one + two + eleven + twelve

Problem: How many positive integer solutions o*n*e+t*w*o+e*l*e*v*e*n+t*w*e*l*v*e=26 ?

o*n*e+t*w*o+e*l*e*v*e*n+t*w*e*l*v*e=26 is equivalent to o*(n*e+t*w)+e^2*l*v(n*e+t*w)=26 is equivalent to (o+e^2*l*v)*(n*e+t*w)=26.

Let (X_1)=o+e^2*l*v,(X_2)=n*e+t*w.

Thus 26=(X_1)×(X_2)=2×13=1×26.

But (X_1) and (X_2) can't be 1 and 26 (or 26 and 1): since we are looking for positive integer solutions, (X_1) can't be 1 because it is the sum of two positive integers; (X_2) can't be 1 for the same reason. So there are only two cases.

First, (X_1)=2, (X_2)=13:

• Because (X_1)=2=1+1=o+e^2*l*v, l=o=v=e=1.

• And (X_2)=n(1)+t*w=13. There are twelve possibilities: n=1and t*w=12, n=2 and t*w=11, ... n=12 and t*w=1.

  • If t*w=1 and n=12, this corresponds to 1 solution to the overall equation.

  • If t*w is prime, each such possibility corresponds to 2 solutions to the overall equation; there are 5 prime numbers ≤12. This yields 10 solutions to the overall equation.

  • If t*w is composite:

    • If t*w is square, i.e. t*w=4or t*w=9, each possibility corresponds to 3 solutions to the overall equation. Overall this yields 6 solutions to the overall equation.

    • If t*w is not square:

      • t*w=6=3×2=1×6 contributes 4 solutions.

      • t*w=8=2×4=1×8 contributes 4 solutions.

      • t*w=10=2×5=1×10 contributes 4 solutions.

      • t*w=12=3×4=6×2=1×12 contributes 6 solutions.

This case has 1+10+6+4+4+4+6=35 solutions in total.

Second, (X_1)=13, (X_2)=2:

• Because (X_2)=2=1+1=n*e+t*w, n=e=w=t=1.

• And (X_1)=o+(1)^2*l*v=13. Above, we have shown that n+t*w=13 has 35 solutions. This is the same case, just relabel the variables.

This case has 35 solutions in total.

Across both cases, there are 35+35=70 solutions in total.