Problem: 5109094x171709440000 = 21!, find x.

Solution

I can immediately think of a couple of question:

  1. What is so special (if, of course, anything at all) about the 8th digit from the left? Can I hide another digit and have a meaningful problem?
  2. What is so special about the number 21? If I compute the factorial of another number, say 10, would I be able to retrieve a hidden digit?

With regard to the first question, all digits look the same to me except for the trailing zeros. Zeros at the end of a number indicate the number of factors of 10 this number has. Let pursue this. Where do these factors of 10 come from? The number being a factorial, it's the product

21! = 1·2·3·4·5·...·19·20·21.

Then there are two factors (10 and 20) that contribute zeros to the result and another two (5 and 15) that combine with other even factors (of which there is a plenty) to add another two 0s. At this point, we began utilizing the fact that the given number is known to have certain factors. However, returning to the question asked, I can't think of a way in which other factors of 21! have affected the 8th digit.

As far as my current understanding of the problem goes, there is nothing special about the 8th digit. I also begin suspecting that the choice of 21 in the formulation of the problem is not very important. Indeed it would not be important if the question was to count the number of trailing 0s.

The important thing is that the number is presented as a product of several factors. This is actually the only piece of information that could be gathered from the original formulation. What do I know about factors? They are divisors of a product. Yes, indeed, I have already used properties of multiples of 5 and 10, right? Do I know of other rules concerning division of a number by other numbers? Actually, in school, we do not study a lot about numbers and their features. The study of Arithmetic concentrates on addition and other operations, on the table of multiplication - nothing very generic. And next we plunged into Algebra where numbers have hardly been mentioned at all. Is it also your recollection? (An aside: this is why I disagree with the dictionary definitions of Mathematics as a study of numbers. There is nothing in our education that remotely suggests any study of numbers as such.)

I am lucky, however, to remember one property of numbers divisible by 3 and 9.

  A number is divisible by 3 (or 9) iff the sum of its digits is divisible by 3 (or 9).

I start feeling excited. The statement above does not say anything specific about the number of digits or their position. This jibes well with the hunch that selection of the 8th digit was arbitrary. But to apply this statement we have to establish whether the number is divisible by 3 (or 9). It's easy to check that all factorials starting with 3! are divisible by 3 whereas, starting with 6!, all factorials are divisible by 9. Therefore, 21! is divisible by 9. Can we use this?

The sum of digits of 21!, as it's presented in the problem, is (61 + x). This is divisible by 9 iff x = 2. This must be the missing digit! Division by 3 could also be used for the 8th digit. However, it would fail for the first one.

Summing up:

  Given n!, n>5, with one digit removed.Then it is possible to recover the digit.

Please do not jump to conclusions. Remember the problem. All we were doing was trying to recover the first digit? May there arise other complications? Sure. There is no way (or, rather, we have not established a way) to tell 0 from 9.

Lastly, if finding a digit as above looked as a magic trick to you, you should know that the principle was and is indeed used in magic tricks. Math Telepathy is one of them.


Related material
Read more...

  • Divisibility Criteria
  • Fermat's Little Theorem
  • Divisibility by 7, 11, and 13
  • Divisibility Criteria (Further Examples)
  • Criteria of divisibility by 9 and 11
  • Division by 81
  • When 3AA1 is divisible by 11?
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