tag:blogger.com,1999:blog-6213919025256134910.post4142076665538968770..comments2023-05-15T08:32:02.241-07:00Comments on The Math Contest: Answers to 1990Curmudgeonhttp://www.blogger.com/profile/04323026187622872114noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6213919025256134910.post-33990378664618922272013-06-26T21:01:27.476-07:002013-06-26T21:01:27.476-07:00Just took a stab at #33 and got 3(pi) minutes. Ca...Just took a stab at #33 and got 3(pi) minutes. Can anyone confirm?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6213919025256134910.post-12385184949256185252013-05-27T22:59:34.738-07:002013-05-27T22:59:34.738-07:00#38 caught my eye because I thought you could grap...#38 caught my eye because I thought you could graph it somehow at first. I then realized if the three absolute values have to add to 5, then each one can be at most 5. So each absolute value only allows for a window of 10 possibilities. I tried the union of these sets (ie. checking all values) and realized quickly it should have been the intersection of these sets, allowing now for only the numbers from 0 to 6 as possibilities. I checked each and found the answer to be x=2,3,4. <br><br>I thought it was interesting that no values are strictly less than 5. There may be a bit more thinking that can further reduce the possibilities to check here, but I didn't see it at first and 6 wasn't too many (especially when I had started checking at 14 possibilities already).mjlhttp://www.blogger.com/profile/07424866855159434391noreply@blogger.comtag:blogger.com,1999:blog-6213919025256134910.post-55593538698618922192013-05-26T12:54:48.953-07:002013-05-26T12:54:48.953-07:00For #31, y = 1990For #31, y = 1990PeggyUnoreply@blogger.comtag:blogger.com,1999:blog-6213919025256134910.post-50193850049027090552013-05-14T12:38:43.682-07:002013-05-14T12:38:43.682-07:0018 is simple counting. Each of the 4 rows must hav...18 is simple counting. Each of the 4 rows must have exactly one "1", slotted in a different column than the others. So should be 7*6*5*4.Markhttp://www.blogger.com/profile/17141745677537493502noreply@blogger.comtag:blogger.com,1999:blog-6213919025256134910.post-3846613546628309322013-05-14T10:40:32.939-07:002013-05-14T10:40:32.939-07:00For #18, I think it is a Counting Principle proble...For #18, I think it is a Counting Principle problem. 7*6*5*4Ericnoreply@blogger.comtag:blogger.com,1999:blog-6213919025256134910.post-46694383332304552572013-05-12T11:47:38.707-07:002013-05-12T11:47:38.707-07:00Answer for #4, I believe, is 119. There are no dup...Answer for #4, I believe, is 119. <br><br>There are no duplicate values possible, so no subtraction is needed.<br><br>There are four choices for 100s (0, 1, 2, or 3), times five choices for 10s (0, 1, 2, 3, or 4), times two for 5s (none or one), times 3 for ones (0, 1, or 2).<br><br>4 x 5 x 2 x 3 (minus the 0, 0, 0, 0 non-positive choice).<br><br>It is equivalent to asking how many factors greater than 1 126,000 has.<br><br>Jonathanjd2718.orghttp://jd2718.org/noreply@blogger.com