{"action":"created","issue":{"url":"https://api.github.com/repos/tomaskrizek/tul-szz-it-nv/issues/15","repository_url":"https://api.github.com/repos/tomaskrizek/tul-szz-it-nv","labels_url":"https://api.github.com/repos/tomaskrizek/tul-szz-it-nv/issues/15/labels{/name}","comments_url":"https://api.github.com/repos/tomaskrizek/tul-szz-it-nv/issues/15/comments","events_url":"https://api.github.com/repos/tomaskrizek/tul-szz-it-nv/issues/15/events","html_url":"https://github.com/tomaskrizek/tul-szz-it-nv/issues/15","id":156147880,"number":15,"title":"Okruh 2 - Základy teorie dělitelnosti","user":{"login":"tomaskrizek","id":10220243,"avatar_url":"https://avatars.githubusercontent.com/u/10220243?v=3","gravatar_id":"","url":"https://api.github.com/users/tomaskrizek","html_url":"https://github.com/tomaskrizek","followers_url":"https://api.github.com/users/tomaskrizek/followers","following_url":"https://api.github.com/users/tomaskrizek/following{/other_user}","gists_url":"https://api.github.com/users/tomaskrizek/gists{/gist_id}","starred_url":"https://api.github.com/users/tomaskrizek/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/tomaskrizek/subscriptions","organizations_url":"https://api.github.com/users/tomaskrizek/orgs","repos_url":"https://api.github.com/users/tomaskrizek/repos","events_url":"https://api.github.com/users/tomaskrizek/events{/privacy}","received_events_url":"https://api.github.com/users/tomaskrizek/received_events","type":"User","site_admin":false},"labels":[],"state":"open","locked":false,"assignee":null,"milestone":null,"comments":3,"created_at":"2016-05-22T11:35:36Z","updated_at":"2016-05-25T07:20:10Z","closed_at":null,"body":"[2. - Základy teorie dělitelnosti](https://github.com/tomaskrizek/tul-szz-it-nv/blob/master/02_zaklady_teorie_delitelnosti/02_zaklady_teorie_delitelnosti.md)\r\n\r\n|     Autor     |     Reviewer    |     Stav    |\r\n| ------------| -----------| ------------- |\r\n| @tomaskrizek  |    | :hourglass: |\r\n\r\n### TODO\r\n\r\n- [ ] Oveřit postup a výsledek u řešení soustavy kongruencí (Čínská věta o zbytku).\r\n- [ ] Rozhodnout, zda uvádět popis / příklad zobecněné čínské věty o zbytku.\r\n"},"comment":{"url":"https://api.github.com/repos/tomaskrizek/tul-szz-it-nv/issues/comments/221493070","html_url":"https://github.com/tomaskrizek/tul-szz-it-nv/issues/15#issuecomment-221493070","issue_url":"https://api.github.com/repos/tomaskrizek/tul-szz-it-nv/issues/15","id":221493070,"user":{"login":"michalmuzicek","id":16854966,"avatar_url":"https://avatars.githubusercontent.com/u/16854966?v=3","gravatar_id":"","url":"https://api.github.com/users/michalmuzicek","html_url":"https://github.com/michalmuzicek","followers_url":"https://api.github.com/users/michalmuzicek/followers","following_url":"https://api.github.com/users/michalmuzicek/following{/other_user}","gists_url":"https://api.github.com/users/michalmuzicek/gists{/gist_id}","starred_url":"https://api.github.com/users/michalmuzicek/starred{/owner}{/repo}","subscriptions_url":"https://api.github.com/users/michalmuzicek/subscriptions","organizations_url":"https://api.github.com/users/michalmuzicek/orgs","repos_url":"https://api.github.com/users/michalmuzicek/repos","events_url":"https://api.github.com/users/michalmuzicek/events{/privacy}","received_events_url":"https://api.github.com/users/michalmuzicek/received_events","type":"User","site_admin":false},"created_at":"2016-05-25T07:20:10Z","updated_at":"2016-05-25T07:20:10Z","body":"A reseni cinske vety je spravne, overil jsem to kalkulackou. \r\nCo se tyce obecne cinske vety, rozhodne bude lepsi mit nastrel co delat, kdyby ty jednotliva mopula nebyla po 2 nesoudelna, Nevim jestli bych primo uvadel i priklad, protoze opravdu pochybuji ze ti daj spocitat priklad kdyz na cele statnice mas 20 minut, to se spis zeptaj jak bys postupoval.\r\nTakze neco ve smyslu:\r\nnajdu nejmensi spolecny nasobek pro dana modula, a ten pak musim rozdelit na nesoudelna nova modula a pak uz pomoci vzorce dostanu serii kongruenci, ktere vyresim, dosadim do vzorce a mam vysledek\r\n"}}
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