1. **Problem statement:** Find the number of ordered triples $(a,b,c)$ of positive integers with $1 \leq a,b,c \leq 50$ such that $$\frac{\mathrm{lcm}(a,c) + \mathrm{lcm}(b,c)}{a+b} = \frac{26c}{27}.$$ 2. **Recall the definition and properties:** The least common multiple satisfies $$\mathrm{lcm}(x,y) = \frac{xy}{\gcd(x,y)}.$$ 3. **Rewrite the equation using the LCM formula:** $$\frac{\frac{ac}{\gcd(a,c)} + \frac{bc}{\gcd(b,c)}}{a+b} = \frac{26c}{27}.$$ Multiply both sides by $a+b$: $$\frac{ac}{\gcd(a,c)} + \frac{bc}{\gcd(b,c)} = \frac{26c}{27}(a+b).$$ 4. **Divide both sides by $c$ (since $c>0$):** $$\frac{a}{\gcd(a,c)} + \frac{b}{\gcd(b,c)} = \frac{26}{27}(a+b).$$ 5. **Set $d = \gcd(a,c)$ and $e = \gcd(b,c)$:** Then $a = d a'$, $c = d c'$, where $\gcd(a', c')=1$, and similarly $b = e b'$, $c = e c''$, $\gcd(b', c'')=1$. Since $c$ is fixed, $c = d c' = e c''$, so $d c' = e c''$. 6. **Rewrite the equation in terms of $a', b', d, e, c$: ** $$\frac{d a'}{d} + \frac{e b'}{e} = \frac{26}{27}(d a' + e b').$$ Simplify left side: $$a' + b' = \frac{26}{27}(d a' + e b').$$ 7. **Rearranged:** $$27(a' + b') = 26(d a' + e b').$$ 8. **Rewrite:** $$27 a' + 27 b' = 26 d a' + 26 e b'.$$ Group terms: $$27 a' - 26 d a' = 26 e b' - 27 b'$$ $$a'(27 - 26 d) = b'(26 e - 27).$$ 9. **Since $a', b'$ are positive integers with $\gcd(a', c')=1$ and $\gcd(b', c'')=1$, and $d,e$ divide $c$, analyze divisibility:** We want positive integers $a', b', d, e$ satisfying $$a'(27 - 26 d) = b'(26 e - 27).$$ 10. **Note the ranges:** $1 \leq a,b,c \leq 50$, so $d,e$ divide $c \leq 50$, and $a', b' \geq 1$. Also, $27 - 26 d$ and $26 e - 27$ must be positive or negative accordingly to keep $a', b'$ positive. 11. **Check possible values of $d$ and $e$ such that $27 - 26 d > 0$ and $26 e - 27 > 0$ or vice versa:** - $27 - 26 d > 0 \implies d < \frac{27}{26} \approx 1.038$, so $d=1$ only. - $26 e - 27 > 0 \implies e > \frac{27}{26} \approx 1.038$, so $e \geq 2$. 12. **Substitute $d=1$:** $$a'(27 - 26 \cdot 1) = a' (1) = b'(26 e - 27).$$ So $$a' = b'(26 e - 27).$$ 13. **Since $a', b'$ are positive integers, $26 e - 27$ must be positive integer, so $e \geq 2$.** 14. **Recall $c = d c' = e c''$, with $d=1$, so $c = c' = e c''$. Since $d=1$, $c = c'$, and $c$ divisible by $e$.** 15. **Summary:** For each $c$, for each divisor $e$ of $c$ with $e \geq 2$, and $d=1$, the equation reduces to $$a' = b'(26 e - 27).$$ 16. **Count solutions:** - $a = d a' = a'$, since $d=1$. - $b = e b'$. - $c$ fixed. - $a,b,c$ all in $[1,50]$. 17. **Constraints:** - $1 \leq a' \leq 50$. - $1 \leq b' \leq \lfloor \frac{50}{e} \rfloor$. - $a' = b'(26 e - 27)$. 18. **For each $e$ dividing $c$ with $e \geq 2$, find all $b'$ such that $$a' = b'(26 e - 27) \leq 50,$$ and $$b' \leq \left\lfloor \frac{50}{e} \right\rfloor.$$ 19. **So $b'$ must satisfy:** $$b' \leq \min\left(\left\lfloor \frac{50}{26 e - 27} \right\rfloor, \left\lfloor \frac{50}{e} \right\rfloor \right).$$ 20. **Calculate for each $c$ from 1 to 50:** - Find divisors $e$ of $c$ with $e \geq 2$. - For each such $e$, count $b'$ from 1 to above minimum. - Sum over all $c$ and $e$. 21. **Also consider the symmetric case swapping $a$ and $b$ (since the equation is symmetric in $a,b$), but here $d=1$ fixed, so no double counting needed.** 22. **Finally, add the case $d=e=1$:** - Then $a'(27 - 26) = b'(26 - 27)$ - $a' = -b'$, impossible for positive integers. 23. **Therefore, total number of solutions is:** $$\sum_{c=1}^{50} \sum_{\substack{e|c \\ e \geq 2}} \min\left(\left\lfloor \frac{50}{26 e - 27} \right\rfloor, \left\lfloor \frac{50}{e} \right\rfloor \right).$$ 24. **Calculate explicitly:** - For $e=2$: $26 \cdot 2 - 27 = 52 - 27 = 25$, $\lfloor 50/25 \rfloor = 2$, $\lfloor 50/2 \rfloor = 25$, min = 2. - For $e=3$: $26 \cdot 3 - 27 = 78 - 27 = 51$, $\lfloor 50/51 \rfloor = 0$, so no solutions. - For $e \geq 3$, $26 e - 27 \geq 51 > 50$, so no solutions. 25. **Thus only $e=2$ contributes.** 26. **Count $c$ with divisor $e=2$:** all even $c$ from 2 to 50, total 25 values. 27. **For each such $c$, number of $b'$ is 2.** 28. **Total solutions:** $$25 \times 2 = 50.$$ **Final answer:** There are \boxed{50} ordered triples $(a,b,c)$ satisfying the given equation.