{"id":1202,"date":"2020-12-18T02:21:00","date_gmt":"2020-12-18T06:21:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=1202"},"modified":"2024-10-25T02:59:54","modified_gmt":"2024-10-25T06:59:54","slug":"sum-and-product-of-all-factors","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=1202","title":{"rendered":"Sum and Product of All Factors"},"content":{"rendered":"\n<p>For a nature number $$n=p_{1}^{q_{1}}\\times p_{2}^{q_{2}}\\times \\cdots \\times p_{m}^{q_{m}}$$ where $p_{i} (1 \\le i \\le m) $ are unique prime numbers.<\/p>\n\n\n\n<p>The total number of positive factors of $n$ is $$f=(q_{1}+1)\\times(q_{2}+1)\\times\\cdots\\times(q_{m}+1)$$ The sum of all positive factors of $n$ is $$\\sum_{d|n}d=\\dfrac{p_{1}^{q_1 +1}-1}{ p_1 -1}\\times \\dfrac{p_{2}^{q_{2}+1}-1}{ p_{2}-1}\\times \\cdots \\times\\dfrac{p_{m}^{q_{m}+1}-1}{p_{m} -1}$$ The product of all positive factors of $n$ is $$\\prod_{d|n}d=n^{\\frac{f}{2}}$$ i.e.  the $\\dfrac{f}{2}th$ power of $n$, where $f$ is the total number of positive factors of $n$.<\/p>\n\n\n\n<p><strong>Question 1<\/strong>: A number is perfect if the sum of all positive factors is twice of the number. What is the sum of the first $4$ perfect numbers found by Euclid?<\/p>\n\n\n\n<p><strong>Question 2<\/strong>: A natural number is multiplicative perfect if the square of the number is the product of all positive factors. What is the sum of the $3$ smallest multiplicative perfect numbers greater than $1$?<\/p>\n\n\n\n<p><strong>Question <\/strong>3: The sum of all positive factors of a natural number is 403. What is this number?<\/p>\n\n\n\n<p><strong>Question <\/strong>4: What is the largest natural number less than $1000$ such that the product of all positive factors is the number raised to the $12$th power?<\/p>\n","protected":false},"excerpt":{"rendered":"<p>For a nature number $$n=p_{1}^{q_{1}}\\times p_{2}^{q_{2}}\\times \\cdots \\times p_{m}^{q_{m}}$$ where $p_{i} (1 \\le i \\le m) $ are unique prime numbers. The total number of positive factors of $n$ is $$f=(q_{1}+1)\\times(q_{2}+1)\\times\\cdots\\times(q_{m}+1)$$ The sum of all positive factors of $n$ is &hellip; <a href=\"https:\/\/mathfun4kids.com\/mlog\/?p=1202\">Continue reading <span class=\"meta-nav\">&rarr;<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13,10],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1202"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=1202"}],"version-history":[{"count":23,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1202\/revisions"}],"predecessor-version":[{"id":1225,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/1202\/revisions\/1225"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=1202"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=1202"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=1202"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}