If 137+ 276 = 435, how much is 731 + 672 ?

Given rule: Add numbers in the base where 137 + 276 = 435 holds.

Find the base \(b\):

\(137_b = 1b^2 + 3b + 7\)
\(276_b = 2b^2 + 7b + 6\)
Sum \(= 3b^2 + 10b + 13\)

\(435_b = 4b^2 + 3b + 5\)

Equate: \(3b^2 + 10b + 13 = 4b^2 + 3b + 5 \;\Rightarrow\; b^2 - 7b - 8 = 0\)

So \(b = 8\) (since base must be > 7). Thus the operations are in base 8 (octal).

Now add in base 8: \(731_8 + 672_8\)

      731₈
    + 672₈
    -------
      (units) 1 + 2 = 3 → write 3
      (eights) 3 + 7 = 10₁₀ = 12₈ → write 2, carry 1
      (sixty-fours) 7 + 6 + 1 = 14₁₀ = 16₈ → write 6, carry 1
      bring down carry → 1
    -------
      1623₈
  

Answer: \(731 + 672 = \mathbf{1623_8}\) (in base 8).