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Struppa's Logic

Logic and Puzzles from Chancellor Daniele Struppa

In my early years, my dad used to pose to me a variety of logic puzzles designed to push my thinking in a way that was engaging to me. I clearly liked it, and the puzzles kept getting harder and harder. In many cases I was not able to solve the problem unaided, and this gave me and my dad an opportunity for elaborate discussions. Some of those discussions are among my most prized memories of my dad.

With this column, I will share with you some of the puzzles that my dad proposed to me, as well as others that I encountered later on in my life. Having become a mathematician I have been immersed in a culture where logic puzzles are a very common form of entertainment. I did not create these puzzles, and probably neither did my dad. Most of the puzzles I will discuss are well known among nerds and math geeks, and I am often unable to give the appropriate origin or the author of the idea underlying the problem.

The puzzles I will choose are those that best illuminate some interesting ways of thinking, which I will try to outline when I offer the solution.

I hope you enjoy trying to solve this problem. I often found that the effort to solve an apparently unsolvable problem is the best way to stretch our mind, and to learn a more creative approach to life problems. Have fun!


This problem is considerably simpler than those I presented in the previous columns, and yet amusing. It can be used as a game to play with friends, and it never fails to astonish.
  • Ask a friend to choose a three digit number, with the only request that the first digit be different from the last one.
  • Tell your friend NOT to share the number with you.
  • Now ask your friend to flip the number (for example, if the original number was 742, the flipped number is now 247).
  • Tell your friend to subtract the smallest of the two numbers from the largest (in this example your friend would calculate 742-247=495).
  • Ask your friend once again to flip this number, so that, in our example, (s)he would have the new number 594.
  • Tell your friend to add the two final results, i.e. 495 594.
  • At this point, ask your friend to focus on the result, pretend to concentrate, and finally yell, triumphantly, that the result he obtained is 1,089! Invariably, (s)he will be astonished that you were able to guess the result.

What makes this trick fascinating, is the fact that you have NO idea of what the initial number is, and therefore it seems impossible that you could guess the outcome of these operations. Of course, the trick relies on the fact that the final result is always 1,089. What I am asking you to explain, is WHY is it always true that the final result is 1,089?

This problem does not require more than what you learned in the first few grades of elementary school, and yet it is beautiful, and somewhat infuriating. Let me add that if the first subtraction results in a two digit number (for example if the chosen number is 453, after flipping we have 354, and the difference is 453-354=99), then you need to put a zero in front of it, before you flip it again. In this example you have 99, which you should write as 099, and after flipping you have 990. Now, the sum of 990 and 99 is, in fact, once again 1,089.


The key for the solution of this problem is to recall (from early elementary school) that the sum of the digits of the first subtraction is always a multiple of 9. Why is that?

That's because the numbers you are subtracting are made with the same digits, and so (by casting out the nines), this is what you obtain. Now note that the last digit of the smallest number is always bigger than the last digit of the largest number (why is that?). So when you subtract, you get a carryover from subtracting the last digits (remember how this is done in second grade...).

Now, since the second digit is the same in both numbers, and since you have a carry over, the second digit in the result of the subtraction will always be 9. if you don't quite believe it, try a few examples, and you will see that this is exactly what happens. Go back and try to understand why.

So...at the end of the first step we get a number of the form A9B, and we know that the sum of its digit is a multiple of 9, which means that A B=9! Now, the last step is asking you to add A9B and B9A. Since A B=9, it should be now clear why the final result is 1,089.


When I finished my Ph.D. and went back to Italy, I became assistant professor at the Scuola Normale Superiore in Pisa, probably one of the greatest institutions in the world. The Scuola only admitted 7 or 8 math students a year, out of several hundred applicants coming from all around Italy. In order to select the students, the applicants were subjected to some of the most difficult problems one can imagine, in a multi-day admission screening process. One of the problems I recall being given to the students is the one I pose below. As you try to solve it, please imagine yourself as a 17 year old kid, with no real math training, except for the formula to solve quadratic equations (which you should remember from your high school times!). Then, it should be no surprise that the Scuola Normale is the alma mater of some of the greatest Italian physicists including Nobel Prize laureates Fermi and Rubbia and mathematicians, like Fields Medal recipient Bombieri and Wolf Prize recipient De Giorgi.
Here is the problem:
An American spaceship lands on Mars and finds a fragment of a Martian textbook in which the equation 5x2-50x 125=0 is being solved. The text shows the answers as x=5, and x=8.
Can you tell us how many fingers Martians have?


If you were a human, and not a Martian, the equation above could be simplified (just divide everything by 5) to x2-10x 25=0, whose only solution is x=5. So, clearly, there is something fishy here!
One thing that should come to mind is that it is widely believed that we use the decimal system because we have ten fingers. So, the only thing one can imagine is that probably the Martians have a different number of fingers, and therefore have chosen a different number base. So, let's assume they have A fingers, then the number 50 really means 5A (just like 50, for humans, means 5 times 10), and 125 really means A2 2A 5 (just like 125, for humans, means 100, plus 2 times ten, plus 5). So, the equation, for a Martian, really means 5x2-(5A)x A2 2A 5=0.
Now, the problem tells us that x=5 is a solution for this equation, which means that 125-25A A2 2A 5 must equal zero.
If you solve this last equation you will obtain that either A is 10, or A is 13. Now, we know that 10 won't work, so it must be that the Martians have 13 fingers. Just to make sure you are right, let's reinterpret the original equation assuming that we have 13 fingers. Then the Martian equation 5x2-50x 125=0 really means for humans (remembering that 50 means 5 times 13, and 125 means 132 plus 2 times 13 plus 5) 5x2-65x 200=0, and you see immediately that x=5 and x=8 are solutions of this equation, as required.


This is a puzzle that I first learned from my dad, but that I realized later was discussed by Martin Gardner in one of his Scientific American columns. I recently gave it to President Doti to help him pass the time while climbing Denali in Alaska. The beauty of this problem is that it appears to be completely impossible, because it seems that there are not enough data. In fact, it can be solved in two ways: either by using some serious analytic geometry (which is how it is usually solved), or by employing a really clever trick that will illustrate an interesting principle. The latter is in fact how my dad (a lawyer with limited knowledge of geometry) was able to find the solution.

Suppose a cylindrical hole is drilled through the center of a sphere as in the picture below:

The length of the hole is six inches. What is the volume of the part of the sphere that remains after the material is removed from the hole?

The only math one needs to know in order to solve this problem is the fact that the volume of a sphere of radius r is (4/3)p r3.


The problem seems unsolvable. We don't know how big the sphere is (what is its radius), and therefore, we don't know how big is the hole. You can try to draw different size circles on paper, and see that the larger the sphere, the larger becomes the cylinder. So, one may think this is an impossible problem. The way my dad argued was the following. Assume that the problem has an answer, otherwise nobody would have posed the problem. The answer then must not depend on the radius of the sphere. If it does not depend on the radius of the sphere, we can assume we have a sphere whose diameter is in fact 6 inches. The only way to drill a cylinder whose length is six inches is to have an infinitesimally small cylinder (almost like a needle piercing the sphere). But then the cylinder has no volume, and the volume remaining is exactly the volume of the sphere. Since its diameter is 6 inches, its radius is 3 inches, and therefore the result is 36p.

The fascinating thing is that in fact the answer of the problem does not depend on the radius of the sphere. If you know enough analytic geometry, you can calculate the volume of the cylinder, and the volume of the sphere. Both depend on the radius. You can also calculate the volume of the two spherical caps that sit on top of the cylinder. This number depends on the radius as well. But now, once you take the volume of the sphere and subtract the volume of the caps and the cylinder, the radius cancels away from the equation, and what you are left with is exactly 36p.

PS. When I gave the problem to President Doti, I stated it incorrectly, so the poor mountain climbers struggled day after day with a really unsolvable problem. I am sure that contributed to additional mountain sickness!