Joy in Repetition

November 11, 2010 at 9:09 pm 4 comments

Today is 11/11, a day of repeated digits, which makes it a good day to share an email with you that I recently received. My coworker Julia blows me out of the water when it comes to being a number geek, and that’s saying something. Of course, her love of mathematics may be destiny — her name is Julia, and her brother’s name is Vitali. This is the message she sent me:

This Thursday, my dad will be 55 years old. On that same day, I’ll be 11,111 days old. It’s a day of repeated digits for [my] family.

I don’t know what compelled her to do the calculations that allowed her to figure that out, but it reminded me of some repdigit problems. I’ll share those in a minute. But first, a question about Julia:

How old was Julia’s dad when she was born?

Okay, that one was pretty easy. Here’s my favorite repdigit problem, which is a little tougher but can still be attempted by most anyone:

What is the smallest positive integer that, when multiplied by 7, gives a positive integer result in which every digit is a 5?

Of course, there’s the really cool repdigit number pattern:

1 × 1 = 1
11 × 11 = 121
111 × 111 = 12,321
1,111 × 1,111 = 1,234,321

Which leads to the question:

What is the value of the product 1,111,111,111 × 1,111,111,111?

And to end all this silliness, just some facts about repdigit polygonal numbers — numbers that repeat the same digit that are also polygonal numbers. Define P(k,n) to be the nth polygonal number with k objects on a side. For instance, P(3,4) = 10, because P(3,4) is the notation for the 4th triangular number (k = 3). Then P(5,4) = 22 is a repdigit polygonal number, and so is P(8,925,662,618,878,671; 387) = 666,666,666,666,666,666,666. Wow.

As it turns out, there’s a formula for these beasts:

P(k,n) = (n/2)(k – 2)(n – 1) + n  (for n, k > 1)

Over and over,
she said the words
’til he could take no more…

– Prince, Joy in Repetition

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Theorems Without Thought Notable and Quotable

4 Comments Add your own

  • 1. David Onder  |  November 12, 2010 at 6:54 pm

    I enjoy finding patterns also which is what makes this problem fun.

    1 x 1 = 1 (total digits in product = 2; total digits in answer = 1)
    11 x 11 = 121 (4; 3)
    111 x 111 = 12321 (6; 5)

    I see a pattern, the number of digits in the answer is 2n-1 where n is the number of digits in the number being squared (at least for the first 9 iterations). Also, the middle digit is the number of digits in the number being squared (at least for the first 9 iterations).

    So, for 111,111,111 x 111,111,111 (which contains 9 digits), the answer is 12,345,678,987,654,321.

    And for the answer to the question: 1,111,111,111 x 1,111,111,111 we need to carry the 1 in the tens place so the answer is 1,234,567,900,987,654,321 (note the carry makes the next number a 10, thus another carry).

    That was fun!

    David

    Reply
    • 2. venneblock  |  November 12, 2010 at 10:27 pm

      Glad you enjoyed it, David!

      The beginning part of your answer (12,345,679, with the 8 missing) reminds me that if you multiply 12,345,679 by a multiple of 9 (9, 18, 27, …, 81), the result is a repdigit number.

      Reply
  • 3. len  |  March 12, 2012 at 3:04 pm

    Hello, I enjoyed this blog and this post … There are more things about these standards:

    1) the numbers that are described only with numerals “1” are called ‘repunit numbers’

    2) the product of “repunit numbers” is called demlo numbers

    3) one question: the pattern that relates the repdigit numbers to multiples of 9 goes for all the results of infinite multiplication table of 9 ??????

    Goodbye … !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    Reply
    • 4. venneblock  |  March 12, 2012 at 8:26 pm

      I understand repunit, Len, since it’s a repdigit number with all 1’s. But where does demlo come from?

      Sorry, I don’t understand the question your asking in #3. Sounds like it could be an interesting investigation, though. Please explain.

      Reply

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The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

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