{"id":191,"date":"2018-10-12T14:29:20","date_gmt":"2018-10-12T19:29:20","guid":{"rendered":"http:\/\/blogs.acu.edu\/mu_sigma\/?p=191"},"modified":"2018-10-26T19:03:37","modified_gmt":"2018-10-27T00:03:37","slug":"problem-6-sums-of-digits-october-12-2018","status":"publish","type":"post","link":"https:\/\/blogs.acu.edu\/mu_sigma\/2018\/10\/12\/problem-6-sums-of-digits-october-12-2018\/","title":{"rendered":"Problem 6 – Sums of Digits – October 12, 2018"},"content":{"rendered":"

Find the Sum of the Digits<\/b><\/span><\/p>\n

This problem is from Hai He, Hunter College of CUNY.<\/i><\/p>\n

Let A<\/b> be the sum of the digits<\/i> of 1999! (that is 1999*1998*1997*..*2*1). Let B<\/b> be the sum of the digits<\/i> of A<\/b>. Let C<\/b> be the sum of the digits<\/i> of B<\/b>. What is the value of C<\/b>?<\/p>\n

Submit your answers to mathpotw@acu.edu. \u00a0Details for submissions can be found here<\/a>.<\/p>\n

Solution to Find the Sum of the Digits<\/b><\/h3>\n

Correct solutions were submitted by Wyatt Witemeyer.\u00a0<\/em><\/p>\n\n\n\n\n\n\n\n
1999! =<\/td>\n1.<\/sup>2…<\/sup>999.<\/sup>1000.<\/sup>1001…<\/sup>1998.<\/sup>1999<\/td>\n<\/tr>\n
=<\/td>\n(1.<\/sup>1999)(2.<\/sup>1998)…<\/sup>(999.<\/sup>1001).<\/sup>1000<\/td>\n<\/tr>\n
<<\/td>\n(1000.<\/sup>1000)(1000.<\/sup>1000)…<\/sup>(1000.<\/sup>1000).<\/sup>1000<\/td>\n<\/tr>\n
=<\/td>\n10001999<\/sup><\/td>\n<\/tr>\n
=<\/td>\n105997<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Therefore 1999! has at most 5997 digits. This implies that A<\/b> (the sum of the digits of 1999!) is at most 5997.<\/sup>9 = 53, 973.<\/p>\n

So B<\/b> (the sum of the digits of A<\/b>) is at most 5 + 4.<\/sup>9 = 41.<\/p>\n

Hence C<\/b> (the sum of the digits of B<\/b>) is at most 4 + 9 = 13.<\/p>\n

But the sum of the digits of a positive integer N<\/b> has the same remainder upon division by 9 as does N<\/b> itself. So 1999! and A<\/b> and B<\/b> and C<\/b> all have the same remainder upon division by 9. But 9 divides 1999!, so 9 must divide C<\/b>. Since C<\/b> is at most 13 and C<\/b> is not 0, then<\/p>\n

C<\/b> = 9.<\/p>\n","protected":false},"excerpt":{"rendered":"

Find the Sum of the Digits This problem is from Hai He, Hunter College of CUNY. Let A be the sum of the digits of 1999! (that is 1999*1998*1997*..*2*1). Let B be the sum of the digits of A. Let C be the sum of the digits of B. What is the value of C? […]<\/p>\n","protected":false},"author":130,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[181534],"tags":[],"class_list":["post-191","post","type-post","status-publish","format-standard","hentry","category-problem-of-the-week"],"_links":{"self":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/191","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/users\/130"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/comments?post=191"}],"version-history":[{"count":3,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/191\/revisions"}],"predecessor-version":[{"id":202,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/191\/revisions\/202"}],"wp:attachment":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/media?parent=191"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/categories?post=191"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/tags?post=191"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}