Math, asked by Sadiq1231, 1 year ago

solve this question......

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Answered by mathematician7
1
let a=6q+r
where 0=<r<6
if r=0
then a=6q
if r=1
then a =6q+1
similarly when r=2,3,4,5 then a=6q+2,6q+3,6q+4,6q+5
but we have to find positive odd integers threfore 6q+1,6q+3,6q+5 are the forms of odd positive integers

mathematician7: thank u vry much for tagging my answer as brainlliest i will try that i would do my best to help everyone:)
Answered by thales
0
the possible remainders are 0,1,2,3,4,5
6(o)=0,6 (1)+1=7, 6 (2)+2=14, 6 (3)+3=21 ,
6 (4)+4=26,6 (5)+5=35,6 (6)+6=42
0,14,26,42 are divisible by 2 therefore they are even
so we can conclude that remaining 6q+1,
6q+3, 6q+5 are odd positive integers

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