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T,3 or, 6q+5, where qINCERTLEVEL-212.Show that the square of any positive integer cannot be of the form 6m+ 2 or 6m+ 5 forany integer m.INCERT EXEMPIAR13. Show that the cuho a |
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Answer» Let a be the positive integer and b = 6.Then, by Euclid’s algorithm, a = 6q + r for some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5.So,a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.(6q)^2= 36q^2= 6(6q^2)= 6m, where m is any integer.(6q + 1)^2= 36q^2+ 12q + 1= 6(6q^2+ 2q) + 1= 6m + 1, where m is any integer.(6q + 2)^2= 36q^2+ 24q + 4= 6(6q^2+ 4q) + 4= 6m + 4, where m is any integer.(6q + 3)^2= 36q^2+ 36q + 9= 6(6q^2+ 6q + 1) + 3= 6m + 3, where m is any integer.(6q + 4)^2= 36q^2+ 48q + 16= 6(6q^2+ 7q + 2) + 4= 6m + 4, where m is any integer.(6q + 5)^2= 36q^2+ 60q + 25= 6(6q^2+ 10q + 4) + 1= 6m + 1, where m is any integer. Hence, The square of any positive integer is of the form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m. |
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