Solving HARD Olympiad Problem With A Neat Trick

Thanks to León from Argentina for suggesting this problem! It comes from the 1995 Mathematical Olympiad, and I will explain how you could solve it in your head in 1 minute.

Amazon's hanging cable problem:

Hardest easy geometry problem:

My book "The Best Mental Math Tricks (rated 4.8/5 stars on 5 reviews)

Picasso joke

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"40 Paradoxes in Logic, Probability, and Game Theory" contains thought-provoking and counter-intuitive results. (rated 4.4/5 stars on 13 reviews)

"The Best Mental Math Tricks" teaches how you can look like a math genius by solving problems in your head (rated 4.8/5 stars on 5 reviews)

"Multiply Numbers By Drawing Lines" This book is a reference guide for my video that has over 1 million views on a geometric method to multiply numbers. (rated 4.3/5 stars on 6 reviews)

Connect with me

[mikekeenanphd]

The answer is no. Because if you could, it would not be possible to prove that without taking up more time than allowed by the exam.

[karuna]

I could assume that the answer is "No", because it is really hard to prove that a 44 digit number is a prime.

[NestorAbad]

Nice problem with a nice key message!


By using modular arithmetic it's also pretty quick to show that any of those arrangements is a multiple of 11: let x_1, x_2, x_3, ..., x_11 be the eleven numbers from 1985 to 1995 in some order. Then, the 44-digit number we get by concatenating them is


x_11·1 + x_10·10^4 + x_9·10^8 + ... + x_1·10^40


Now, as 10≡-1 (mod 11) then 10^(4n)≡1 (mod 11) and the above sum (mod 11) is congruent to


x_11 + x_10 + x_9 + ... + x_1 (mod 11)


but as these eleven numbers are consecutive (in some order), they form a complete system of remainders (mod 11), that is, the sum is congruent to


0+1+2+3+4+5+6+7+8+9+10 (mod 11) ≡ (10·11)/2 (mod 11) ≡ 55 (mod 11) ≡ 0 (mod 11),


as we wanted to show.

[wschmrdr]

Never knew that about 11. I was looking for 3, though.

[alexholker]

If the answer was "yes, " it would take more than one minute to determine that it was true. Therefore, the answer is "no."

[phoenixshade]

Never knew that divisibility by 11 rule, but this has successfully implanted it in the permanent mental repository.

[GIFPES]

At the very beginning, I watched the most incredible lesson of humility I have ever seen in a math teacher when you say that there are many problems you could not solve by yourself. Not all mathematicians are so humble. I really loved your honesty. The greatest teacher must know that some things are not known by him, in order to be able to learn and be even wiser!

[Simio_Da_Tundra]

Awesome! Tip for any puzzle involving digits in big number like this: usually the answer has to do with the divisibility rules for 3 and 11, since those are easier than others (say, for 7)

[HeroDarkStorn]

I don't think "You captured my essence" is something anyone said to Picasso ever.

[pietergeerkens]

Even simpler:

Note that for eleven consecutive numbers, every residual modulo 11, from 0 to 10, occurs exactly once.

The sum of those 11 numbers, 0 + 1 + 2 + ... + 9 + 10 = 10*11/2 = 55 which is divisible by eleven.

Therefore ANY ELEVEN consecutive four digit numbers will be divisible by eleven for any arrangement, using the "casting out elevens" divisibility rule you mentioned.

Further: Since 11 is prime. for any integer a and any four digit number b such that

1000 < b

and

b + 10a < 9999

the eleven four digit numbers

b, b+a, b+2a, ... b+9a, b+10a

will also be divisible by eleven in every arrangement.

Finally:

"Casting out elevens" modulo 100, taking each pair of digits moving right-to-left from the decimal point, works for the very same reason that "casting out nines" works for individual digits taken one at a time right-to-left from the decimal point, because 99 = 11*9 = 100-1 just as 9 = 10-1. And this works for any base, for every divisor of the number one less than the base.

[PlutoTheSecond]

There's actually another divisibility by 11 rule that makes this slightly easier. Just break the number into blocks of two digits and then add up all the blocks (if there's an odd number of digits, pad the left with a 0). For example, 1991 is divisible by 11 because 19 + 91 = 110 is divisible by 11. However, 243 is not divisible by 11 because 02 + 43 = 45 is not divisible by 11.

Then, however you arrange the 11 consecutive numbers, the sum is 19 * 11 + (85 + 86 + 87 + … + 94 + 95). 19 * 11 is obviously divisible by 11, and (85 + … + 95) = 90 * 11 because it's easy to calculate the average of n consecutive numbers, and 90 * 11 is obviously divisible by 11. Therefore, the entire number must be divisible by 11.

[loginacces]

Thank you for the speech there, I went to a hard time on some math situations, and I realise that I should keep improving on what I like, math.

[wladislawortlieb]

Thank you.
I'm 17th yo russian guy, i'm not even interested in math a years ago, but recently i start learning physics so hard and i understand, that without math i can not to study physics seriously. And when i start learning math seriously, i fell in love with this science)
I'll be mathemetican. I love it. I'm not so gifted as mostly of guys which do math very well, but i can't imagine my life without math.

[pandabearguy]

For once I solved it myself quickly. Actually just learned about the divisibilty tricks, of which 11 was one of them, a few months ago.

[CauchyIntegralFormula]

Since 10000=1 mod 9999, any arrangement is congruent to the sum of the whole numbers, mod 9999. That sum is 1990*11, which shares a factor of 11 with 9999, so the concatenated integer will always be divisible by 11. (This method would also have picked up if the concatenated number always had a factor of 3 or 101.)

[mijmijrm]

love the inro and love the revelation of the unknown-to-me divisibility-by-11 test.

[anuragsinghbisht]

A very great with a very great message ! Everyone should really learn this fact!

[MattStMarie-bmsq]

I see a few problems with your answer. 1. I can arrange the numbers within other numbers such as in an anagram.
2. They didn't say they all had to be in the same base you could have used any base of 10 and above. Also the base of the prime number could be different.

[jedagelijksebraintraining]

1991 (middle number) is divisible by 11, the other numbers are 11n+/- x. Added together you get 11m. What is also divisible by 11.

[jeminkachhadiya]

It is also divisible by 3 with sum of 249, which is divisible by 3.