 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/2/f/solution

Use modular arithmetic to keep all intermediate numbers small:

34 \equiv 2, 45 \equiv 5, 55 \equiv 7, 89 \equiv 1 (mod 8) .

Thus

34 \cdot 45 \cdot 55 + 89 \equiv 2 \cdot 5 \cdot 7 + 1 = 71 \equiv 7 \equiv - 1 (mod 8)

. Cubing gives

(34 \cdot 45 \cdot 55 + 89)^{3} \equiv (- 1)^{3} \equiv 7 (mod 8); remainder 7.

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