show that the square of an odd positive integer is of the form 6q+1 or 6q+3 for some integer of 3
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Answer: a=bq+r,0<=r<b
Where b=6
Let us take r=0,1,2,3,4,5
a=6q+0=6q-even integer
a=6q+1-odd integer
a=6q+2-even integer
a=6q+3-odd integer
a=6q+4-even integer
a=6q+5-odd integer
Therefore 6q+1,6q+3,6q+5 are odd integer
Hence proved
Step-by-step explanation:
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