## Archive for November 11, 2010

### Joy in Repetition

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*