This problem was given to 8th graders in Russia for admission to a selective high school, but I admit that it stumped me! The problem was also presented with only text. But to get you started, I will provide the following figure.

Can you figure it out? Give the problem a try, and when you’re ready, watch the video for a solution.

I thank Michael K. for sending the problem and deriving the solution presented in the video.

**Can You Solve This HARD 8th Grade Geometry Problem From Russia? The Octagon In The Parallelogram**

Or keep reading.

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**Answer To The Octagon In A Parallelogram**

There are many ways to approach the problem. One approach involving higher mathematics goes like follows.

Apply an affine transformation (a linear transformation plus a possible vector translation) to transform the entire shape so the parallelogram becomes the unit square at the origin. The ratio of areas is invariant under affine transformations, so we can solve for the ratio of an octagon in the unit square and it will be the same answer as the original problem.

There are several methods to solve the octagon in the square at CAS Musings.

This is the easiest solution I’ve found: Google docs octagon in square.

There are also several methods to solve the problem at Cut The Knot, but most of them depends on the following fact: the centroid of a triangle (where the medians intersect) divides each median into a 2:1 ratio.

But an 8th grader will definitely not know about affine transformations, and might not know about medians and centroids (in America geometry is usually taught in 9th or 10th grade).

So how can an 8th grader solve this problem? Michael K. came up with a remarkable solution that depends only on similar shapes and knowing the area of a triangle (or a trapezoid). These concepts are certainly taught by 8th grade.

The key is seeing the pattern of similar shapes.

**The Solution**

The problem was sent to me by email by Michael K. who also derived the solution.

Let *p* be the area of the parallelogram and *x* be the area of the octagon. We want to solve for *x*/*p*.

Connect the midpoints of opposite sides of the parallelogram. This divides the octagon into 4 regions of equal area (*x*/4) and also the parallelogram into 4 smaller parallelograms with area *p*/4.

In the upper right small parallelogram, repeat the steps again: connect the midpoints of opposite sides. Then repeat for the lower left small parallelogram: connect the midpoints of the opposite sides.

(For better graphics please watch the video in which I also do animations to show the “zooming” into the smaller regions: **The Octagon In The Parallelogram Problem**)

We end up partitioning a quarter of the octagon into four regions

–a small parallelogram with area *p*/64 (we scaled by 1/4 three times)

–two regions with trapezoids that have area (3/4)(*p*/64) (the trapezoid is the area of the small parallelogram minus a triangle with area 1/4 of the small parallelogram)

–one region that is a scaled copy of the quarter of the octagon! This region is scaled by 1/4 two times, so its area is (1/4)(1/4)(*x*/4) = *x*/64. (For rigor, you can prove the similarity of the quarter of an octagon (a kite-shaped figure). One method is to connect the lower left corner to the upper right corner to split the shape into triangles, and you can prove the triangles of the small kite and the large kite are similar.)

We can add up these areas to find the area of a quarter of the octagon. And we also know that area is *x*/4. So we equate the expressions to get:

*x*/4 = *p*/64 + 2(3/4)*p*/64 + *x*/64

We can solve this to find:

*x*/*p* = 1/6

Amazing! The octagon’s area is 1/6 the parallelogram’s area.

As we have solved for a parallelogram, the same result applies for any parallelogram (including rhombuses, rectangles, and squares).

**Sources and solutions for octagon in the square**

Cut The Knot (1): *The diagrams are in an applet (blocked by most browsers), so it’s hard to figure out the solutions, but a careful reader can work through it!

https://www.cut-the-knot.org/Curriculum/Geometry/OctagonInParallelogram.shtml

Cut The Knot (2): *The diagrams are in an applet (blocked by most browsers), so it’s hard to figure out the solutions, but a careful reader can work through it!

https://www.cut-the-knot.org/Curriculum/Geometry/OctagonByOverlap.shtml

Math StackExchange octagon in square

https://math.stackexchange.com/questions/978384/area-of-octagon-constructed-in-a-square

CAS Musings octagon in square compilation

https://casmusings.wordpress.com/2014/11/13/squares-and-octagons-a-compilation/

CAS Musings octagon in square trigonometry solution

https://casmusings.wordpress.com/2014/11/08/squares-and-octagons/

FiveTriangles octagon in square solution 1

https://docs.google.com/document/d/1AvrWFOf1TeUhaJPFDJLH2BNCKu2QLSIsorjngJ6JoB0/edit#heading=h.kmcbdznv9nix

FiveTriangles octagon in square solution 2

https://docs.google.com/document/d/1MHSp0RFVHBUQLPmLpXeSvd_jiJKF3-EdTJxIaxlnJBo/edit#heading=h.5k5c3iq9cfjy

David Radcliffe octagon in square solution

http://gotmath.com/doc/octagon.pdf

Mike Lawler octagon in square solution

https://twitter.com/mikeandallie/status/531765555754446848/photo/1

CoMaC octagon in square similar triangles

http://apfstatic.s3.ap-south-1.amazonaws.com/s3fs-public/14-comac_alternate-solution-octagon-problem.pdf

Home plate is made as a 17 inch square with two corners removed. The top edge is 17 inches, and its two adjacent sides are 8.5 inches. The remaining two sides are defined to be 12 inches each and they meet at a right angle.

But there’s something wrong with this figure. It’s mathematically impossible!

Can you figure out why? Let’s assume that all angles and the dimensions of 17 and 8.5 are correct. What should be the length of the remaining two sides? Can you figure it out?

Watch the video for a solution.

**Baseball’s Home Plate Is Impossible Mathematically**

Or keep reading.

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**Answer To Baseball’s Home Plate Is Impossible Mathematically**

Let’s connect the two points that are opposite the sloping lines. This side has a length of 17 inches, and it divides home plate into an upper rectangle and a lower isosceles right triangle.

The isosceles right triangle has a hypotenuse of 17 inches. If each leg of the triangle has a length *x*, then by the Pythagorean Theorem the sum of the squares of the legs has to be the square of the hypotenuse.

*x*^{2} + *x*^{2} = 17^{2}

2*x*^{2} = 17^{2}

*x*^{2} = 17^{2}/2

*x* = 17/√2 ≈ 12.02 inches

Technically the sloping sides should be slightly longer than 12 inches!

While the error is quite small, baseball is a game of inches. Maybe the next iteration of the rulebook can include the language the sides should be “approximately 12 inches” to be completely mathematically correct.

But this rule for home plate has been in the books since 1900. So it stands to reason baseball will continue with this tradition, even if it’s slightly mathematically incorrect.

**Sources**

Rule 2.02 in 2017 MLB handbook

http://mlb.mlb.com/mlb/downloads/y2016/official_baseball_rules.pdf

Home base shall be marked by a five-sided slab of whitened rubber. It shall be a 17-inch square with two of the corners removed so that one edge is 17 inches long, two adjacent sides are 8½ inches and the remaining two sides are 12 inches and set at an angle to make a point. It shall be set in the ground with the point at the intersection of the lines extending from home base to first base and to third base; with the 17-inch edge facing the pitcher’s plate, and the two 12-inch edges coinciding with the first and third base lines. The top edges of home base shall be beveled and the base shall be fixed in the ground level with the ground surface. (See drawing D in Appendix 2.)

*Note: drawing D in Appendix 2 indicates the 12 inch sides span a right angle of 90 degrees.

MLB Field Dimensions

http://m.mlb.com/glossary/rules/field-dimensions

Home plate is a 17-inch square of whitened rubber with two of the corners removed so that one edge is 17 inches long, two adjacent sides are 8 1/2 inches each and the remaining two sides are 12 inches each and set at an angle to make a point. The 17-inch side faces the pitcher’s plate, and the two 12-inch edges coincide with the first- and third-base lines. The back tip of home plate must be 127 feet, 3 and 3/8 inches away from second base.

Now I know

http://nowiknow.com/why-every-baseball-game-breaks-the-rules/

Via NY Post

http://nypost.com/2016/10/05/americas-pastime-is-based-on-impossible-math/

MathWorld

http://mathworld.wolfram.com/HomePlate.html

Math on the McKenzie

http://mathonthemckenzie.blogspot.com/2013/05/pythagorean-geometry-confounds-home.html

The Mathematical Tourist

http://mathtourist.blogspot.com/2010/06/pythagoras-at-plate.html

This problem comes from Elchanan Mossel. Mathematician Gil Kalai posted the puzzle to his blog Combinatorics and more and asked readers to test their intuition by selecting an answer choice from a list of options.

The blog frequently covers topics from graduate level mathematics, so it would be fair to say its readers are exceptional mathematically. But amazingly over 50 percent chose the same incorrect answer, and only about 18 percent got the correct answer. So it’s fair to say this is a counter-intuitive probability puzzle. Can you figure it out?

Watch the video for a solution.

**This Dice Puzzle Stumps Most Mathematicians! The Even Number Rolls To 6 Paradox**

Or keep reading.

*I find it laughable many discussions on this problem also include a discussion of whether “die” is the singular of “dice.” Modern usage is that “dice” is a correct singular form. See The Oxford English Dictionary definition and usage for dice:

https://en.oxforddictionaries.com/definition/dice.

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**Answer To Elchanan Mossel’s Dice Probability Puzzle**

At the time I drafted this post, the poll on Gil Kalai’s blog showed over 50 percent gave the incorrect answer of 3 rolls.

The correct answer is 1.5 rolls, which was given by about 18 percent of readers.

Why did people pick 3? The probability of rolling a 6, given that you rolled an even number is:

Pr(6 | even roll) = Pr(6 | roll 2, 4, or 6) = 1/3

The rolls of a dice follow a geometric distribution, so the expected number of rolls to a particular outcome is then the reciprocal of the mean. So many people thought:

E(rolls to 6 | even rolls) = 1/Pr(6 | even roll) = 1/(1/3) = 3

But this is wrong! This would be the correct answer to the following: roll a three-sided dice with only the even numbers 2, 4, 6. How many rolls does it take for a 6 to appear? That does take 3 rolls on average.

This puzzle is asking a different question: it is asking about rolling a standard dice until you get a 6, and then only considering sequences of rolls with even numbers (2s and 4s) before the roll of 6.

The slight difference changes the conditional expectation, so let’s now derive the answer.

**The Solution**

In the event that we roll only even numbers until we roll a 6, we must only roll 2s and 4s until we eventually roll a 6.

In other words, the outcome of the rolls is a sequence of 2s and 4s that eventually terminate in a 6. We are interested in the average length of the sequence, including the final 6.

But there’s nothing special about 6. Since each roll shows with equal chance, the sequence could also terminate in a 1, or a 3, or a 5, and it would also have the same average length.

So we ask: what is the expected number of rolls until we get a 1, 3, 5, or 6, given that all previous rolls involved 2s or 4s?

You might think “intuitively” as follows, as explained by Ben Orlin on Math With Bad Drawings. The event of rolling 1, 3, 5, or 6 is 4/6, so the expected rolls to that is the reciprocal, or 1/(4/6) = 6/4 = 3/2 = 1.5. This is the correct answer. But since our intuition already led us to the wrong answer, let’s use another method to be sure this is the correct answer.

We can solve this by capitalizing on the memoryless property of the geometric distribution.

Consider the very first roll. There is a 4/6 chance we roll a 1, 3, 5, or 6, and the experiment ends with 1 roll. But there is also a 2/6 chance we roll a 2 or 4. In that case, we have used 1 roll. As we start out second roll, the dice has no memory of the first roll–it is like we start anew. So the expected rolls to end is now 1 more than when we started. So we have the equations:

*L* = E(total rolls until 1, 3, 5, 6 | previous rolls are 2, 4)

*L* = E(*L* | first roll 1, 3, 5, 6) + E(*L* | first roll not 1, 3, 5, 6)

*L* = (1)Pr(first roll 1, 3, 5, 6) + (*L* + 1)Pr(first roll not 1, 3, 5, 6)

*L* = (1)(4/6) + (*L* + 1)(2/6)

(4/6)*L* = 1

*L* = 6/4 = 3/2 = 1.5

Conditional on the event you only roll an even number, it only takes an expected 1.5 rolls to see a 6. This is because in most of the cases you will roll a 6 on the very first roll which brings the average down.

**Alternate solution**

We can also solve the problem from the definition of a conditional expectation.

Let’s first solve for the denominator.

Then we substitute the result and expand the expectation in the numerator.

Finally we can solve this series by subtracting 1/3 the sum and then solving for the sum of a geometric series.

(Technically we need to make sure the sum converges first–we would need to solve for the partial sum and then take the limit as the number of terms goes to infinity. But I’ll take a shortcut–this is an arithmetico-geometric series with a common ratio less than 1, so it converges. So knowing the series converges, we can solve for the value of the sum.)

So the answer is again 3/2 = 1.5 rolls.

**Sources**

Gil Kalai blog test your intuition (puzzle and poll)

https://gilkalai.wordpress.com/2017/09/07/tyi-30-expected-number-of-dice-throws/

Gil Kalai blog test your intuition solution

https://gilkalai.wordpress.com/2017/09/08/elchanan-mossels-amazing-dice-paradox-answers-to-tyi-30/

Puzzling StackExchange

https://puzzling.stackexchange.com/questions/54974/elchanan-mossel-s-dice-puzzle

Math With Bad Drawings – A probability puzzle that you’ll get wrong

https://mathwithbaddrawings.com/2017/09/25/a-probability-puzzle-that-youll-get-wrong/

*The Puzzling StackExchange post was removed (I am not sure why). Here is another discussion:

https://math.stackexchange.com/questions/2463768/understanding-the-math-behind-elchanan-mossel-s-dice-paradox

And another explanation:

http://www.yichijin.com/files/elchanan.pdf

“Jess wants 200 ribbons of length 110 centimeters for a party. However, the ribbons were sold at 25 meters per tape. How many tapes will Jess need?”

Source: Yahoo News (this might not be the exact wording but it has the key details)

Many adults are finding this question is tricky. And I admit I fumbled in solving it too. Can you figure it out?

Watch the video for a solution.

**Adults Can’t Solve This Maths Problem For 12 Year Olds In Singapore**

Or keep reading.

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**Answer To Singapore Ribbon Question**

Here is the mistake that I and many people made.

*Wrong approach*

Each ribbon is 110 centimeters, which is 1.1 meters. Jess needs 200 ribbons, so Jess needs 200(1.1) = 220 meters of ribbon.

As each tape has 25 meters, Jess needs 220/25 = 8.8 tapes. As 8 tapes are not enough, Jess needs 9 tapes.

But this is wrong! The correct answer is 10.

*Correct answer*

While 9 tapes provide enough total ribbon length 9(25 m) = 225 m > 220 m, there is a problem. The calculation overlooks the wasted material in each tape.

Here is the correct approach. Each tape has 25 meters, so each tape can make 25/1.1 = 22.7272… ribbons. In other words, each tape can make 22 whole ribbons (using 24.2 meters) and there will be 0.8 meters leftover which goes to waste.

(The problem does not state if we can join the excess material between tapes, but it would be reasonable to assume we cannot as we have no way to join ribbon material easily).

Now, Jess needs to make 200 ribbons. As each tape has material for 22 ribbons, Jess needs 200/22 = 9.0909… tapes. As 9 tapes are enough each since 9(22) = 198, we need to round up to 10 tapes.

Therefore, Jess needs 10 tapes.

**Sources**

The New Paper Question

http://www.tnp.sg/news/singapore/parents-irked-tricky-psle-maths-question

The New Paper Answer

http://www.tnp.sg/news/singapore/psle-ribbon-answer

Yahoo News Singapore

https://sg.news.yahoo.com/2017s-psle-maths-ribbon-question-needlessly-tricky-012631653.html

Mothership

https://mothership.sg/2017/10/can-you-solve-this-tricky-2017-psle-math-question-on-ribbons/

6 + 66 + 666 + … + 66…6 = ?

Each subsequent term in the series has an extra 6, and the last term has the digit 6 repeated 666 times.

Watch the video for a solution.

**Can You Solve The Sum Of 6s? An “Evil” Math Problem**

Or keep reading.

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**Answer To An Evil Summation**

The answer is:

(2/3)((10^{667} – 10)/9 – 666)

= 740740…740296

The result has 666 digits, of which the last 3 are 296, and the other 663 digits are the string 740 repeated 221 times.

I will present 3 solutions to this problem, which are very similar variations of the same idea. The main concept needed to solve this problem is the sum of a geometric series.

**Sum of a geometric series formula**

Consider a geometric series where each term is multiplied by a common ratio *r* and the series begins with the term 1. Let’s consider *S*_{k}, the sum of the first *k* terms.

*S*_{k} = 1 + *r* + *r*^{2} + … + *r*^{k – 1}

To find the sum, we multiply both sides by *r*, which “shifts” the terms one spot.

*r**S*_{k} = *r* + *r*^{2} + … + *r*^{k}

Now we subtract the two equations. Almost all terms cancel except the first term of *S*_{k} and the last term of *r**S*_{k}.

*S*_{k} – *r**S*_{k} = 1 – *r*^{k}

Now we can factor *S*_{k} and solve:

*S*_{k}(1 – *r*) = 1 – *r*^{k}

*S*_{k} = (1 – *r*^{k})/(1 – *r*)

If the series instead started with the term *a* instead of 1, then we would multiply each term by that factor, and so the eventual sum is also multiplied by the factor *a*. The general formula for a geometric series is:

*S*_{k} = *a*(*r*^{k} – 1)/(*r* – 1) = *a*(1 – *r*^{k})/(1 – *r*)

For an infinite series, if |*r*| < 1, we can take the limit as *k* goes to infinity to arrive at the well-known formula:

*a*/(1 – *r*)

**Solution 1: geometric series, twice**

Let’s solve the problem in which the last term has the digit 6 repeated *n* times:

6 + 66 + 666 + … + 66…6

First we factor out 6 from each term.

6 + 66 + 666 + … + 66…6

= 6(1 + 11 + 111 + … + 11…1)

Now write the expanded form of each term.

6((1) + (10 + 1) + (100 + 10 + 1) + … + (10^{n – 1} + 10^{n – 2} + … + 1))

Each group of parentheses is a geometric series with a starting term *a* = 1 and a common ratio *r* = 10. So we use the formula for the sum of a geometric series to find each group of *k* terms has a sum:

*a*(*r*^{k} – 1)/(*r* – 1)

(1)(10^{k} – 1)/(10 – 1)

= (10^{k} – 1)/9

Substituting this formula into each group, we get:

6((1) + (10 + 1) + (100 + 10 + 1) + … + (10^{n – 1} + 10^{n – 2} + … + 1))

= 6((10^{1} – 1)/9 + (10^{2} – 1)/9 + (10^{3} – 1)/9 + … + (10^{n} – 1)/9)

Now we can factor out 1/9.

(6/9)((10^{1} – 1) + (10^{2} – 1) + (10^{3} – 1) + … + (10^{n} – 1))

We can group the *n* terms of -1 which will have a sum of –*n*. We can also group what’s left, which is another geometric series of powers of 10.

(6/9)(- 1(*n*) + (10^{1} + 10^{2} + 10^{3} + … + 10^{n})

Using the sum of a geometric series formula again, we get:

(6/9)(-*n* + (10^{n + 1} – 10)/9)

= (6/9)((10^{n + 1} – 10)/9 – *n*)

The problem had *n* = 666, so we can substitute to find the sum:

(6/9)((10^{667} – 10)/9 – 666)

= (2/3)(9999…90)/9 – 666)

= (2/3)(111…10) – 666)

= (2/3)(111…10 – 666)

= (2)(370370…370 – 222)

= 740740…740 – 444

= 740740…740296

The number has a total of 666 digits, and the string 740 is repeated 221 times.

You can verify the result on WolframAlpha.

A nice thing about this problem is we can quickly generalize to any other digit besides 6. Notice that we have found:

1 + 11 + 111 + … + 11…1 = (1/9)((10^{n + 1} – 10)/9 – *n*)

If we wanted to sum 2s, or 3s, or another digit *k*, we can get the formula by multiplying both sides of the equation by *k*.

*k*(1 + 11 + 111 + … + 11…1) = (*k*/9)((10^{n + 1} – 10)/9 – *n*)

**Solution 2: magical substitution**

This method is a slight shortcut on the first solution. First, factor out 6 from each term.

6 + 66 + 666 + … + 66…6

= 6(1 + 11 + 111 + … + 11…1)

In solution method 1, we wrote each term in expanded form and then used a geometric series. But there is a faster way to solve this.

We can multiply the formula by 1 = 9/9. The factor 9 we will distribute inside the sum, and the 1/9 factor will stay outside the sum.

(9/9)6(1 + 11 + 111 + … + 11…1)

= (1/9)6(1(9) + 11(9) + 111(9) + … + 11…1(9))

= (1/9)6(1(9) + 11(9) + 111(9) + … + 11…1(9))

= (6/9)(9 + 99 + 999 + … + 99…9)

Now we do another trick: each term is 1 less than a power of 10.

(6/9)(9 + 99 + 999 + … + 99…9)

= (6/9)((10 – 1) + (100 – 1) + (1000 – 1) + … + (10^{n+1} – 1))

We can now group the powers of 10 as a geometric series, and we can group the *n* terms of -1 as another factor.

= (6/9)((-1)*n* + 10 + 100^{2} + 10^{3} + … + 10^{n+1})

Now the proof continues just like in solution 1. You’ll notice this solution method is slightly shorter as you only need to apply the sum of the geometric series once. If you work through enough competition math questions, you will use tricks like this all the time, and you will need to use these kinds of tricks to solve the problem quickly.

But this solution may be off-putting for students that don’t “see” the magical substitutions. So I’m fine with solving the problem using more steps, if it means more people can understand the solution and apply its methods.

In fact, I solved the problem in an even harder way.

**Solution 3: arithmetico–geometric sequence**

I first factored out the 6 from each term and considered the expanded form.

6 + 66 + 666 + … + 66…6

= 6(1 + 11 + 111 + … + 11…1)

= 6((1) + (10 + 1) + (100 + 10 + 1) + … + (10^{n – 1} + 10^{n – 2} + … + 1))

Then I figured I could count the number of 1s, 10s, 100s, and so on. As each of the *n* groups has a 1, there are *n* terms of 1. Then every group except the first has a 10, so there are *n* – 1 groups with 10. The pattern continues where each factor of 10 is in 1 fewer groups, so there are *n* – *k* groups with the factor 10^{k}. So we can write the sum as:

6((1) + (10 + 1) + (100 + 10 + 1) + … + (10^{n – 1} + 10^{n – 2} + … + 1))

= 6(*n*(1) + (*n* – 1)10 + (*n* – 2)100 + … + 1(10^{n – 1}))

= 6(*n*(10^{0}) + (*n* – 1)10^{1} + (*n* – 2)10^{2} + … + 1(10^{n – 1}))

Notice the terms *n*, *n* – 1, …, 1 form an arithmetic series where each term is decreased by 1. And notice the terms 10^{0}, 10^{1}, … , 10^{n – 1} form a geometric series in which each term is multiplied by 10.

At first I was stuck on this step and thought I made a mistake. I had never learned how to solve this kind of sum so I thought about starting over. But part of mathematics is experimenting.

It turns out this hybrid kind of series is called an artihmetico-geometric series. And we can actually solve for the sum much like we can the sum of a geometric series. We multiply the partial sum by the common ratio of the geometric series (10) and then subtract the two equations (subtract the first from the second) as follows:

*S*_{n} = 6(*n*(10^{0}) + (*n* – 1)10^{1} + (*n* – 2)10^{2} + … + 1(10^{n – 1}))

10*S*_{n} = 6(*n*(10^{1}) + (*n* – 1)10^{2} + (*n* – 2)10^{3} + … + 1(10^{n}))

9*S*_{n} = 6(-*n*(10^{0}) + 10^{1} + 10^{2} + … + 10^{n – 1} + 10^{n})

Now we can use the sum of a geometric series on the powers of 10.

9*S*_{n} = 6(-*n* + (10^{n+1} – 10)/9)

Finally we divide both sides of the equation by 9.

*S*_{n} = (6/9)[(10^{n+1} – 10)/9 – *n*]

**Thanks to all patrons! Special thanks to:**

Marlon Forbes

Shrihari Puranik

Kyle

If you like my videos, you can support me at Patreon and get exclusive rewards: http://www.patreon.com/mindyourdecisions

**Appendix: proof 10 ^{n} – 9n – 1 is divisible by 81 for n a natural number**

The general formula always evaluates to an integer, and I received a really nice write-up proving this. Check out three proofs of this:

**Sources**

https://alokgoyal1971.com/2017/08/20/puzzle-192-too-many-6s/

*A Moscow Math Circle, Week by Week Problem Sets* by Sergey Dorichenko

https://alokgoyal1971.com/2017/09/02/solution-to-puzzle-192-too-many-6s/

https://www.quora.com/What-is-the-sum-of-1+11+111+1111+11111

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