ITPEC FE Subject B October 2024 Question 4
ITPEC FE Subject B October 2024 — Question 4 of 20
Filling in blanks for a prime factorization procedure using trial division.
The program performs trial division: starting from i = 2, it repeatedly checks whether num is divisible by i. If yes, it outputs i as a factor, divides num by i, and checks whether more factors remain. If not, it increments i.
Blank A must check divisibility: num mod i = 0. When true, i is a prime factor — the program outputs i and divides num by i.
Blank B appears in two places:
1. After outputting a factor — if num > 1, more factors remain, so print the "×" separator. If num = 1, factorization is complete — no separator needed.
2. Loop condition (`while`) — continue the loop while num > 1. Once num reaches 1, all prime factors have been extracted.
Trace with input 12:
| Step | `i` | `num` | `num mod i` | Action | Output |
|---|---|---|---|---|---|
| 1 | 2 | 12 | 0 | divide → num = 6, check 6 > 1 → true | 2× |
| 2 | 2 | 6 | 0 | divide → num = 3, check 3 > 1 → true | 2× |
| 3 | 2 | 3 | 1 | not divisible → i = 3 | — |
| 4 | 3 | 3 | 0 | divide → num = 1, check 1 > 1 → false | 3 |
| 5 | — | 1 | — | while (1 > 1) → false → exit | — |
Final output: 2×2×3 ✓
Why not others:
- (a) A = `num < 1` — never true for positive input; no factor would ever be extracted
- (b) A = `num > 1`, B = `num mod i = 0` — A should test divisibility, not magnitude; B as a while-condition (while num mod i = 0) would exit the loop whenever i doesn't divide num, breaking the algorithm
- (d) A = `num mod i ≠ 0` — inverted logic; would divide num by i precisely when i is not a factor
Key rule: In trial division, the inner check is always "does i divide num evenly?" (mod = 0), and the outer loop continues while num > 1.
AI-generated — may contain errors
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