 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/4h/b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

3 m = 4 l

and

2 n = 5 k

are equivalent to

m = 4 r, l = 3 r, n = 5 s, k = 2 s

for nonnegative integers

r, s

. Hence the language is

{a^{4 r} b^{5 s} c^{2 s} d^{3 r} : r, s \geq 0},

S \to aaaa S ddd ∣ T, T \to bbbbb T cc ∣ ϵ .

It is therefore context-free.

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