{"id":5365,"date":"2025-11-12T22:09:00","date_gmt":"2025-11-13T02:09:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=5365"},"modified":"2025-12-08T01:09:16","modified_gmt":"2025-12-08T05:09:16","slug":"integers-with-3-prime-factors-or-more","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=5365","title":{"rendered":"Integers with 3 Prime Factors or More"},"content":{"rendered":"\n

Let $n=pqrc$, where $p$, $q$, and $r$ are three distinct prime numbers, $p<q<r$, and $c$ is a positive integer. For any two distinct integers $1\\le x<y\\le n-1$, there exists $s$ which is a proper factor of $n$, $1<s<n$, such that $s \\nmid x$ and $s \\nmid y$.\ud83d\udd11<\/a><\/p>\n

<\/p>\n\n\n\n

Proof:<\/strong> Let $F$ be the set of prime factors of $n$. Since $n=pqrc$, $p$, $q$, and $r$ are three distinct prime numbers, we have $|F|\\ge 3$.<\/p>\n\n\n\n

Assume, for contradiction, that for every proper factor $m$ of $n$, $1<m<n$, we have $m\\ |\\ x$ or $m\\ |\\ y$.<\/p>\n\n\n\n

For each prime number $f\\in F$, consider $s_f=\\dfrac{n}{f}$. Since $f\\ge 2$, we have $1<s_f<n$. Therefore, $s_f$ is a proper factor of $n$. Hence, for each $f\\in F$, either $s_f\\ |\\ x$ or $s_f\\ |\\ y$.<\/p>\n\n\n\n

Let $$F_x={\\ f\\in F : s_f\\ |\\ x\\ },\\ \\ \\ \\ \\ \\ \\ F_y={\\ f\\in F : s_f\\ |\\ y\\ }$$ We have $F=F_x\\cup F_y$.<\/p>\n\n\n\n

If $|F_x|\\ge 2$, then there are two distinct prime factors, $f$ and $f’$ in $F_x$,
such that $s_f\\ |\\ x$ and $s_{f’}\\ |\\ x$. Since $f$ and $f’$ are two distinct prime factors of $n$, then $n=f\\cdot f’\\cdot M$, where $M$ is a positive integer. Then
$$s_f=\\dfrac{n}{f}=\\dfrac{f\\cdot f’\\cdot M}{f}=f’\\cdot M \\ \\ \\ \\ \\ \\ \\
s_{f’}=\\dfrac{n}{f’}=\\dfrac{f\\cdot f’\\cdot M}{f’}=f\\cdot M$$
Because $f$ and $f’$ are distinct prime factors of $n$, we have
$$lcm(s_f, s_{f’})=lcm(f’\\cdot M, f\\cdot M)=f\\cdot f’\\cdot M=n$$<\/p>\n\n\n\n

Because $s_f\\ |\\ x$ and $s_{f’}\\ |\\ x$, we have $lcm(s_f,s_{f’})\\ |\\ x$, i.e $n\\ |\\ x$. Therefore $x\\ge n$, contradicting $x\\le n-1$. Therefore $|F_x|\\le 1$. Similarly, $|F_y|\\le 1$.<\/p>\n\n\n\n

Thus, $|F|=|F_x\\cup F_y|\\le |F_x|+|F_y|\\le 1+1=2$, contradicting $|F|\\ge 3$.<\/p>\n\n\n\n

Therefore, the assumption is false, the claim is proved.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Let $n=pqrc$, where $p$, $q$, and $r$ are three distinct prime numbers, $p<q<r$, and $c$ is a positive integer. For any two distinct integers $1\\le x<y\\le n-1$, there exists $s$ which is a proper factor of $n$, $1<s<n$, such that … Continue reading →<\/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":[16],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5365"}],"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=5365"}],"version-history":[{"count":4,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5365\/revisions"}],"predecessor-version":[{"id":5370,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5365\/revisions\/5370"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5365"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5365"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5365"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}