 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-1/3h/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

.02 + .03 = .05, .04 + .04 = .08, .05 + .08 = .13, .13 + .14 = .27, .27 + .28 = .55

gives, for example,

x_{1} : 0, x_{2} : 10, x_{3} : 110, x_{4} : 11110, x_{5} : 11111, x_{6} : 11101, x_{7} : 11100.

Any consistent interchange of siblings is equivalent.

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