{"id":66,"date":"2009-07-09T14:00:36","date_gmt":"2009-07-09T18:00:36","guid":{"rendered":"http:\/\/www.duncansonelectric.com\/blog\/?page_id=66"},"modified":"2011-01-22T07:32:10","modified_gmt":"2011-01-22T11:32:10","slug":"solution-to-the-12-coin-problem","status":"publish","type":"page","link":"http:\/\/www.duncansonelectric.com\/blog\/solution-to-the-12-coin-problem\/","title":{"rendered":""},"content":{"rendered":"

<\/p>\n

SOLUTION TO THE 12 COIN PROBLEM<\/strong><\/span><\/div>\n

IF<\/span><\/strong> one had 12 seemingly identical coins, with 11 being of the exact same weight and 1 being either heavier or lighter than the other 11,<\/p>\n

THEN<\/span><\/strong> using only a balance, not a scale, and with only 3 measurements allowed, how could one determine which of these 12 seemingly identical coins was different and whether it were heavier or lighter than the other 11?<\/p>\n

Here is the solution<\/span>:<\/strong><\/p>\n

<\/span><\/p>\n

Step 1<\/span><\/strong>: Balance coins 1, 2, 3 & 4 against coins 5, 6, 7 & 8: (1,2,3,4) | (5,6,7,8)<\/span><\/strong><\/p>\n

The first possible outcome from Step 1:<\/span><\/strong><\/p>\n

If Step 1<\/span><\/strong> yields Heavy on Left, then either one of 1, 2, 3 or 4 is Heavy or one of 5, 6, 7 or 8 is Light.<\/p>\n

Step 2<\/span><\/strong>: Balance 1, 5, & 6 against 2, 7, & 9: (1,5,6) | (2,7,9)<\/span><\/strong><\/p>\n

If Step 2<\/span><\/strong> yields Heavy on the Left, then either 1 is Heavy or 7 is Light.<\/p>\n

Step 3<\/span><\/strong>: Balance 1 & 7 against 9 & 10: (1,7) | (9,10)
\n<\/span><\/strong>Heavy on Left: 1 is Heavy
\nLight on Left: 7 is Light
\nEqual is not possible<\/p>\n

If Step 2<\/span><\/strong> yields Light on Left, then either 2 is Heavy or 5 or 6 is Light.<\/p>\n

Step 3<\/span><\/strong>: Balance 5 against 6: (5) | (6)
\n<\/strong><\/span>Heavy on Left: 6 is Light
\nLight on Left: 5 is Light
\nEqual: 2 is Heavy<\/p>\n

If Step 2<\/span><\/strong> yields Equal, then either 3 or 4 is Heavy or 8 is Light.<\/p>\n

Step 3<\/span><\/strong>: Balance 3 against 4: (3) | (4)
\n<\/span><\/strong>Heavy on Left: 3 is Heavy
\nLight on Left: 4 is Heavy
\nEqual: 8 is Light<\/p>\n

Back to Step 1 – The second possible outcome from Step 1:<\/span><\/strong><\/p>\n

Step 1<\/span><\/strong>: Again, balancing 1, 2, 3 & 4 against 5, 6, 7 & 8: (1,2,3,4) | (5,6,7,8)<\/span><\/strong><\/p>\n

If Step 1<\/span><\/strong> yields Light on Left, then either one of 1, 2, 3 or 4 is Light or one of 5, 6, 7 or 8 is Heavy.<\/p>\n

Step 2<\/span><\/strong>: Balance 1, 5, & 6 against 2, 7, & 9: (1,5,6) | (2,7,9)<\/span><\/strong><\/p>\n

If Step 2<\/span><\/strong> yields Heavy on Left, then either 5 or 6 is Heavy or 2 is Light.<\/p>\n

Step 3<\/span><\/strong>: Balance 5 against 6: (5) | (6)
\n<\/span><\/strong>Heavy on Left: 5 is Heavy
\nLight on Left: 6 is Heavy
\nEqual: 2 is Light<\/p>\n

If Step 2<\/span><\/strong> yields Light on Left, then either 1 is Light or 7 is Heavy.<\/p>\n

Step 3<\/span><\/strong>: Balance 1 & 7 against 9 & 10: (1,7) | (9,10)
\n<\/span><\/strong>Heavy on Left: 1 is Light
\nLight on Left: 7 is Heavy
\nEqual is not possible<\/p>\n

If Step 2<\/span><\/strong> yields Equal then either 3 or 4 is Light or 8 is Heavy.<\/p>\n

Step 3<\/span><\/strong>: Balance 3 against 4: (3) | (4)
\n<\/span><\/strong>Heavy on Left: 4 is Light
\nLight on Left: 3 is Light
\nEqual: 8 is Heavy<\/p>\n

Back to Step 1 – The last possible outcome from Step 1:<\/span><\/strong><\/p>\n

Step 1<\/span><\/strong>: Again, balancing 1, 2, 3 & 4 against 5, 6, 7 & 8: (1,2,3,4) | (5,6,7,8)<\/span><\/strong><\/p>\n

If Step 1<\/span><\/strong> yields Equal Left to Right balance, then one of 9, 10, 11 or 12 is either Light or Heavy.<\/p>\n

Step 2<\/span><\/strong>: Balance 9 & 10 against 11 & 1: (9,10) | (11,1)<\/span><\/strong><\/p>\n

If Step 2<\/span><\/strong> yields Heavy on Left then either 9 or 10 is Heavy or 11 is Light.<\/p>\n

Step 3<\/span><\/strong>: Balance 9 against 10: (9) | (10)
\n<\/span><\/strong>Heavy on Left: 9 is Heavy
\nLight on Left: 10 is Heavy
\nEqual: 11 is Light<\/p>\n

If Step 2<\/span><\/strong> yields Light on Left then either 9 or 10 is Light or 11 is Heavy.<\/p>\n

Step 3<\/span><\/strong>: Balance 9 against 10: (9) | (10)
\n<\/span><\/strong>Heavy on Left: 10 is Light
\nLight on Left: 9 is Light
\nEqual: 11 is Heavy<\/p>\n

If Step 2<\/span><\/strong> yields Equal Left Right balance, then 12 is either Heavy or Light.<\/p>\n

Step 3<\/span><\/strong>: Balance 12 against 1: (12) | (1)
\n<\/span><\/strong>Heavy on Left: 12 is Heavy
\nLight on Left: 12 is Light
\nEqual is not possible<\/p>\n","protected":false},"excerpt":{"rendered":"

SOLUTION TO THE 12 COIN PROBLEM IF one had 12 seemingly identical coins, with 11 being of the exact same weight and 1 being either heavier or lighter than the other 11, THEN using only a balance, not a scale, and with only 3 measurements allowed, how could one determine which of these 12 seemingly […]<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":121,"comment_status":"open","ping_status":"open","template":"","meta":[],"_links":{"self":[{"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/pages\/66"}],"collection":[{"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/comments?post=66"}],"version-history":[{"count":36,"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/pages\/66\/revisions"}],"predecessor-version":[{"id":69,"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/pages\/66\/revisions\/69"}],"wp:attachment":[{"href":"http:\/\/www.duncansonelectric.com\/blog\/wp-json\/wp\/v2\/media?parent=66"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}