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Can You Win Riddler Jenga?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. Two puzzles are presented each week: the Riddler Express for those of you who want something bite-size and the Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in the next column. Please wait until Monday to publicly share your answers! If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

This week marks the third of four CrossProductâ„˘ puzzles. This time, there are seven three-digit numbers â€” each belongs in a row of the table below, with one digit per cell. The products of the three digits of each number are shown in the rightmost column. Meanwhile, the products of the digits in the hundreds, tens and ones places, respectively, are shown in the bottom row.

280
168
162
360
60
256
126
183,708 245,760 117,600

Can you find all seven three-digit numbers and complete the table?

The solution to this Riddler Express can be found in the following column.

## Riddler Classic

In the game of Jenga, you build a tower and then remove its blocks, one at a time, until the tower collapses. But in Riddler Jenga, you start with one block and then place more blocks on top of it, one at a time.

All the blocks have the same alignment (e.g., east-west). Importantly, whenever you place a block, its center is picked randomly along the block directly beneath it. For example, the following animation shows Riddler Jenga towers that were randomly constructed before ultimately collapsing when the fifth, 10th and 15th blocks were placed. The block highlighted in red is the one above which the blocks were no longer balanced.

On average, how many blocks must you place so that your tower collapses â€” that is, until at least one block falls off?

(Note: This problem is not asking for the average height of the tower after any unbalanced blocks have fallen off. It is asking for the average number of blocks added in order to make the tower collapse in the first place.)

The solution to this Riddler Classic can be found in the following column.

## Solution to last weekâ€™s Riddler Express

Congratulations to đź‘ŹDavid Mercado đź‘Ź of Fairfield, CT, winner of last weekâ€™s Riddler Express.

In last weekâ€™s CrossProductâ„˘ puzzle, there were six three-digit numbers, with products shown in the following table:

210
144
54
135
4
49
6,615 15,552 420

As always, a good place to start was to write out the possible ways each rowâ€™s product could be the product of three digits:

• 210 was 5Ă—6Ă—7.
• 144 was 2Ă—8Ă—9, 3Ă—6Ă—8, 4Ă—4Ă—9 or 4Ă—6Ă—6.
• 54 was 1Ă—6Ă—9, 2Ă—3Ă—9 or 3Ă—3Ă—6.
• 135 was 3Ă—5Ă—9.
• 4 was 1Ă—2Ă—2 or 1Ă—1Ă—4.
• 49 was 1Ă—7Ă—7.

Next, many solvers worked out the prime factorization of the three columnsâ€™ products, so they could figure out which digit went into which column.

• 6,615 was 33Ă—5Ă—72.
• 15,552 was 26Ă—35.
• 420 was 22Ă—3Ă—5Ă—7.

Since the middle columnâ€™s product didnâ€™t have a factor of 7, that meant the last rowâ€™s number had to be 717. There was just one 7 left in the puzzle: in the first column and the first row. Also, since there were no factors of 5 in the second column, that meant the first rowâ€™s number was 765. Now there was just one 5 left in the puzzle: in the first column and the fourth row. And because the last column had just one factor of 3 (i.e., not two factors of 3), the number in the fourth row was 593.

Now that all the 5s and 7s were accounted for, all that remained were factors of 2 and 3. To make the numbers work out, the three remaining digits of the middle column somehow had to multiply to 288. The only two ways this was possible with three digits were 4Ă—8Ă—9 or 6Ă—6Ă—8. Since there was no way to get 6, 6 and 8 from the three remaining rows, those digits had to be 4, 8 and 9. Moreover, an 8 was only possible in the second row, meaning the 9 came in the third row and the 4 in the fifth row, which meant the number in the fifth row had to be 141. Finally, because the first columnâ€™s product had only factors of 3 remaining and the third column had only factors of 2, the number in the second row was 982 and the number in the third row was 392.

Here was an alternative animated approach, courtesy of solver Andrew Heairet:

## Solution to last weekâ€™s Riddler Classic

Congratulations to đź‘Ź Justin Ahmann đź‘Ź of Bloomington, Indiana, đź‘Ź Jenny Mitchell đź‘Ź of Nashville, Tennessee and đź‘Ź Peter Exterkate đź‘Ź of Sydney, Australia, winners of last weekâ€™s Riddler Classic.

Last week, you looked at self-intersecting polygons, whose sides cross over each other. (These are distinct from the simple polygons you learned about in school, whose sides do not intersect each other.)

For example, itâ€™s possible to draw a polygon for which every side intersects exactly two other sides (not counting the vertices, of course). The polygon with the fewest sides that still meets this criterion is the pentagram:

But was it possible to draw a polygon where each side intersected exactly three other sides? And if so, what was the minimum number of sides this polygon can have?

Despite the disbelief of several readers, it was indeed possible to draw such a polygon! The solutions and drawings you submitted were endlessly creative, replete with zig, zags and symmetries.

Many solvers found different 18-gons where each side intersected three other sides. These figures varied quite a bit, and included David Devoreâ€™s three concentric triangles, each of whose sides is crossed by inward and outward zags:

But it was possible to do even better! Jenny Mitchell was among this weekâ€™s winners for her submission of this 14-sided polygon:

Sure enough, 14 was the least number of sides anyone was able to find. Unless, of course, you look beyond Euclidean geometry, as Steve Curry did. Steve found that you could draw a 6-sided polygon on a sphere in which each side intersected three other sides, shown below. Meanwhile, Jenny found a 2-sided polygon (yes, a â€śbi-gonâ€ť) in a cylindrical geometry.

Finally, letâ€™s return to two dimensions for a moment. Solvers Rohan Lewis and James Anderson both explored the minimum number of sides for polygons in which each side intersected exactly k other sides for different values of k. Starting from k = 1, it appeared that this sequence was 6, 5, 14 (this weekâ€™s solution), 7, 10, etc. At the time of this writing, no such sequence exists on OEIS. Knowing Riddler Nation, a new sequence will surely be authored there sooner rather than later.

And if this sort of riddle wasnâ€™t for you, I hope youâ€™ll let bi-gons be bi-gons.

## Want more riddles?

Well, arenâ€™t you lucky? Thereâ€™s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. Itâ€™s called â€śThe Riddler,â€ť and itâ€™s in stores now!

## Want to submit a riddle?

Email Zach Wissner-Gross at riddlercolumn@gmail.com.

## Footnotes

1. Important small print: In order to đź‘Ź win đź‘Ź, I need to receive your correct answer before 11:59 p.m. Eastern time on Monday. Have a great weekend!

Zach Wissner-Gross leads development of math curriculum at Amplify Education and is FiveThirtyEight’s Riddler editor.