(Untitled)

[eightbit]

A few days ago the NYT Magazine featured a nice article about Freeman Dyson, a scientist at IAS. I liked it, and found the problem from this quote (on page 8) to be a nice diversion for a little while. The story is great, too.

At Jason, taking problems to Dyson is something of a parlor trick. A group of scientists will be sitting around the

Comments

[arvindn]

fantastic story :-)

here's the shortest way i can think of to get the answer. still very, very impressive to do it in your head.

(10^n y + x) = 2 * (10x+y)

giving 10^n ≡ 2 (mod 19)

now 1/2 is 10 mod 19, or 10^(-1) is 2. if you already know that 10 is a primitive root mod 19 since multiplication modulo a prime is cyclic, 10 is a primitive root, and it follows that n >= 17, making 18 digits the smallest possible.

[eightbit]

Nice!

My solution was nearly the same, but I didn't get there so directly. For some reason I was distracted by the required bounds on x and y.

[rusty42]

0