From 6b6bc11e44f66d5df2c4dc330ce9e48318ebc0d6 Mon Sep 17 00:00:00 2001
From: Walker30263 <gangulysurjo@gmail.com>
Date: Thu, 11 Jan 2024 18:47:04 -0500
Subject: [PATCH] Create 243.js

---
 243.js | 95 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 95 insertions(+)
 create mode 100644 243.js

diff --git a/243.js b/243.js
new file mode 100644
index 0000000..2c40ada
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+++ b/243.js
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+/*
+A fraction is resilient iff its numerator and denominator are coprime.
+Thus, the resilience of a denominator can be related to Euler's totient function (phi(n) returns the number of numbers <= n which are coprime to n ) (Read https://infinitedescent.xyz/ to learn more about the number theory used in this solution)
+
+R(d) = phi(d)/(d-1)
+
+phi(d) = d * (1-1/p_1)*(1-1/p_2)*...(1-1/p_n) where p_1, p_2, ..., p_n are unique prime numbers that are factors of d (Read Page 288 of the aforementioned textbook for a detailed proof of this formula (Theorem 7.3.59))
+
+Thus, we are trying to find the smallest number d such that (d * (1-1/p_1)*(1-1/p_2)*...(1-1/p_n))/(d-1) is less than 15499/94744
+
+For approximation purposes, we can assume that d and (d-1) are going to be almost equal, and can focus on thinking 
+about how to minimize (1-1/p_1)*(1-1/p_2)*...(1-1/p_n)
+
+(1-1/p_n) is always going to be a number less than 1 but greater than 0, so the more (1-1/p_n) terms we have, the smaller the value of the product is going to be.
+Thus, we want to multiply the first (1-1/p_n)s together (where the p_n's are the first x primes) until we get a R(product) that is less than 15499/94744.
+
+After we do this, we'll have an array of the first x primes, and we'll know that
+R(product of first x-1 primes) < 15499/94744 < R(product of first x primes)
+
+Instead of checking R(d) for all numbers in the range (product of first x-1 primes) to (product of first x primes),
+we can only check multiples of (product of first x-1 primes) in that range, because they won't have any additional prime factors
+*/
+
+const target = 15499/94744;
+
+let primes = [];
+let currentNumber = 2;
+
+while (resilience(primes.reduce((accumulator, currentValue) => accumulator * currentValue, 1)) >= target) {
+    let isPrime = true;
+
+    //since every composite number must have a prime factor less than or equal to its square root...
+    for (let j = 0; j < primes.length; j++) {
+        if (currentNumber % primes[j] === 0) {
+            isPrime = false;
+            break;
+        }
+
+        if (primes[j] > Math.floor(Math.sqrt(currentNumber))) {
+            break;
+        }
+    }
+
+    if (isPrime) {
+        primes.push(currentNumber);
+    }
+
+    currentNumber++;
+}
+
+let productOfAll = primes.reduce((accumulator, currentValue) => accumulator * currentValue, 1);
+let productOfAllExceptLast = primes.slice(0, primes.length - 1).reduce((accumulator, currentValue) => accumulator * currentValue, 1);
+
+for (let i = productOfAllExceptLast; i <= productOfAll; i += productOfAllExceptLast) {
+    if (resilience(i) < target) { //if we go below our target, we have our answer!
+        console.log(i); //our answer!
+        break;
+    }
+}
+
+//helper function
+function resilience(d) {
+    return phi(d)/(d-1);
+}
+
+//helper function
+function phi(n) {
+    let input = n; //for the end
+
+    let primes = [];
+
+    outerLoop: while (n !== 1) {
+        innerLoop: for (let i = 2; i <= Math.floor(Math.sqrt(n)) + 1; i++) {
+            if (i === Math.floor(Math.sqrt(n)) + 1) {
+                primes.push(n);
+                break outerLoop;
+            }
+
+            if (n % i === 0) {
+                if (!primes.includes(i)) {
+                    primes.push(i);
+                }
+
+                while (n % i === 0) {
+                    n /= i;
+                }
+
+                break innerLoop;
+            }
+        }
+    }
+
+    //using the formula for phi(n)
+    return Math.round(primes.reduce((accumulator, currentValue) => accumulator *= (1 - (1 / currentValue)), input)); // rounding to make up for floating point rounding errors
+}
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