To check precision, we can use the following pair of examples:

2147419572 seconds is 282.53499 Mercury years, rounding to 28253 Mercury centi-years.

2147419724 seconds is 282.53501 Mercury years, rounding to 28254 Mercury centi-years.

An Earth centi-year is 315576 seconds, so a Mercury centi-year is 0.2408467 × 315576 ≈ 76005.4382 seconds.

What happens when we use integer division by 76005?

2147419572 ÷ 76005 gives 28253, remainder 50307

2147419724 ÷ 76005 gives 28253, remainder 50459

We might round the answers up because the remainders exceed half of 76005. Either way, one of the answers won’t be ideal.

The problem is with 76005, we are only using 17 bits of precision.

Instead of dividing, can we multiply by 1/76005.4382 ?

To use integer arithmetic, we will need to scale up 1/76005.4382 by some 2^n, and then remove the bottom n bits from the multiplication result.

Let’s try n=48

2^48 * 1/76005.4382 = 3703353120

This fits in 32 bits, provided we use unsigned arithmetic.

We don’t have negative ages in space customs forms, so let’s proceed.

When we multiply using MIPS, the 64 bit result is split across 2 registers.

2147419572 * 3703353120 gives HI 1851621310, LO 888586880

2147419724 * 3703353120 gives HI 1851621441, LO 1157545344

We need to remove the bottom 48 bits. 32 of those bits are LO, the other 16 are the bottom half of HI.

Let’s look at the top and bottom 16 bits of HI.

1851621310 splits as 28253 32702, in hexadecimal 0x6e5d 0x7fbe

1851621441 splits as 28253 32833, in hexadecimal 0x6e5d 0x8041

To round up when those bottom 16 bits are 2^15 or more, we can start by adding 2^15, and then we’ll discard the bottom 16 bits.

1851621310 + 2^15 = 1851654078, splits as 28253 65470, in hexadecimal 0x6e5d 0xffbe

1851621441 + 2^15 = 1851654209, splits as 28254 00065, in hexadecimal 0x6e5e 0x0041

We have the results 28253 and 28254 that we were seeking.