DJ Trousdale

Many numbers can be expressed as the sum of two palindromes. 389 for instance is 383 + 6 as well as 323 + 66.

What three-digit number cannot be expressed as the sum of two palindromes?

(For the sake of this problem, palindromes with leading zeroes [like 030 and 0550] are not allowed, and neither are negative numbers or non-integers. Single digits are allowed.)

Contrived situation time!

Five bands, inconveniently named 1st, 2nd, 3rd, 4th and 5th, are scheduled to play at a festival in a certain order. One manager was looking over this schedule. It had two columns, each with a 1st, 2nd, 3rd, 4th and 5th, but no column titles. So he couldn’t tell which column was the band names, and which one was the order in which they played.

He asked the scheduling manager, who told him that it made no difference which was which.

How many different possible orders are there, in which the bands play?

Extra credit: Same problem, but with 8 bands.

Okay, this has been bugging me for days. See if you can figure it out.

This problem is the same as the last one, except with five surrounding circles instead of three.

Also, as noted in the comments, finding the ratio of two radii is made simpler by assuming the second radius is 1. So:

Update: I found an answer, but it’s an ugly cubic root. Nothing too interesting.

StevenRoy gave a very nice solution to the 6742816 problem here.

I thought about how it might be extended to three digit numbers. While that’s a much larger field of possibilities, the rules of the problem still restrict them quite a bit. C++ gave me this list of three-digit numbers that survived one multiplication:

2214
31515
35115
4114
499324
555125
61212
61954
62112
71428
72228
74128
899648

I continued them and found:

2214832
3151525
351155250
4114416
4993243
555125
612124864
619545
621122416
7142864
7222832
7412816
8996486

So it appears 351, 612 and 621 tie for the longest three-digit starters for this problem.

Where that might be considered a “strict” multiplier chain, let’s call the following one a “relaxed” one:

“You can make a chain by starting with a number, multiplying its digits, multiplying the result’s digits, etc. The chain ends when it has decreased to one digit. What starter number for x digits results in the longest chain?”

This is tough to solve with anything other than brute force, so with my new Javascript powers, I made this. (It may take a minute to generate)

Both answers for the three digit (and their permutations, by consequence) have digits that add to 22. :D

Fascinated by the emerging patterns, I did a google search and found that this kind of thing had already been investigated and was called “Multiplicative persistence.” Good band name perhaps?

I didn’t find much on the internet about the 6742816 problem though. Mine! :D

So back to that

Finding four-digit starters for strict multiplier chains definitely needs computer help. Here’s (hopefully) the program that generates the numbers that work for the first two multiplications.

These definitely work for at least one step. Notice the 767720588 – it contains a zero and works, but it won’t get much further. If there aren’t any longer ones, this may take it.
I’ve spent literally all day on this and don’t feel any closer to a solution. :< Anyone want to tackle it?

Update! This script should do the trick. Check out the source code too, for a good laugh. :D

Also, this script solves it for X digits. Right now it’s set to 5, and I haven’t tried 6; it would probably take minutes to process it. I would hazard a guess though, that the longest chains in 6+ digits will start with 315, 612, and 621 (which all have digits that sum to 9.)

This is a difficult, abstract thinker.

Part 1: What property does the number 6742816 have, shared only by less than a hundred other numbers?

Part 2: There is a ten-digit number, the longest number that shares 6742816’s property. What is it?

Can you find six distinct digits (none of them 0) such that

a^b + c^d = e^f?

I only know of one solution.