Putnam math competition

This is really cool. Tonight, I joined the kids here who will give this math test a shot in a few months. We had an overflow crowd at our first meeting.

The professor who's teaching us the ropes in weekly sessions is a great guy.

Here's one problem I did, from the promo poster:

Find the smallest natural number n which has the following properties:
(i) its decimal representation has a 6 as its last digit, and
(ii) if the last digit 6 is erased and placed in front of the remaining digits, the resulting number is four times as large as the original number n.

If n starts with 6, and n = 4k, then k must start with a 1. So:

n = 61 ..,

k = 1 ... 6

Now the cool part: some recursive long multiplication.

Multiplying 6*4 gives you 24, so 4 is the last digit of n. This means that 4 is also the second-to last digit of k. So 4*4 +2 = 18. 8 is the second-to-last digit of n, and third-to-last digit of k. We know that 1 is the second digit of n, so eventually, we have to get to a 1.

    2
1....6
    x4
61...4

   1
1...46
    x4
61..84

  3
1..846
    x4
61.384

25
153846
    x4
615384

This is true. So I think that 153846 is the smallest n that ends in 6 where shifting the digits right gives you 4n.
I love this stuff.

Hello world

This is my first real blog. I started my freshman year at Stanford a week ago today, and it’s been fantastic. There are too many great things to do and not enough hours in a day to do them. So much is going on, in fact, that I think it’s high time for me to write a blog.

I’ll write about Stanford, about programming, about math, and about whatever looks fun or interesting to me.

Onwards…