tag:blogger.com,1999:blog-15628310.post6984375603956919976..comments2021-04-15T02:41:46.574-04:00Comments on Question of the day: Employee NumberAnonymoushttp://www.blogger.com/profile/18153935609499338685noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-15628310.post-50838668930629304892009-09-17T05:21:31.716-04:002009-09-17T05:21:31.716-04:00I started with 5 in the 5th slot knowing that woul...I started with 5 in the 5th slot knowing that would be the only end number divisible by 9 if there is no 0 and worked my way up. <br /><br />I thought if 5 was in the 5th slot then what number in the 50's is divisible by 6. I got 54 and then continued on thinking what number in the 40's is divisible by 7. 42 and 49. I tried both and eventually discovered the last five digits by concentrating on only two digits at a time. Knowing the four digits left, 1 7 8 and 2, and knowing that the second number in the badge number had to be even to be divisible by two I tried out 2 and 8. Once I discovered 2 wouldn't be able to make a number divisible by three with the numbers I had left, I just worked my way up from the 8 in the 2nd position and got my final answer. <br /><br />187254963<br /><br />I tested it along the way using only the last two digits and it worked. 18 is divisible by 2, 87 is divisible by 3, and so on. But when I tried the entire badge number it failed once. 187/3=62.3333 arrg. <br /><br />good puzzle though. Thanks!~Lissahttps://www.blogger.com/profile/06834140401817766254noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-39522100917737307632009-09-01T13:13:33.280-04:002009-09-01T13:13:33.280-04:00Dang excel -- rounded up that decimal. Shoul'...Dang excel -- rounded up that decimal. Shoul've looked better. good puzzleUSAF Pilothttps://www.blogger.com/profile/01326339691636594880noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-11390087727874602932009-09-01T10:56:45.419-04:002009-09-01T10:56:45.419-04:00BTW, USAF Pilot: 987654321 comes close but doesn&#...BTW, USAF Pilot: 987654321 comes close but doesn't work. 9876543/7 = 1410934.71. Every other number works, though.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-3660981921295174352009-09-01T10:42:53.847-04:002009-09-01T10:42:53.847-04:00The number is 381654729.
If the number is ABCDEFG...The number is 381654729.<br /><br />If the number is ABCDEFGHI:<br /><br />Then B, D, F and H are even numbers since they are divisible by even numbers. The rest are odd.<br /><br />ABCDE can be divided evenly by 5, so E is 5.<br /><br />ABCD is divisible by 4, therefore, CD can also be divided evenly by 4, and since C is an odd number, D can only be 2 or 6.<br /><br />ABCDEF can be divided by 6 (by 2 and by 3). Since ABC can be divided by 3, DEF can be also. So, DEF is 258 or 654.<br /><br />Keep going from here to get the rest.Anonymoushttps://www.blogger.com/profile/18153935609499338685noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-56265489271601715152009-09-01T09:26:09.056-04:002009-09-01T09:26:09.056-04:00I guess I should have said the "larger" ...I guess I should have said the "larger" numbers would be more difficult to place, not the "harder" numbers.USAF Pilothttps://www.blogger.com/profile/01326339691636594880noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-7673448292599171592009-09-01T09:24:07.677-04:002009-09-01T09:24:07.677-04:00So I created an excel program with some knowns:
E...So I created an excel program with some knowns:<br /><br />Even digits in place #: 2,4,6,8<br />Odd digits in place #: 1,3,7,9<br />A 5 must go in position 5<br /><br />Then i created rules for the division, i.e.:<br /><br />First 2 digits: even, divisible by 2<br />First 3 digits: odd, summation div by 3<br />First 4 digits: even, last two digits div by 4<br />First 5 digits: must end in 5<br />First 6 digits: even, summation div by 3<br />First 8 digits: even, last 3 digits div by 8<br /><br />Then I figured the harder numbers would be more dificult to place than the smaller numbers so I started with 9.<br /><br />Then 8, so far so good, then 7, looking good, then 6 all working then 5, success, then 4,3,2,1. All worked. So your badge number could be:<br /><br />987654321USAF Pilothttps://www.blogger.com/profile/01326339691636594880noreply@blogger.comtag:blogger.com,1999:blog-15628310.post-67739572758633348322009-08-31T15:25:22.377-04:002009-08-31T15:25:22.377-04:00I think you can work out the second digit has to b...I think you can work out the second digit has to be 8 due to the fact that the first and third digits have to be odd, and all three of the fisrt digits have to add up to a number that is divisible by 3 (in order to be divisible by 3)<br /><br />Of course I'm working backwards after seeing your answer, but mathematically you should be able to see that the first 3 digits are 381 or 183.<br /><br />After that, the third and fourth digit have to be divisible by 4 in order for the whole four digits to be divisible by 4, so we're looking at 381X or 183X, which gives us 3812, 3816, or 1832, 1836... and I don't really know where to go from there and it's lunch time.Jeffnoreply@blogger.comtag:blogger.com,1999:blog-15628310.post-82297696896742604032009-08-31T13:19:11.862-04:002009-08-31T13:19:11.862-04:00381654729
Trial and error, though I'm sure th...381654729<br /><br />Trial and error, though I'm sure there is a better way to figure it out. Well, I did know that the 5th digit had to be 5, and the 2nd, 4th, 6th, and 8th, had to be some combination of even numbers...but the rest was guess work for me. :-PAnonymousnoreply@blogger.com