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@@ -7,24 +7,24 @@
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"source": [
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"<div style=\"text-align: right\">Peter Norvig<br>Mar 2024</div> \n",
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"\n",
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"# Stubborn Number Endings\n",
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"# Stubborn Number Endings (e.g. 5² = 25)\n",
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"\n",
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"[Francis Su](https://www.francissu.com/)'s book *Mathematics for Human Flourishing* mentions the fact that numbers that end in \"5\" have a square that also ends in \"5\". \n",
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"[Francis Su](https://www.francissu.com/)'s fine book *[Mathematics for Human Flourishing](https://www.francissu.com/flourishing)* mentions the fact that numbers that end in \"5\" have a square that also ends in \"5\". \n",
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"\n",
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"For example, 5² = 25, 15² = 225, and 25² = 625. \n",
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"\n",
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"This leads to some questions:\n",
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"\n",
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"- Is there an easy way to calculate the square of a number ending in \"5\"?\n",
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"- Is there an easy way to calculate the square of a number ending in \"5\" without using a calculator?\n",
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"- What should we call this property of \"square has same ending\"?\n",
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"- Are there other digits besides 5 that have this property?\n",
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"- Can we prove the property, not just show some examples?\n",
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"- Are there multi-digit endings that have this property?\n",
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"- Can we prove it, not just show some examples?\n",
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"\n",
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"\n",
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"## Is there an easy way to calculate the square of a number ending in \"5\"?\n",
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"\n",
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"Let's make a table of {number: square} pairs and try to see a pattern:"
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"Let's make a table of {number: square} pairs and look for a pattern:"
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]
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},
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{
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@@ -111,36 +111,34 @@
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"\n",
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"## What should we call this property?\n",
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"\n",
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"Let's define an **ending** as the rightmost-digits (zero or more) of a number in decimal notation. \n",
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"Let's define an **ending** as the rightmost-digits of an integer in decimal notation. \n",
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"\n",
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"Then we can say that an ending is **stubborn** if every number with that ending has a square with the same ending.\n",
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"Then we can say that an ending is **stubborn** if every integer with that ending has the same ending when squared.\n",
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"\n",
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"## Are there other digits besides 5 that are stubborn?\n",
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"\n",
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"We could work this out in our heads, or with paper and pencil, or we can compute it with an expression:"
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"We could work this out in our heads, or with paper and pencil, or we can compute it:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 15,
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"id": "c86f99ed-48fa-4954-9467-b72d67252d73",
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"execution_count": 3,
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"id": "ddcdafa8-bfa6-4707-951c-6683701e06c2",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"{'0', '1', '5', '6'}"
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"{0, 1, 5, 6}"
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]
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},
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"execution_count": 15,
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"execution_count": 3,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"digits = '0123456789'\n",
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"\n",
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"{e for e in digits if str(int(e) ** 2).endswith(e)}"
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"{e for e in range(10) if str(e ** 2).endswith(str(e))}"
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]
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},
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{
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@@ -152,23 +150,27 @@
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"\n",
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"## Can we prove stubborness?\n",
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"\n",
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"We have seen that 0² ends in 0, 1² ends in 1, 5² ends in 5, and 6² ends in 6. And we have checked some numbers with those endings, for example, 245² ends in 5. But can we **prove** that **every** number ending in 0, 1, 5, or 6 has a square that ends in the same digit?\n",
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"We have seen that 0² ends in 0, 1² ends in 1, 5² ends in 5, and 6² ends in 6. And we have checked some numbers with those endings, for example, 245² ends in 5. But can we **prove** that **every** integer ending in 0, 1, 5, or 6 has a square that ends in the same digit?\n",
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"\n",
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"Some notation: I'll use quote marks, as in \"*se*\" to mean the string of staring digits \"*s*\" followed by the string of ending digits \"*e*\".\n",
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"Some notation: I'll use quote marks, as in \"*se*\" to mean the string of starting digits \"*s*\" followed by the string of ending digits \"*e*\".\n",
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"\n",
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"With a little bit of algebra we can see that if *s* is any string of digits and *e* is a single ending digit, then:\n",
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"\n",
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"\"*se*\"² = (10⋅*s* + *e*)² = (10⋅*s*)² + 2⋅(10⋅*s* ⋅ *e*) + *e*² = 10 ⋅ (10⋅*s*² + 2⋅*s*⋅*e*) + *e*²\n",
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"\"*se*\"² = (10⋅*s* + *e*)² = (10⋅*s*)² + 2⋅(10⋅*s*⋅*e*) + *e*² = 10 ⋅ (10⋅*s*² + 2⋅*s*⋅*e*) + *e*² = *e*² (mod 10)\n",
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"\n",
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"This is ten times some integer, plus *e*², so \"*se*\"² ends in the digit *e* if and only if *e*² ends in *e*, and we know that is true for 0, 1, 5, and 6, and for no other digits.\n",
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"In other words, \"*se*\"² is 10 times some integer plus *e*², so \"*se*\"² ends in the digit *e* if and only if *e*² ends in *e*. \n",
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"\n",
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"We know that is true for 0, 1, 5, and 6, and for no other digits.\n",
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"\n",
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"## Are there multi-digit endings that are stubborn?\n",
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"\n",
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"The algebraic argument above can be extended to work with an ending string *e* that is *k* digits long:\n",
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"\n",
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"\"*se*\"² = (10<sup>*k*</sup>⋅*s* + *e*)² = (10<sup>*k*</sup>⋅*s*)² + 2⋅(10<sup>*k*</sup>⋅*s* ⋅ *e*) + *e*² = 10<sup>*k*</sup> ⋅ (10<sup>*k*</sup>⋅*s*² + 2⋅*s*⋅*e*) + *e*²\n",
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"\"*se*\"² = (10<sup>*k*</sup>⋅*s* + *e*)² = (10<sup>*k*</sup>⋅*s*)² + 2⋅(10<sup>*k*</sup>⋅*s* ⋅ *e*) + *e*² = 10<sup>*k*</sup> ⋅ (10<sup>*k*</sup>⋅*s*² + 2⋅*s*⋅*e*) + *e*² = *e*² (mod 10)\n",
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"\n",
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"This is 10<sup>*k*</sup> times some integer, plus *e*², so again \"*se*\"² ends in *e* if and only if *e*² ends in *e*. To put it another way, to test whether *e* is stubborn, all we have to do is square *e* and see if the result ends in *e*. There is one complication: we'd like to say that \"00\" is stubborn, because any number ending in \"00\", when squared, also ends in \"00\", for example 200² = 40000. But 00² is zero, which we write as \"0\", not as \"00\" or \"0000\". To make sure that \"00\" is considered stubborn, I will define the predicate function `stubborn(ending)` to test the square of \"1\" followed by the ending (I could have chosen any other starting digit string and the result would be the same). \n"
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"Again, \"*se*\"² ends in *e* if and only if *e*² ends in *e*. To test whether *e* is a stubborn ending (with all possible starting digit strings), all we have to do is square *e* and see if the result ends in *e*. \n",
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"\n",
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"There is one **complication**: we'd like to say that \"0\" and \"00\" are two distinct stubborn endings, but 0 and 00 are the same integer. To distinguish them, I will describe endings as strings, not integers. I will define the predicate function `stubborn(ending: str)`:"
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]
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},
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{
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@@ -179,8 +181,10 @@
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"outputs": [],
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"source": [
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"def stubborn(ending: str) -> bool:\n",
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" \"\"\"Does the square of any number with this ending also end with this ending?\"\"\"\n",
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" return str(int(\"1\" + ending) ** 2).endswith(ending)"
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" \"\"\"Does every integer with this ending, when squared, also end with this ending?\"\"\"\n",
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" start = \"1\" # Arbitrary choice of start; could be any non-zero digit string\n",
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" square = str(int(start + ending) ** 2)\n",
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" return square.endswith(ending)"
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]
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},
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{
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@@ -193,7 +197,7 @@
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},
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{
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"cell_type": "code",
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"execution_count": 17,
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"execution_count": 5,
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"id": "57e64e31-0674-4849-bf72-f6c6b437009a",
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"metadata": {},
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"outputs": [
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@@ -203,13 +207,16 @@
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"['00', '01', '25', '76']"
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]
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},
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"execution_count": 17,
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"execution_count": 5,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"[s + e for s in digits for e in digits if stubborn(s + e)]"
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"digits = '0123456789'\n",
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"endings = [s + e for s in digits for e in digits]\n",
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"\n",
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"list(filter(stubborn, endings))"
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]
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},
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{
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@@ -217,12 +224,12 @@
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"id": "61a2b8e2-2bdf-458f-adaf-ffde56125862",
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"metadata": {},
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"source": [
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"We could easily continue on in this way, enumerating all three-, four- or even six-digit endings, and checking each one to see if it is stubborn. There are only a million six-digit endings. But there are a quadrillion 15-digit endings, so it would take a *long* time to check all of those. And 100-digit endings? Forget about it. Instead we need to rely on a simplification:\n",
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"We could easily continue on in this way, enumerating all three-, four-, or more-digit endings, and checking each one to see if it is stubborn. There are only a million possible six-digit endings; we could quickly check them all. But there are a quadrillion 15-digit endings, so it would take a *long* time to check all of those. And 100-digit endings? Forget about it. Instead we need to rely on an optimization:\n",
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"- Any two-digit stubborn ending ('00', '01', '25', '76') has to end in a one-digit-stubborn ending ('0', '1', '5', '6').\n",
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"- In general, any *d*-digit stubborn ending has to end in a (*d*-1)-digit stubborn ending.\n",
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"- So, to find the stubborn endings of length 100, I don't need to generate and test all 10<sup>100</sup> endings, I only need to consider the stubborn endings of length 99 and check each one of them. (If this simplification is not obvious, stop and convince yourself it is true.)\n",
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"- So, to find the stubborn endings of length 100, I don't need to generate and test all 10<sup>100</sup> endings, I only need to consider the stubborn endings of length 99, prepend each of the ten digits to each of these endings to get a list of 100-digit endings, and then check each one for stubborness. \n",
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"\n",
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"Using this simplification we can efficiently compute all stubborn-endings of a given length *d* as follows (caching greatly improves efficiency):"
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"Using this optimization we can efficiently compute all stubborn-endings of a given length *d* as follows (caching greatly improves efficiency):"
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]
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},
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{
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@@ -232,17 +239,16 @@
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"metadata": {},
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"outputs": [],
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"source": [
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"from functools import lru_cache\n",
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"from functools import cache\n",
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"\n",
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"@lru_cache(None)\n",
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"def stubborn_endings(d: int) -> list:\n",
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"@cache\n",
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"def stubborn_endings(d: int) -> list[str]:\n",
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" \"\"\"A list of all stubborn endings of length `d` digits.\"\"\"\n",
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" if d == 0:\n",
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" return [''] # The empty ending is the sole stubborn ending of length 0.\n",
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" return [''] # The empty ending is the only stubborn ending of length 0.\n",
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" else:\n",
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" return [(s + e) for e in stubborn_endings(d - 1) \n",
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" for s in digits \n",
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" if stubborn(s + e)]"
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" endings = [s + e for e in stubborn_endings(d - 1) for s in digits]\n",
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" return list(filter(stubborn, endings))"
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]
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},
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{
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@@ -262,7 +268,16 @@
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{
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"data": {
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"text/plain": [
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"['000', '001', '625', '376']"
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"{0: [''],\n",
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" 1: ['0', '1', '5', '6'],\n",
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" 2: ['00', '01', '25', '76'],\n",
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" 3: ['000', '001', '625', '376'],\n",
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" 4: ['0000', '0001', '0625', '9376'],\n",
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" 5: ['00000', '00001', '90625', '09376'],\n",
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" 6: ['000000', '000001', '890625', '109376'],\n",
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" 7: ['0000000', '0000001', '2890625', '7109376'],\n",
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" 8: ['00000000', '00000001', '12890625', '87109376'],\n",
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" 9: ['000000000', '000000001', '212890625', '787109376']}"
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]
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},
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"execution_count": 7,
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@@ -271,19 +286,22 @@
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}
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],
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"source": [
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"stubborn_endings(3)"
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"{d: stubborn_endings(d) for d in range(10)}"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 8,
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"id": "d1dc31eb-e5dd-405e-b16d-eceef5c9493c",
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"id": "38c26d56-d916-481e-bc04-51af7a52b1e8",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"['0000', '0001', '0625', '9376']"
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"['000000000000000000000000000000000000000000000000000000000000',\n",
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" '000000000000000000000000000000000000000000000000000000000001',\n",
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" '863811000557423423230896109004106619977392256259918212890625',\n",
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" '136188999442576576769103890995893380022607743740081787109376']"
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]
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},
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"execution_count": 8,
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@@ -292,73 +310,31 @@
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}
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],
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"source": [
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"stubborn_endings(4)"
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"stubborn_endings(60)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 9,
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"id": "81730e6b-7218-4fca-9107-2a9539f62ea3",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"['00000', '00001', '90625', '09376']"
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]
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},
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"execution_count": 9,
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}
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],
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"source": [
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"stubborn_endings(5)"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 10,
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"id": "288506c8-6dc1-40d5-842a-eb65ed94ae98",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"['000000', '000001', '890625', '109376']"
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]
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},
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"execution_count": 10,
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}
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"source": [
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"stubborn_endings(6)"
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"execution_count": 11,
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"id": "73491ea5-a4c8-4221-8fc7-d6ce44d17bb4",
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{
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"data": {
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"text/plain": [
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"['0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000',\n",
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" '0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001',\n",
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" '3953007319108169802938509890062166509580863811000557423423230896109004106619977392256259918212890625',\n",
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" '6046992680891830197061490109937833490419136188999442576576769103890995893380022607743740081787109376']"
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"['000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000',\n",
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" '000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001',\n",
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" '055462996814764263903953007319108169802938509890062166509580863811000557423423230896109004106619977392256259918212890625',\n",
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" '944537003185235736096046992680891830197061490109937833490419136188999442576576769103890995893380022607743740081787109376']"
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"execution_count": 11,
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"source": [
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"stubborn_endings(100)"
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"stubborn_endings(120)"
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]
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},
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{
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@@ -366,26 +342,35 @@
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"id": "37a45341-8dac-4720-be7d-388fa47eec1a",
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"metadata": {},
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"source": [
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"## More questions!\n",
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"\n",
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"This leads to a few new questions.\n",
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"\n",
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"## Can each stubborn ending be extended exactly one way?\n",
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"\n",
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"We know that each stubborn ending of length *d* has to build on the endings of length *d - 1*, but is it always the case that for any length *d* there will be exactly four endings? Might there be a case where two digits work with one of the endings, or no digits?\n",
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"We know that each stubborn ending of length *d* has to build on the endings of length *d* - 1, and so far, there has always been exactly one way (out of the ten possible digits) to extend each of the length *d* - 1 endings. Will that always be true? Might there be a case where two digits work with one of the endings, or no digits?\n",
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"\n",
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"I can show that there are always 4 endings up to length *d* = 2000, but I leave it to you to describe a proof that this will always be true."
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"I can show that for all lengths *d* up to 2000, the result of `stubborn_endings(d)` is always four strings that end with '0', '1', '5', and '6' in that order:"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 12,
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"execution_count": 18,
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"id": "654d53af-e58f-4ed2-874b-c4573f3a9a7a",
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"metadata": {},
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"outputs": [],
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"outputs": [
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{
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"data": {
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"text/plain": [
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"Counter({('0', '1', '5', '6'): 2000})"
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]
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},
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"execution_count": 18,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
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"for d in range(1, 2000):\n",
|
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" assert len(stubborn_endings(d)) == 4 "
|
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"Counter(tuple(ending[-1] for ending in stubborn_endings(d))\n",
|
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" for d in range(1, 2001))"
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]
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},
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{
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@@ -393,37 +378,47 @@
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"id": "537556a3-d570-4a50-93bf-9632b40f9b47",
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"metadata": {},
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"source": [
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"One cool implication: if it is true that a stubborn ending of any length can always be extended, that means there is an infinitely long integer whose square ends with itself!\n",
|
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"I leave it up to the reader to find a proof ofr numbers beyond 2000.\n",
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"\n",
|
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"## What digits are used to extend each ending?\n",
|
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"\n",
|
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"There doesn't seem to be a pattern, and all digits seemingly get used roughly evenly. \n",
|
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"\n",
|
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|
"I don't have a theory of which digit comes next, but I can count how many times each digit appears in the 2000-digit endings that end in \"5\" and \"6\", and see that each of the ten digits appears about 200 times:"
|
|
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|
"I don't have a theory of which digit is used to extend each stubborn ending, but I can count:"
|
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]
|
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},
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{
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"cell_type": "code",
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"execution_count": 13,
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"execution_count": 11,
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"id": "4dbd7ace-8b41-4ea7-b71f-8f7e318243e1",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"Counter({'0': 205,\n",
|
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" '3': 173,\n",
|
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"{'0': Counter({'0': 1999}),\n",
|
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" '1': Counter({'0': 1999}),\n",
|
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" '5': Counter({'8': 214,\n",
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" '2': 208,\n",
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" '6': 197,\n",
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" '9': 198,\n",
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" '5': 197,\n",
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" '4': 206,\n",
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" '8': 214,\n",
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" '0': 205,\n",
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" '7': 205,\n",
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" '1': 197})"
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" '9': 198,\n",
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" '6': 197,\n",
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" '1': 197,\n",
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" '5': 196,\n",
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" '3': 173}),\n",
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" '6': Counter({'1': 214,\n",
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" '7': 208,\n",
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" '5': 206,\n",
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" '9': 205,\n",
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" '2': 205,\n",
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" '0': 198,\n",
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" '3': 197,\n",
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" '8': 197,\n",
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" '4': 196,\n",
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" '6': 173})}"
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]
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},
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"execution_count": 13,
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"execution_count": 11,
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"metadata": {},
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"output_type": "execute_result"
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}
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@@ -431,47 +426,23 @@
|
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"source": [
|
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"from collections import Counter\n",
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"\n",
|
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|
"zero, one, five, six = stubborn_endings(2000)\n",
|
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"\n",
|
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|
"Counter(five)"
|
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"{e[-1]: Counter(e[:-1]) for e in stubborn_endings(2000)}"
|
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]
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},
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{
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"cell_type": "code",
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"execution_count": 14,
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"id": "96bdab37-c5a3-4597-b4ca-fb560e4c3163",
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"cell_type": "markdown",
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"id": "35e84df9-0b76-4d8d-aa13-2c4e73a86643",
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"metadata": {},
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"outputs": [
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{
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"data": {
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"text/plain": [
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"Counter({'9': 205,\n",
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" '6': 174,\n",
|
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" '7': 208,\n",
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" '3': 197,\n",
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" '0': 198,\n",
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" '4': 196,\n",
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" '5': 206,\n",
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" '1': 214,\n",
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" '2': 205,\n",
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" '8': 197})"
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]
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},
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"execution_count": 14,
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"metadata": {},
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"output_type": "execute_result"
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}
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],
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"source": [
|
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|
"Counter(six)"
|
|
|
|
|
"This is saying that the stubborn endings \"0\" and \"1\" are always extended by prepending a \"0\", but for \"5\" and \"6\", it seems to be pretty random; each of the ten digits appears roughly 200 times."
|
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|
]
|
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "Python 3 (ipykernel)",
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"display_name": "Python [conda env:base] *",
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"language": "python",
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"name": "python3"
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"name": "conda-base-py"
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},
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"language_info": {
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"codemirror_mode": {
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@@ -483,7 +454,7 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
|
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"version": "3.8.15"
|
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"version": "3.13.9"
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}
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},
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"nbformat": 4,
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