Not Remarkably Rich

Annette, Bernice, and Claudia are three remarkable women, each having some remarkable characteristics.

1. Just two are remarkably intelligent, just two are remarkably beautiful, just two are remarkably artistic, and just two are remarkably rich.
2. Each has no more than three remarkable characteristics.
3. Of Annette it is true that:
   if she is remarkably intelligent, she is remarkably rich.
4. Of Bernice and Claudia it is true that:
   if she is remarkably beautiful, she is remarkably artistic.
5. Of Annette and Claudia it is true that:
   if she is remarkably rich, she is remarkably artistic.

Who is not remarkably rich?

(Source: Test Your Logic: 50 Puzzles in Deductive Reasoning by George J. Summers)

This looks pretty difficult because, in themselves, implications with false antecedents don't tell us anything about the truth of their consequents. This means, for example, that if Annette is not remarkably intelligent, whether or not she is remarkably rich is still up in the air. Fortunately, there really is enough structure here to determine the truth. One can start with the assumption that Annette is remarkably intelligent. If she is remarkably intelligent then she is also remarkably rich. If she is remarkably rich, then she is also remarkably artistic. Additionally, because two of the women have to be remarkably beautiful, and none of them have more than three remarkable traits, the two remarkably beautiful women must be Bernice and Claudia. This implies in turn that they are both remarkably artistic:

	Intelligent	Beautiful	Artistic	Rich
Annette	✓		✓	✓
Bernice		✓	✓
Claudia		✓	✓

Uh-oh! We already have a problem. Fortunately, the problem stems entirely from the single assumption that Annette is remarkably intelligent; nothing further needed to be added and so this assumption can in no sense be seen as "protected", which is a good thing, because we already have some solid truth about what is the case: Annette is not remarkably intelligent and, because two of the women must be remarkably intelligent, Bernice and Claudia are. All further calculations must then start with these facts:

	Intelligent	Beautiful	Artistic	Rich
Annette	✗
Bernice	✓
Claudia	✓

After I reached this point, I had to pause for a while to figure out my next step. It turned out that it was fruitful to assume that Annette is remarkably rich, which in turn implied that she is remarkably artistic:

	Intelligent	Beautiful	Artistic	Rich
Annette	✗		✓	✓
Bernice	✓
Claudia	✓

This is a good assumption to try because it means that exactly one of Bernice and Claudia can (and must) be remarkably beautiful, but not both, because remarkable beauty in either also implies remarkable artistry for the same woman, and that would leave us with three remarkably artistic women, which is a no-no. Moreover, because only one of Bernice and Claudia can be remarkably beautiful, this assumption also means that Annette has to pick up the slack for the remaining slot, which may as well be added now:

	Intelligent	Beautiful	Artistic	Rich
Annette	✗	✓	✓	✓
Bernice	✓
Claudia	✓

Of Bernice and Claudia, which is remarkably beautiful? What if it's Bernice? If Bernice is remarkably beautiful, then she is remarkably artistic. This confers three remarkable qualities on Bernice and so she can't be remarkably rich. This means Claudia must be remarkably rich and, if Claudia is remarkably rich, then she is also remarkably artistic:

	Intelligent	Beautiful	Artistic	Rich
Annette	✗	✓	✓	✓
Bernice	✓	✓	✓	✗
Claudia	✓	✗	✓	✓

Déjà vu! We have a problem again. As before, all three women are now remarkably artistic, which is not possible. This means that, under the current assumption, Claudia must be remarkably beautiful. Can it work or do we have to go back to the drawing board? If Claudia is remarkably beautiful, then she is also remarkably artistic. That means Claudia now has three remarkable traits and cannot have a fourth of being remarkably rich. Bernice must then be remarkably rich and here is our completed tableau:

	Intelligent	Beautiful	Artistic	Rich
Annette	✗	✓	✓	✓
Bernice	✓	✗	✗	✓
Claudia	✓	✓	✓	✗

This tableau indeed fulfills all the criteria set out in the beginning. Exactly two of the women have each of the four traits and none has more than three. Remembering that only implications with true antecedents need to be checked, it is equally true that, if Claudia is remarkably beautiful, she is remarkably artistic and that, if Annette is remarkably rich, she is remarkably artistic. Which leaves Claudia without remarkable wealth, but I suppose if she really wanted it she could go into consulting or modeling or the like with her other assets.

Final answer: Claudia is not remarkably rich.

De Morgan and Another

Augustus De Morgan, the mathematician who died in 1871, used to boast that he was $x$ years old in the year $x^2$. Jasper Jenkins [possibly not a real person], wishing to improve on this, told me in 1925 that he was $a^2 + b^2$ in $a^4 + b^4$; that he was $2m$ in the year $2m^2$; and that he was $3n$ years old in the year $3n^4$. Can you give the years in which De Morgan and Jenkins were respectively born?

(Source: 536 Curious Problems and Puzzles by Henry Ernest Dudeney)

Squaring away the year of birth for De Morgan is pretty easy. The maximum value possible for $x$ is the floor of the square root of 1871, which is 43. 43 squared is 1849, which seems like a reasonable value. This would mean that De Morgan was born in 1806, which is correct.

The second part seems harder because it contains information that is extraneous to getting the solution (though I suppose one might consider the extra statements "bonus rounds"). But really all that one need focus on is the last statement, that he was $3n$ years old in the year $3n^4$. In other words:

\[ y + 3n = 3n^4 \]

Or:

\[ y = 3n(n^3 - 1) \]

Where $y$ is the year of his birth. This one is the easiest to work with because there is only one variable and the cubic term blows up quickly, meaning only a few values of $n$ need to be tried. Reformulating a little, one has to try different values of the function:

\[ y(n) = 3n(n^3 - 1) \]

Starting at zero, these are:

\begin{align*}
y(0) &= 0 \\
y(1) &= 0 \\
y(2) &= 42 \\
y(3) &= 234 \\
y(4) &= 756 \\
y(5) &= 1860 \\
y(6) &= 3870
\end{align*}

Like I said, the cubic term has let us dash through Antiquity, jump over the Early Middle Ages and land right in the Victorian era before going almost two millennia into the future. (By comparison, the naive guess-and-check method used here, but with the second statement that involves only a squared term, requires thirty-two iterations, starting from zero, to get the right answer.) 1860 is indeed the right answer and it can be verified by noting that, in this instance, the third statement means that he was 15 ($3n$) in 1875 ($3n^4$).

Final answer: De Morgan was born in 1806; Jenkins, in 1860.

The Logic Question Almost Everyone Gets Wrong

Jack is looking at Anne, but Anne is looking at George. Jack is married, but George is not. Is a married person looking at an unmarried person? [Possible answers: yes, no, cannot be determined]

(Source: an article in The Guardian by Alex Bellos)

This is the sort of question like the one about the cost of a bat and a ball where a reflexive answer will tend to be wrong. Because Anne's marital status has not been given upfront, as it has with Jack and George, one might immediately suspect that it cannot be determined. But consider that "being married" is an obviously binary predicate. So look at both possibilities: if Anne is married, then a married person (Anne) is looking at an unmarried one (George) and if Anne is unmarried then a married person (Jack) is still nonetheless looking at an unmarried person (Anne). Anne's marital status is irrelevant, so the answer is yes.

Will You Crack the Code?

(Source: all over the place; no idea about exact origin)

First, note that an implicit assumption in the problem is that no digits will be repeated in the solution. Then start with the first triplet, 682. One (and presumably only one) digit is correct and well-placed. It can't be six because then the description of the second triplet could not be true. Nor can it be eight, because the fourth triple, 738, is described as having nothing correct. This means that two is correct and in the right place.

Next, focus on the last two triplets. The "intersection" of the two, as it were, is seven and eight, that is, the two digits they have in common. Because everything is wrong in the fourth triplet, this means that the one correct, but wrongly placed digit is zero.

It is already known that the zero can't occupy the third position of the triplet, because two already has that. Additionally, the third triplet and its description tell us the middle position cannot be occupied by zero, leaving us with 0-2. The only task remaining is to find out the middle digit.

All but two candidates can be ruled out. Zero and two have already been used up. Three, seven and eight are explicitly ruled out. Six was implicitly ruled out in the first paragraph of this solution. That leaves one and four. If the missing digit is one, then the description of the second triplet is false. All that leaves is four.

Final answer: 042. (Don't panic?)

Regex Crossword

I'm going to phone this one in, sort of, because it's an entire site of puzzles. Regex Crossword is a twist on the traditional crossword concept. In each cell you have to enter in a character that satisfies two or more regular expressions that apply to the cell in question. They start out really trivial but start to get very difficult later on. This is the last one I solved and holy shit was it ever tough:

I'm hooked. Go check these out.

The Merchants and the Coin Purse

Three merchants saw in the road a purse [containing money].
One said, "If I secure this purse, I shall become twice as rich as both of you together."
Then the second said, "I shall become three times as rich."
Then the third said, "I shall become five times as rich."
What is the value of the money in the purse, as also the money on hand [with each of the three merchants]?

(Source: The Penguin Book of Curious and Interesting Puzzles by David Wells, this puzzle by way of the Bakhshali manuscript)

Let $p$ be the value of the contents of the purse and $m_1$, $m_2$ and $m_3$ be the values of the cash on hand held by the first, second and third merchants, respectively. If you read the problem right, this problem becomes a system of linear equations:

\begin{align*}
p + m_1 &= 2(m_2 + m_3) \\
p + m_2 &= 3(m_1 + m_3) \\
p + m_3 &= 5(m_2 + m_3)
\end{align*}

You may notice that there are four unknowns and only three equations, making the system underdetermined. (But that's just as well given that my knowledge of medieval Indian currency is effectively nonexistent.) That being said, to the extent that it can be solved—I did so in Maxima—the answer is $p = 3$, $m_1 = \frac{r_1}{5}$, $m_2 = \frac{3 r_1}{5}$, $m_3 = r_1$, where $r_1$ is effectively an arbitrary constant. But if we hone in on the solution that has the smallest values possible without allowing fractional coins, then the merchants, in order, hold one, three and five coins of equal value, respectively, and the purse contains 15 coins of that value. This means, for example, that by securing the purse, the first merchant would then hold 16 coins, twice as many as three held by the second and the five held by the third added together.

Extracting the Cherry

This puzzle is a golden oldie, with a simple but elusive answer.

The cocktail cherry is inside the glass, which is formed from four matches. Your task is to move at most two of the matches, so that the cherry is then outside the glass. You can turn the glass sideways or upside down if you wish, but the shape must remain the same.

Move two matches to extract the cherry.

(Source: Professor Stewart's Cabinet of Mathematical Curiosities by Ian Stewart)

The solution I came up with was to shift the horizontal match far enough to the right that its non-flammable end touches that of the lower vertical match and then take the left vertical match at the top and set it perpendicular to the flammable end of the horizontal match, so that it forms a new "side" of the glass. (Of course this process can also by started by shifting the horizontal match to the left and the answer will be symmetrical to the one obtained doing things how I did them.)

If that description was confusing—I wouldn't blame you for finding it thus—here's how it's portrayed in the solutions: