If the side lengths of the prism are $a,b,c$, then the volume being equal to the surface area implies that
\n\\[
\nabc = 2ab+2ac+2bc.
\n\\]Let\u2019s assume the side lengths are ordered, so $a \\ge b \\ge c$. We therefore have: $ab \\ge ac \\ge bc$. So $abc = 2ab+2ac+2bc \\leq 6ab$. Simplifying, we obtain $c \\leq 6$. We can now consider each case separately:<\/p>\n

If $c=6$, our equation reduces to $ab=3a+3b$. Rearranging and factoring, this is $(a-3)(b-3)=9$. Since $c=6$ was assumed to be the smallest side length, each of the factors $(a-3)$ and $(b-3)$ is at least $3$. So the only solution is $(6,6,6)$.\n<\/li>

If $c=5$, our equation reduces to $3ab=10a+10b \\le 20a$. Therefore, $3b \\le 20$ and so $b \\le 6$. Since $c\\le b$, this means we only have to check $b=5$ and $b=6$. Checking each case, we find the only solution is $(10,5,5)$.\n<\/li>

If $c=4$, our equation reduces to $ab=4a+4b$. Rearranging and factoring, this is $(a-4)(b-4)=16$. Each way of factoring $16$ as a product of two factors yields a different solution. These are: $16\\times 1$, $8\\times 2$, and $4\\times 4$. The resulting $(a,b,c)$ triples are: $(20,5,4)$, $(12,6,4)$, $(8,8,4)$.\n<\/li>

If $c=3$, our equation reduces to $ab=6a+6b$. Rearranging and factoring, this is $(a-6)(b-6)=36$. Each way of factoring $36$ as a product of two factors yields a different solution. These are: $36\\times 1$, $18\\times 2$, $12\\times 3$, $9\\times 4$, $6\\times 6$. The resulting triples are: $(42,7,3)$, $(24,8,3)$, $(18,9,3)$, $(15,10,3)$, $(12,12,3)$.\n<\/li>

If $c=2$, our equation reduces to $0=4a+4b$, which has no solutions.\n<\/li>

If $c=1$, our equation reduces to $0=ab+2a+2b$, which also has no solutions.\n<\/li><\/ul>\n
And that\u2019s it! Overall, the problem has exactly 10 solutions. They are:
\n$(6,6,6)$, $(10,5,5)$, $(20,5,4)$, $(12,6,4)$, $(8,8,4)$, $(42,7,3)$, $(24,8,3)$, $(18,9,3)$, $(15,10,3)$, $(12,12,3)$.<\/p>\n
Note: I assumed that \u201cwhole number\u201d in this context means a positive integer. If we allow for side lengths of zero, then any degenerate prism with side lengths $(a,0,0)$ will have both area and volume equal to zero, so there are infinitely many solutions.<\/p>\n
Another note: If there were some way to upper-bound $a$, then we could do an exhaustive numerical search to find all solutions because there would only be finitely many triples $(a,b,c)$ to search. Alas, I have not been able to find an upper bound on $a$. I suspect it might not be possible, since if we set $a\\to\\infty$, the problem reduces to $bc=2b+2c$, which is the 2D version of the problem (a rectangle whose area equals its perimeter)!\n<\/p><\/div>\n<\/body>","protected":false},"excerpt":{"rendered":"
This week\u2019s Riddler Classic is simple to state, but tricky to figure out: What whole number dimensions can rectangular prisms have so that their volume (in cubic units) is the same as their surface area (in square units)? Here is my solution: [Show Solution] If the side lengths of the prism are $a,b,c$, then the … Continue reading “Prismatic puzzle”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[7],"tags":[36,2],"class_list":["post-2710","post","type-post","status-publish","format-standard","hentry","category-riddler","tag-number-theory","tag-riddler"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t