## 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 *n*th 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*

Entry filed under: Uncategorized. Tags: Prince, repdigit, repetition.

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

2.venneblock | November 12, 2010 at 10:27 pmGlad 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.

3.len | March 12, 2012 at 3:04 pmHello, 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 … !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

4.venneblock | March 12, 2012 at 8:26 pmI understand

repunit, Len, since it’s a repdigit number with all 1’s. But where doesdemlocome from?Sorry, I don’t understand the question your asking in #3. Sounds like it could be an interesting investigation, though. Please explain.