If (L)M represents a number L in the base-M number system, then which of the…

Concept — a number written in base M is a positional sum: the digits to the left of the radix point carry the weights M0, M1, M2, … while the digits to its right carry M−1, M−2, M−3, … . Changing the base never changes the value; it only re-expresses that value against a different set of weights.

So a decimal value is converted in two independent halves. The integer part is converted by repeated division by the new base, reading the remainders bottom-up; the fractional part is converted by repeated multiplication by the new base, reading the whole-number overflows top-down. The two halves are then rejoined at the radix point, and an equality of this kind is true only when both halves match.

Application — convert (127.125)10 into each stated base and compare it with the claim.

1. Base 2 — 127 = 64 + 32 + 16 + 8 + 4 + 2 + 1 uses every weight from 26 down to 20, so the integer part is 1111111. The fraction 0.125 = 1/8 = 2−3 is a single weight three places past the point, so it is .001. Rejoining them: (127.125)10 = (1111111.001)2.

2. Base 4 — 127 = 1×43 + 3×42 + 3×41 + 3×40 = 64 + 48 + 12 + 3, so the integer part is 1333. Repeated multiplication of the fraction by 4 gives 0.125 × 4 = 0.5 (overflow 0), then 0.5 × 4 = 2.0 (overflow 2), so it is .02. Rejoining them: (127.125)10 = (1333.02)4.

3. Base 8 — 127 = 1×82 + 7×81 + 7×80 = 64 + 56 + 7, so the integer part is 177. The fraction 0.125 = 1/8 = 8−1 is a single weight one place past the point, so it is .1. Rejoining them: (127.125)10 = (177.1)8.

4. Base 16 — 127 = 7×161 + 15×160 = 112 + 15, and the hexadecimal digit standing for fifteen is F, so the integer part is 7F. The fraction 0.125 = 2/16 is two sixteenths, so it is .2. Rejoining them: (127.125)10 = (7F.2)16.

Cross-check — because 4 = 22, 8 = 23 and 16 = 24, each of these forms can also be read straight off the binary expansion 1111111.001 by grouping bits outward from the radix point and padding the outer ends with zeros. All three groupings reproduce the values stated in the question.

Result — every one of the four stated equalities converts correctly in both halves, so the selection that covers A, B, C and D is the one that holds.