Dec 24 2024
Mars’s ancient atmosphere might be locked in clay

Despite increasing evidence that water flowed on Mars billions of years ago, scientists have been mystified by what happened to the thick, carbon dioxide–rich atmosphere that must have once kept that water from freezing. 

Now two MIT geologists think they know. Geology professor Oliver Jagoutz and Joshua Murray, PhD ’24, propose that much of this missing atmosphere could be locked up in the planet’s clay-covered crust.

While water was present on Mars, they suggest, the liquid could have trickled through certain rock types and set off a slow chain of reactions that progressively drew carbon dioxide out of the atmosphere and converted it into methane, a form of carbon that could be stored in the clay for eons. 

This schematic illustrates the progressive alteration of iron-rich rocks on Mars as the rocks interact with water containing CO2 from the atmosphere. Over several billion years, this process could have stored enough CO2 in the clay surface, in the form of methane, to explain most of the CO2 that went missing from the planet’s early atmosphere.
COURTESY OF THE RESEARCHERS

The researchers applied their knowledge of interactions between rocks and gases on Earth to how similar processes could play out on Mars. They found that the quantity of clay covering the Martian surface could hold up to 1.7 bar of CO2, which would be equivalent to around 80% of the planet’s early atmosphere. “In some ways, Mars’s missing atmosphere could be hiding in plain sight,” Murray says.

The researchers think it’s possible that this sequestered carbon could one day be recovered and converted into propellant to fuel future missions between Mars and Earth. 

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Dec 21 2024
Puzzle Corner September/October 2024 bonus solutions

Here are solutions for the three bonus problems that appeared in the September/October 2024 Puzzle Corner column we guest edited. Solutions for S/O2, S/O4, and S/O6 are below; those for S/O1, S/O3, and S/O5 can be found here.

S/O2. Frank notes that a repunit Rk is a decimal integer consisting of the digit 1 repeated k times, with k > 0. For example, R1 = 1, R2 = 11, R3 = 111, etc. Let N be any integer not divisible by 2 or 5. Prove that there is a repunit divisible by N.

David Dewan came up with this solution (which Richard Lipes also found on Wikipedia): Let N be any integer not divisible by 2 or 5. Consider repunits R1 = 1, R2 = 11, R3 = 111, …, RN+1 and their residues modulo N. There are at most N different residues, so the set of N + 1 residues modulo N must contain at least one repeat. Assume RA mod N and RB mod N with B > A are the same. Then (RB – RA) mod N = (RB-A10A) mod N = 0. Since N and 10 are relatively prime, N divides RB-A.

S/O4. Frank offers this sudoku problem:

Many readers tackled it successfully. For anyone who got stuck, here’s the answer key:

“Ten years ago I was doing what is now called AI,” Richard Marks ’58 writes. He noted that trying all possible iterations of a sudoku problem “will tie up your Cray for a week. So I personally wrote the rules and coded this little AI program that solves any sudoko by the time you have released the “Solve It” button. Since I wrote the program from scratch, I guess you can say I solved S/O4.”

S/O6. On behalf of the MIT Chess Club, Justin Zhou ’25 asked how White can play and mate in two (see below).

Frank Model ’63 says there are two cases to consider: when Black can castle, and when Black cannot. If Black can castle, then Black’s last move had to be Pc7-c5. In this case White can take en passant, Pb5-c6. If Black castles, then White plays Pb7#. If Black can castle, but makes some other move, then White plays Rf8#. 

If Black cannot castle, then White plays Ke6. Black cannot escape mate on the next move when White plays Rf8#.

Steve Gordon noted that this problem illustrates the three special chess moves. The following was adapted from his analysis with his algebraic chess notation. 

Black last moved either a king, rook, or c5. If c5, then 1. bxc6 (en passant a.k.a e. p.). If Black can still castle queenside (1. Kc8 & Rd8, a.k.a O-O-O), then 2. b7#. If Black cannot, its king is still trapped on rank 8, so after any black move, 2. Rf8#. If Black last moved a king or rook, en passant is not possible for White, but Black can’t castle either, so 1. Ke6 also traps the black king on rank 8, and after any Black move, 2. Rf8#. Note: To create this puzzle, the white bishop on g7 resulted from an underpromotion on h8.

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