If the square of any
Positive integer
divided by 6, the remainder can not be
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Let a be any positive integer then for aand7 by Euclid's division lemma we have,
a=7q+r, where r=0,1,2,3,4,5,6
Case 1: when r=0
a=7q
a
2
=49q
2
Case2:Whenr=1
a=7q+1
a
2
=(7q+1)
2
=49q
2
+1+14q
=7(7q
2
+2q)+1
Case3:Whenr=2
a=7q+2
a
2
=(7q+2)
2
=49q
2
+4+28q
a
2
=7(7q
2
+4q)+4
Case4:Whenr=3
a=7q+3
a
2
=(7q+3)
2
=49q
2
+9+42q
a
2
=7(7q
2
+6q+1)+2
Case5:Whenr=4
a=7q+4
a
2
=(7q+4)
2
=49q
2
+16+56q
a
2
=7(7q
2
+8q+2)+2
Case6:Whenr=5
a=7q+5
a
2
=(7q+5)
2
=49q
2
+25+70q
a
2
=7(7q
2
+10q+3)+4
Case7:Whenr=6
a=7q+6
a
2
=(7q+6)
2
=49q
2
+36+84q
a
2
=7(7q
2
+12q+5)+1
Hence, from case 1 to case 7 we can say that when the square of any positive integer is divided by 7 then it never leaves 5or6
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