Răspuns:
1.din teorema impartirii cu rest =>R < I
I=6=> R=0,1,2,3,4,5
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2. I=8 => R=0,1,2,3,4,5,6,7
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5. I=5 => R=0,1,2,3,4
a:5=60 rest 0=>a=5×60+0=>a=300
a:5=60 rest 4=>a=5×60+4=>a=304
nr sunt:300,301,302,303,304
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6. I=3=>R=0,1,2
a:3=9 rest 0=>a=3×9+0=>a=27
a:3=9 rest 1=>a=28
a:3=9 rest 2=>a=29
nr sunt: 27,28,29