If the machine is decimal, the bytes of rX represent 1235(100^3) + 3(100) + 1 = 1,235,000,301, and the divisor is 2(100) + 0 = 200. The quotient is then 6,175,001 remainder 101, and so the digital bytes of rA become 6 17 50 01.

But if the machine is binary, the bytes of rX represent 1235(64^3) + 3(64) + 1 = 323,748,033, and the divisor is 2(64) + 0 = 128. The quotient is 2529281 remainder 13, which can be written 9(64^3) + 41(64^2) + 32(64) + 1, and so the binary bytes of rA become 9 41 32 1. But 9(64) + 41 = 617, so bytes 2 and 3 can be interpreted as 617 regardless of whether the machine is decimal or binary. However, the less significant bytes are different.

Similarly, the last byte of rX is 1 for the decimal and binary calculations: the remainder is 1(100) + 1 in the decimal calculation, or 1 in the binary calculation. Either way, the least significant byte is 1. So the 617 in bytes 2-3 of rA, and the 1 in byte 5 of rX, are the same regardless, but all the other bytes are different, hence the question marks in all other byte slots.