ITPEC FE Subject B April 2026 Question 16

64 — the trailing byte carries the low 6 bits of the code point, so both blanks split the value at 2⁶.

In the pattern 110xxxxx 10xxxxxx the six rightmost x positions belong to the second byte:

• •second byte — codePoint mod 64 isolates those low 6 bits and is added to the base 128
• •first byte — integer part of (codePoint ÷ 64) shifts the remaining bits down and is added to the base 192

Check it against the worked example, the Registered Sign with code point 174:

• •174 mod 64 = 46, and 128 + 46 = 174, the byte 10101110
• •174 ÷ 64 = 2 discarding the remainder, and 192 + 2 = 194, the byte 11000010

That reproduces the stated result of 194 and 174.

Answer (c)

Why not others:
- (b) 32 — that is 2⁵, the payload width of the first byte; the boundary between the two bytes falls after the sixth bit, not the fifth
- (d) 256 — a whole byte, but only 6 bits of the code point fit into the trailing byte
- (a) 2 — would move a single bit into the second byte
- (e) and (f) — array elements rather than a divisor; the calculation would depend on the value it is still building

Key rule: Every trailing byte in UTF-8 carries exactly 6 payload bits, so mod 64 and ÷ 64 split the code point in any multi-byte case — only the number of leading bytes changes.

AI-generated — may contain errors