{"id":6860,"date":"2022-09-05T13:59:35","date_gmt":"2022-09-05T08:29:35","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-finding-solutions-for-a-given-pair-number-of-factors-number-of-prime-factors\/"},"modified":"2022-09-05T13:59:35","modified_gmt":"2022-09-05T08:29:35","slug":"solved-finding-solutions-for-a-given-pair-number-of-factors-number-of-prime-factors","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-finding-solutions-for-a-given-pair-number-of-factors-number-of-prime-factors\/","title":{"rendered":"[Solved] Finding solutions for a given pair (number of factors, number of prime factors)"},"content":{"rendered":"

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Let p<\/em>1<\/sub>, p<\/em>2<\/sub>, p<\/em>3<\/sub>, \u2026 p<\/em>k<\/em><\/sub> be the prime factors of some positive integer n<\/em>. By the fundamental theorem of arithmetic<\/a>, n<\/em> can be represented as p<\/em>1<\/sub>e<\/em>1<\/sub><\/sup>\u2022p<\/em>2<\/sub>e<\/em>2<\/sub><\/sup>\u2022p<\/em>3<\/sub>e<\/em>3<\/sub><\/sup>\u2022 \u2026 p<\/em>k<\/em><\/sub>e<\/em>k<\/em><\/sub><\/sup> for some positive integers e<\/em>1<\/sub>, e<\/em>2<\/sub>, e<\/em>3<\/sub>, \u2026 e<\/em>k<\/em><\/sub>. Furthermore, any such set of such positive integers represents one positive integer in this way.<\/p>\n

Every factor of n<\/em> can be represented as p<\/em>1<\/sub>f<\/em>1<\/sub><\/sup>\u2022p<\/em>2<\/sub>f<\/em>2<\/sub><\/sup>\u2022p<\/em>3<\/sub>f<\/em>3<\/sub><\/sup>\u2022 \u2026 p<\/em>k<\/em><\/sub>f<\/em>k<\/em><\/sub><\/sup> for some integers f<\/em>i<\/em><\/sub> where 0 \u2264 f<\/em>i<\/em><\/sub> \u2264 e<\/em>i<\/em><\/sub>.<\/p>\n

Each f<\/em>i<\/em><\/sub> can have e<\/em>i<\/em><\/sub>+1 values from 0 to e<\/em>i<\/em><\/sub>, so the number of factors of n<\/em> is (e<\/em>1<\/sub>+1)\u2022(e<\/em>2<\/sub>+1)\u2022(e<\/em>3<\/sub>+1)\u2022 \u2026 (e<\/em>k<\/em><\/sub>+1).<\/p>\n

Since each e<\/em>i<\/em><\/sub> must be positive, each e<\/em>i<\/em><\/sub>+1 must be at least 2. Now we can see that, if n<\/em> has x<\/em> factors, then x<\/em> is expressible as a product of k<\/em> integers each at least 2. Conversely, if x<\/em> is expressible as a product of k<\/em> integers each at least 2, that product gives us values for e<\/em>i<\/em><\/sub>, which gives us a positive integer n<\/em> that has x<\/em> factors and k<\/em> prime factors.<\/p>\n

Therefore, a number with x<\/em> factors and k<\/em> prime factors exists if and only if x<\/em> is expressible as a product of k<\/em> integers each at least 2.<\/p>\n

To test this, simply divide x<\/em> by prime numbers, e.g.<\/em>, divide by 2 as many times as possible without a remainder, then by 3, then by 5, and so on. Once x<\/em> has been divided by k<\/em>\u22121 factors and the result is greater than 1, then we know x<\/em> is expressible as a product of k<\/em> integers each at least 2 (the last may be composite rather than a prime, such as 2\u20223\u20223\u20223\u202235). If x<\/em> reaches 1 before or just as we have divided it by k<\/em>\u22121 factors, then no such number exists.<\/p>\n

Addendum<\/h2>\n

Thinking about it further, it is not necessary to use prime numbers as candidates when testing x<\/em> for factors. We can simply iterate through the integers, testing whether each candidate f<\/em> is a factor of x<\/em>. Trying to filter these by first testing whether f<\/em> is prime would take more time than simply testing whether f<\/em> is a factor of x<\/em>. (This is not to say some more sophisticated method of testing x<\/em> for prime factors, such as using a prepared table of many primes, would not be faster.) So the following code suffices.<\/p>\n

#include <stdint.h>\n\n\n\/*  Return true if and only if there exists a positive integer A with exactly\n    x factors and exactly k prime factors.\n*\/\nstatic _Bool DoesAExist(uintmax_t x, uintmax_t k)\n{\n    \/*  The only positive integer with no prime factors is 1.  1 has exactly\n        one factor.  So, if A must have no prime factors (k is zero), then an A\n        exists if and only if x is one.\n    *\/\n    if (k == 0) return x == 1;\n\n    \/*  A number with exactly one prime factor (say p) has at least two factors\n        (1 and p) and can have any greater number of factors (p^2, p^3,...).\n        So, if k is one, then an A exists if and only if x is greater than one.\n    *\/\n    if (k == 1) return 1 < x;\n\n    \/*  Otherwise, an A exists only if x can be expressed as the product of\n        exactly k factors greater than 1.  Test this by dividing x by factors\n        greater than 1 until either we have divided by k-1 factors (so one\n        more is needed) or we run out of possible factors.\n\n        We start testing factors with two (f = 2).\n\n        If we find k factors of x during the loop, we return true.\n\n        Otherwise, we continue as long as there might be more factors.  (When\n        f*f <= x is false, we have tested the current value of x by every\n        integer from two to the square root of x.  Then x cannot have any more\n        factors less than x, because if there is any factorization x = r*s,\n        either r or s must be less than or equal to the square root of x.)\n\n        For each f, as long as it is a factor of the current value of x and we\n        need more factors, we divide x by it and decrement k to track the\n        number of factors still required.\n    *\/\n    for (uintmax_t f = 2; f*f <= x; ++f)\n        while (x % f == 0)\n        {\n            \/*  As long as f is a factor of x, remove it from x and decrement\n                the count of factors still needed.  If that number reaches one,\n                then:\n\n                    If the current value of x exceeds one, it serves as the\n                    last factor, and an A exists, so we return true.\n\n                    If the current value of x exceeds one, there is no\n                    additional factor, but we still need one, so no A exists,\n                    so we return false.\n            *\/\n            x \/= f;\n            if (--k <= 1) return 1 < x;\n        }\n\n    \/*  At this point, we need k more factors for x, and k is greater than one\n        but x is one or prime, so x does not have enough factors.  So no A with\n        the required properties exists, and we return false.\n    *\/\n    return 0;\n}\n\n\n#include <stdio.h>\n\n\nint main(void)\n{\n    printf(\"False test cases:\\n\");\n    printf(\"%d\\n\", DoesAExist(0, 0));   \/\/  False since each positive integer has at least one factor (1).\n    printf(\"%d\\n\", DoesAExist(2, 0));   \/\/  False since no number has two factors and no prime factors.\n\n    printf(\"%d\\n\", DoesAExist(0, 1));   \/\/  False since each positive integer has at least one factor (1).\n    printf(\"%d\\n\", DoesAExist(1, 1));   \/\/  False since each positive integer with a prime factor has at least two factors (one and the prime).\n    printf(\"%d\\n\", DoesAExist(2, 2));   \/\/  False since each number with two prime factors (p and q) has at least four factors (1, p, q, and pq).\n    printf(\"%d\\n\", DoesAExist(3, 2));   \/\/  False since each number with two prime factors (p and q) has at least four factors (1, p, q, and pq).\n\n    printf(\"%d\\n\", DoesAExist(8, 4));\n\n    printf(\"%d\\n\", DoesAExist(6, 3));\n    printf(\"%d\\n\", DoesAExist(22, 3));\n\n    printf(\"%d\\n\", DoesAExist(24, 5));\n    printf(\"%d\\n\", DoesAExist(88, 5));\n    printf(\"%d\\n\", DoesAExist(18, 4));\n    printf(\"%d\\n\", DoesAExist(54, 5));\n    printf(\"%d\\n\", DoesAExist(242, 4));\n    printf(\"%d\\n\", DoesAExist(2662, 5));\n\n    printf(\"True test cases:\\n\");\n    printf(\"%d\\n\", DoesAExist(1, 0));   \/\/  True since 1 has one factor and zero prime factors.\n    printf(\"%d\\n\", DoesAExist(2, 1));   \/\/  True since each prime has two factors.\n    printf(\"%d\\n\", DoesAExist(3, 1));   \/\/  True since each square of a prime has three factors.\n    printf(\"%d\\n\", DoesAExist(4, 1));   \/\/  True since each cube of a prime has three factors.\n    printf(\"%d\\n\", DoesAExist(4, 2));   \/\/  True since each number that is the product of two primes (p and q) has four factors (1, p, q, and pq).\n\n    printf(\"%d\\n\", DoesAExist(8, 2));\n    printf(\"%d\\n\", DoesAExist(8, 3));\n\n    printf(\"%d\\n\", DoesAExist(6, 2));\n    printf(\"%d\\n\", DoesAExist(22, 2));\n\n    printf(\"%d\\n\", DoesAExist(24, 2));\n    printf(\"%d\\n\", DoesAExist(24, 3));\n    printf(\"%d\\n\", DoesAExist(24, 4));\n    printf(\"%d\\n\", DoesAExist(88, 2));\n    printf(\"%d\\n\", DoesAExist(88, 3));\n    printf(\"%d\\n\", DoesAExist(88, 4));\n    printf(\"%d\\n\", DoesAExist(18, 2));\n    printf(\"%d\\n\", DoesAExist(18, 3));\n    printf(\"%d\\n\", DoesAExist(54, 2));\n    printf(\"%d\\n\", DoesAExist(54, 3));\n    printf(\"%d\\n\", DoesAExist(54, 4));\n    printf(\"%d\\n\", DoesAExist(242, 2));\n    printf(\"%d\\n\", DoesAExist(242, 3));\n    printf(\"%d\\n\", DoesAExist(2662, 2));\n    printf(\"%d\\n\", DoesAExist(2662, 3));\n    printf(\"%d\\n\", DoesAExist(2662, 4));\n}\n<\/code><\/pre>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
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solved Finding solutions for a given pair (number of factors, number of prime factors) <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Let p1, p2, p3, \u2026 pk be the prime factors of some positive integer n. By the fundamental theorem of arithmetic, n can be represented as p1e1\u2022p2e2\u2022p3e3\u2022 \u2026 pkek for some positive integers e1, e2, e3, \u2026 ek. Furthermore, any such set of such positive integers represents one positive integer in this way. Every … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[457,324,1920,1921,1919],"class_list":["post-6860","post","type-post","status-publish","format-standard","hentry","category-solved","tag-algorithm","tag-c","tag-factors","tag-largenumber","tag-primes"],"yoast_head":"\n[Solved] Finding solutions for a given pair (number of factors, number of prime factors) - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-finding-solutions-for-a-given-pair-number-of-factors-number-of-prime-factors\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Finding solutions for a given pair (number of factors, number of prime factors) - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Let p1, p2, p3, \u2026 pk be the prime factors of some positive integer n. 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