A249067.ipynb

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 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "A function to work through the $n$ times table until every possible first digit has been seen."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 96,
   "metadata": {},
   "outputs": [],
   "source": [
    "def time_until_all_seen(n):\n",
    "    seen = set()\n",
    "    c = 0\n",
    "    t = n\n",
    "    while len(seen)<9:\n",
    "        c += 1\n",
    "        s=str(t)[0]\n",
    "        seen.add(s)\n",
    "        t += n\n",
    "    return c"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Look how bad the 13 times table is! But 112 is even worse."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 95,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "62\n",
      "81\n"
     ]
    }
   ],
   "source": [
    "print(time_until_all_seen(13))\n",
    "print(time_until_all_seen(112))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now do the same thing as above, but keep track of how many times each pair of leading digits is seen"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 98,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " 1  1  0  2  1  1  1  0  1  1\n",
      " 1  0  1  1  1  0  2  1  1  1\n",
      " 0  1  1  1  0  1  1  1  0  2\n",
      " 1  1  1  0  1  1  1  0  1  1\n",
      " 1  0  2  1  1  1  0  1  1  1\n",
      " 0  1  1  1  0  2  1  1  1  0\n",
      " 1  1  1  0  1  1  1  0  2  1\n",
      " 1  0  0  0  0  0  0  0  0  0\n",
      " 0  1  0  0  0  0  0  0  0  0\n"
     ]
    }
   ],
   "source": [
    "def count_times_seen(n):\n",
    "    times_seen = defaultdict(lambda:0)\n",
    "    seen = set()\n",
    "    c = 0\n",
    "    t = n\n",
    "    while len(seen)<9:\n",
    "        c += 1\n",
    "        s=str(t)[0]\n",
    "        s_more = str(t)[:2]\n",
    "        times_seen[s_more] += 1\n",
    "        seen.add(s)\n",
    "        t += n\n",
    "    for a in range(1,10):\n",
    "        bits = []\n",
    "        for b in range(0,10):\n",
    "            s = str(a)+str(b)\n",
    "            bits.append('{:>2}'.format(times_seen[s]))\n",
    "        print(' '.join(bits))\n",
    "\n",
    "count_times_seen(13)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 99,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "111112 to 112499 take 81 steps\n"
     ]
    }
   ],
   "source": [
    "from itertools import count\n",
    "first = 111112\n",
    "for i in count(first):\n",
    "    c = time_until_all_seen(i)\n",
    "    if c!=81:\n",
    "        break\n",
    "print('{} to {} take 81 steps'.format(first,i-1))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "How long do numbers of the form $\\left\\lceil \\frac{1}{9} \\cdot 10^n \\right\\rceil$ take?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 100,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1 9\n",
      "2 45\n",
      "12 42\n",
      "112 81\n",
      "1112 81\n",
      "11112 81\n",
      "111112 81\n",
      "1111112 81\n",
      "11111112 81\n",
      "111111112 81\n"
     ]
    }
   ],
   "source": [
    "from math import ceil\n",
    "for i in range(10):\n",
    "    n = ceil(10**i/9)\n",
    "    print(n, time_until_all_seen(n))"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The last thing to do is to find the record-setters:"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 102,
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "1 9\n",
      "2 45\n",
      "13 62\n",
      "112 81\n"
     ]
    }
   ],
   "source": [
    "n = 1\n",
    "best = 0\n",
    "while best<81:\n",
    "    c = time_until_all_seen(n)\n",
    "    if c>best:\n",
    "        print(n,c)\n",
    "        best = c\n",
    "    n += 1"
   ]
  }
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