Solve thesw question

Solve thesw question Short Answer 27. (4 Maiks 32. 33.

$$\begin{gathered} Dear Student, \\ Solution20) \\ Let a be positive integer. Then by euclid\text{'}s division lemma, we have, \\ a=bq+r, where 0\le r<b \\ Now, for b=3, we get, \\ a=3q+r, where 0\le r<3.....(i) \\ SO, the nos. are of form 3q,3q+1 and 3q+2. \\ Therefore, {\left(3q\right)}^2=9q^2=3(3q^2)=3m, where m is an integer. \\ {\left(3q+1\right)}^2=9q^2+6q+1=3(3q^2+2q)+1=3m+1, where m is an integer. \\ and \\ {\left(3q+2\right)}^2=9q^2+12q+4 whichcannot be expressed in the form of 3m+2. \\ Thus, square of any +ve integer cannot be expressed in the form of 3m+2. \end{gathered}$$
$$\begin{gathered} Sol 22) \\ Let three consecutive +ve integers be n,n+1 and n+2. \\ Whenever a no. is divided by 3, the remainder obtained is either 0, or 1 or 2. \\ therefore, n=3p or 3p+1 or 3p+2 , where p is any integer. \\ Now, \\ if n=3p, then n is divisible by 3. \\ if n=3p+1, then n+2=3p+1+2=3p+3=3(p+1) is divisible by 3. \\ if n=3p+2, then n+1=3p+1+2=3p+3=3(p+1) is divisible by 3. \\ Thus, one of numbers among n, n+1 and N=2 is always divisibleby 3. \end{gathered}$$
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