Question

show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 Where Q is some integers​

Answer 1

$\tt Let \: a \: be \: an \: given \: no. \\ \tt on \: dividing \: a \: by \: 6,we \: get \: q \: as \\ \tt the \: quotient \: and \: r \: as \: the \: remainder \\ \tt such \: that \: a = 6q + r,r = 0,1,2,3,4,5 \\ \tt when \: r = 0 \\ \tt a = 6q, \: even \: no.\\ \tt when \: r = 1 \\ \tt a = 6q + 1, \: odd \: no.\\ \tt when \: r = 2\\ \tt a = 6q + 2, \: even \: no. \\ \tt when \: r = 3 \\ \tt a = 6q + 3, \: odd \: no.\\ \tt when \: r = 4 \\ \tt a = 6q + 4, \: even \: no.\\ \tt when \: r = 5 \\ \tt a = 6q + 5, \: odd \: no. \\ \tt so \: any \: positive \: odd \: integer \: is \: of \: form \\ \tt6q + 1,6q + 3,6q + 5. \\ \\ \tt \dag HENCE \: PROVED \dag$

Answer 2

Answer:

It is the correct answer.

Step-by-step explanation:

Hope this attachment helps you.