I recently had occasion to learn that 108 and 1008 are both divisible by 18.<\/p>\r\n
Which led me to wonder whether any integer consisting of 1 followed by some number of 0s followed by 8 is divisible by 18.<\/p>\r\n
And in fact it turns out to be, and I came up with two fun proofs of that fact.<\/p>\r\n
The quick test to see whether an integer is divisible by 9 is to add up the digits. If the digits add up to 9 (or a multiple of 9), then the original number is divisible by 9.<\/p>\r\n
And 1 + 8 = 9, so any integer consisting of a 1, an 8, and some 0s is divisible by 9.<\/p>\r\n
Also, all integers ending in 8 are even. And if a number is even, then it\u2019s divisible by 2, by definition. And a number that\u2019s divisible by both 9 and 2 is divisible by 18.<\/p>\r\n
Thus, all numbers of the form 10\u202608 (where the \u201c\u2026\u201d indicates any number of 0s) are divisible by 18.<\/p>\r\n
Even though I\u2019ve known the digits-add-up-to-9 thing forever, I wanted to prove it to myself. So here goes:<\/p>\r\n
Saying that an integer is divisible by 9 is the same as saying it\u2019s equivalent to 0 mod 9. You can think of a given positive integer as consisting of a set of numbers that are less than ten, each multiplied by a different power of 10; for example, 1,233 is 1 x 10^3 + 2 x 10^2 + 3 x 10^1 + 3 x 10^0. But every power of 10 is 1 mod 9, so any positive integer of the form n x 10^m is equivalent to n mod 9. (For example, 100 is 1 mod 9 (because it\u2019s 99 + 1), so 200 is 2 mod 9 (because it\u2019s (1 mod 9) plus (1 mod 9))).<\/p>\r\n
So the original integer is equivalent to 0 mod 9 if and only if the sum of the digits is equivalent to 0 mod 9.<\/p>\r\n
Any integer that consists entirely of a string of 9s is obviously divisible by 9. (It\u2019s equal to 9 times a number that\u2019s a string of 1s.)<\/p>\r\n
And if you take an integer of the form 99\u202699 and add 9 to it, you get an integer of the form 10\u202608.<\/p>\r\n
Since 99\u202699 is divisible by 9, adding 9 to it gives you another number that\u2019s still divisible by 9, and since 10\u202608 is even, it\u2019s also divisible by 18.<\/p>\r\n\n","protected":false},"excerpt":{"rendered":"","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[124],"tags":[],"class_list":["post-17559","post","type-post","status-publish","format-standard","hentry","category-math"],"acf":[],"_links":{"self":[{"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/posts\/17559","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/comments?post=17559"}],"version-history":[{"count":6,"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/posts\/17559\/revisions"}],"predecessor-version":[{"id":17565,"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/posts\/17559\/revisions\/17565"}],"wp:attachment":[{"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/media?parent=17559"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/categories?post=17559"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kith.org\/jed\/wp-json\/wp\/v2\/tags?post=17559"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}