2} = 1 \]
   This gives \( d = \frac{20}{1-1} \) which is undefined.

2. \( m = 3 \):
   \[ n = \frac{3(3) - 1}{2} = 4 \]
   This gives \( d = \frac{20}{3-1} = 10 \).

3. \( m = 5 \):
   \[ n = \frac{3(5) - 1}{2} = 7 \]
   This gives \( d = \frac{20}{5-1} = 5 \).

4. \( m = 7 \):
   \[ n = \frac{3(7) - 1}{2} = 10 \]
   This gives \( d = \frac{20}{7-1} = \frac{20}{6} = \frac{10}{3} \) which is not an integer.

5. \( m = 9 \):
   \[ n = \frac{3(9) - 1}{2} = 13 \]
   This gives \( d = \frac{20}{9-1} = \frac{20}{8} = 2.5 \) which is not an integer.

6. \( m = 11 \):
   \[ n = \frac{3(11) - 1}{2} = 16 \]
   This gives \( d = \frac{20}{11-1} = 2 \).

7. \( m = 13 \):
   \[ n = \frac{3(13) - 1}{2} = 19 \]
   This gives \( d = \frac{20}{13-1} = \frac{20}{12} = \frac{5}{3} \) which is not an integer.

8. \( m = 15 \):
   \[ n = \frac{3(15) - 1}{2} = 22 \]
   This gives \( d = \frac{20}{15-1} = \frac{20}{14} = \frac{10}{7} \) which is not an integer.

9. \( m = 17 \):
   \[ n = \frac{3(17) - 1}{2} = 25 \]
   This gives \( d = \frac{20}{17-1} = \frac{20}{16} = 1.25 \) which is not an integer.

10. \( m = 19 \):
    \[ n = \frac{3(19) - 1}{2} = 28 \]
    This gives \( d = \frac{20}{19-1} = \frac{20}{18} = \frac{10}{9} \) which is not an integer.

11. \( m = 21 \):
    \[ n = \frac{3(21) - 1}{2} = 31 \]
    This gives \( d = \frac{20}{21-1} = \frac{20}{20} = 1 \).

12. \( m = 23 \):
    \[ n = \frac{3(23) - 1}{2} = 34 \]
    This gives \( d = \frac{20}{23-1} = \frac{20}{22} = \frac{10}{11} \) which is not an integer.

13. \( m = 25 \):
    \[ n = \frac{3(25) - 1}{2} = 37 \]
    This gives \( d = \frac{20}{25-1} = \frac{20}{24} = \frac{5}{6} \) which is not an integer.

14. \( m = 27 \):
    \[ n = \frac{3(27) - 1}{2} = 40 \]
    This gives \( d = \frac{20}{27-1} = \frac{20}{26} = \frac{10}{13} \) which is not an integer.

15. \( m = 29 \):
    \[ n = \frac{3(29) - 1}{2} = 43 \]
    This gives \( d = \frac{20}{29-1} = \frac{20}{28} = \frac{5}{7} \) which is not an integer.

16. \( m = 31 \):
    \[ n = \frac{3(31) - 1}{2} = 46 \]
    This gives \( d = \frac{20}{31-1} = \frac{20}{30} = \frac{2}{3} \) which is not an integer.

17. \( m = 33 \):
    \[ n = \frac{3(33) - 1}{2} = 49 \]
    This gives \( d = \frac{20}{33-1} = \frac{20}{32} = \frac{5}{8} \) which is not an integer.

18. \( m = 35 \):
    \[ n = \frac{3(35) - 1}{2} = 52 \]
    This gives \( d = \frac{20}{35-1} = \frac{20}{34} = \frac{10}{17} \) which is not an integer.

19. \( m = 37 \):
    \[ n = \frac{3(37) - 1}{2} = 55 \]
    This gives \( d = \frac{20}{37-1} = \frac{20}{36} = \frac{5}{9} \) which is not an integer.

20. \( m = 3