\displaystyle \textbf{Question 1. }\text{Prove that the product of two consecutive positive integers is divisible by }2.
\displaystyle \text{Answer:}
\displaystyle \text{Let the two consecutive positive integers be }n\text{ and }n+1.
\displaystyle \text{Then their product}=n(n+1)
\displaystyle \text{Among two consecutive integers, one is always even.}
\displaystyle \therefore n(n+1)\text{ is even.}
\displaystyle \therefore n(n+1)\text{ is divisible by }2.
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\displaystyle \textbf{Question 2. }\text{Prove that the product of three consecutive positive integers is divisible by }6.
\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive positive integers be }n,\ n+1,\ n+2.
\displaystyle \text{Their product}=n(n+1)(n+2)
\displaystyle \text{Among three consecutive integers, one is always divisible by }3.
\displaystyle \text{Also, among three consecutive integers, at least one is even.}
\displaystyle \therefore n(n+1)(n+2)\text{ is divisible by both }2\text{ and }3.
\displaystyle \therefore n(n+1)(n+2)\text{ is divisible by }2\times3=6.
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\displaystyle \textbf{Question 3. }\text{Show that any positive odd integer is of the form }
\displaystyle 6q+1\text{ or }6q+3\text{ or }6q+5, \text{where }q\text{ is some integer.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be any positive integer.}
\displaystyle \text{By Euclid's division lemma, for divisor }6,\text{ we have}
\displaystyle a=6q+r,\text{ where }0\le r<6
\displaystyle \therefore r=0,1,2,3,4,5
\displaystyle \therefore a=6q,\ 6q+1,\ 6q+2,\ 6q+3,\ 6q+4,\ 6q+5
\displaystyle \text{Now }6q,\ 6q+2,\ 6q+4\text{ are even integers.}
\displaystyle \therefore \text{Odd positive integers are of the form }6q+1,\ 6q+3,\text{ or }6q+5.
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\displaystyle \textbf{Question 4. }\text{Show that one and only one of }n,\ n+1\text{ and }n+2\text{ is divisible by }3,
\displaystyle \text{where }n\text{ is a positive integer.}
\displaystyle \text{Answer:}
\displaystyle \text{By Euclid's division lemma, }n=3q+r,\text{ where }0\le r<3.
\displaystyle \therefore r=0,1,2
\displaystyle \text{Case 1: If }r=0,\text{ then }n=3q.
\displaystyle \therefore n\text{ is divisible by }3.
\displaystyle \text{Case 2: If }r=1,\text{ then }n=3q+1.
\displaystyle \therefore n+2=3q+3=3(q+1)
\displaystyle \therefore n+2\text{ is divisible by }3.
\displaystyle \text{Case 3: If }r=2,\text{ then }n=3q+2.
\displaystyle \therefore n+1=3q+3=3(q+1)
\displaystyle \therefore n+1\text{ is divisible by }3.
\displaystyle \therefore \text{One and only one of }n,\ n+1\text{ and }n+2\text{ is divisible by }3.
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\displaystyle \textbf{Question 5. }\text{Show that the square of an odd positive integer is of the form }
\displaystyle 8q+1, \text{ for some integer }q.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be any odd positive integer.}
\displaystyle \therefore a=2m+1,\text{ for some integer }m.
\displaystyle a^{2}=(2m+1)^{2}
\displaystyle =4m^{2}+4m+1
\displaystyle =4m(m+1)+1
\displaystyle \text{Since }m\text{ and }m+1\text{ are consecutive integers, one of them is even.}
\displaystyle \therefore m(m+1)\text{ is divisible by }2.
\displaystyle \text{Let }m(m+1)=2q,\text{ for some integer }q.
\displaystyle \therefore a^{2}=4(2q)+1
\displaystyle =8q+1
\displaystyle \therefore \text{The square of an odd positive integer is of the form }8q+1.
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\displaystyle \textbf{Question 6. }\text{Prove that the square of any positive integer is of the form }
\displaystyle 4q\text{ or }4q+1  \text{ for some integer }q.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be any positive integer.}
\displaystyle \text{Then }a\text{ is either even or odd.}
\displaystyle \text{Case 1: If }a\text{ is even, then }a=2m.
\displaystyle \therefore a^{2}=(2m)^{2}=4m^{2}
\displaystyle \text{Let }q=m^{2}.
\displaystyle \therefore a^{2}=4q
\displaystyle \text{Case 2: If }a\text{ is odd, then }a=2m+1.
\displaystyle \therefore a^{2}=(2m+1)^{2}
\displaystyle =4m^{2}+4m+1
\displaystyle =4(m^{2}+m)+1
\displaystyle \text{Let }q=m^{2}+m.
\displaystyle \therefore a^{2}=4q+1
\displaystyle \therefore \text{The square of any positive integer is of the form }4q\text{ or }4q+1.
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\displaystyle \textbf{Question 7. }\text{For any positive integer }n,\text{ prove that }n^{3}-n\text{ is divisible by }6.
\displaystyle \text{Answer:}
\displaystyle n^{3}-n=n(n^{2}-1)
\displaystyle =n(n-1)(n+1)
\displaystyle =(n-1)n(n+1)
\displaystyle \text{These are three consecutive integers.}
\displaystyle \text{Among three consecutive integers, one is divisible by }3.
\displaystyle \text{Also, among three consecutive integers, at least one is even.}
\displaystyle \therefore (n-1)n(n+1)\text{ is divisible by }2\times3=6.
\displaystyle \therefore n^{3}-n\text{ is divisible by }6.
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\displaystyle \textbf{Question 8. }\text{Show that the square of any positive integer cannot be of the form }
\displaystyle 6m+2 \text{ or }6m+5\text{ for any integer }m.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be any positive integer.}
\displaystyle \text{By Euclid's division lemma, }a=6q+r,\text{ where }0\le r<6.
\displaystyle \therefore r=0,1,2,3,4,5
\displaystyle \text{Now,}
\displaystyle \text{If }a=6q,\text{ then }a^{2}=36q^{2}=6(6q^{2})
\displaystyle \text{If }a=6q+1,\text{ then }a^{2}=36q^{2}+12q+1=6(6q^{2}+2q)+1
\displaystyle \text{If }a=6q+2,\text{ then }a^{2}=36q^{2}+24q+4=6(6q^{2}+4q)+4
\displaystyle \text{If }a=6q+3,\text{ then }a^{2}=36q^{2}+36q+9=6(6q^{2}+6q+1)+3
\displaystyle \text{If }a=6q+4,\text{ then }a^{2}=36q^{2}+48q+16=6(6q^{2}+8q+2)+4
\displaystyle \text{If }a=6q+5,\text{ then }a^{2}=36q^{2}+60q+25=6(6q^{2}+10q+4)+1
\displaystyle \therefore a^{2}\text{ is of the form }6m,\ 6m+1,\ 6m+3,\text{ or }6m+4.
\displaystyle \therefore a^{2}\text{ cannot be of the form }6m+2\text{ or }6m+5.
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\displaystyle \textbf{Question 9. }\text{Show that the cube of a positive integer is of the form }6q+r,\text{ where }q
\displaystyle \text{is an integer and }r=0,1,2,3,4,5.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be any positive integer.}
\displaystyle \text{By Euclid's division lemma, }a=6m+r,\text{ where }0\le r<6.
\displaystyle \therefore r=0,1,2,3,4,5
\displaystyle \text{Now, }a^{3}\text{ is also a positive integer.}
\displaystyle \text{Again, by Euclid's division lemma, for divisor }6,
\displaystyle a^{3}=6q+r,\text{ where }0\le r<6.
\displaystyle \therefore r=0,1,2,3,4,5
\displaystyle \therefore \text{The cube of a positive integer is of the form }6q+r,
\displaystyle \text{where }q\text{ is an integer and }r=0,1,2,3,4,5.
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\displaystyle \textbf{Question 10. }\text{Show that one and only one out of }n,\ n+4,\ n+8,\ n+12
\displaystyle \text{ and }n+16 \text{ is divisible by }5,\text{ where }n\text{ is any positive integer.}
\displaystyle \text{Answer:}
\displaystyle \text{By Euclid's division lemma, }n=5q+r,\text{ where }0\le r<5.
\displaystyle \therefore r=0,1,2,3,4
\displaystyle \text{Case 1: If }r=0,\text{ then }n=5q.
\displaystyle \therefore n\text{ is divisible by }5.
\displaystyle \text{Case 2: If }r=1,\text{ then }n=5q+1.
\displaystyle \therefore n+4=5q+5=5(q+1)
\displaystyle \therefore n+4\text{ is divisible by }5.
\displaystyle \text{Case 3: If }r=2,\text{ then }n=5q+2.
\displaystyle \therefore n+8=5q+10=5(q+2)
\displaystyle \therefore n+8\text{ is divisible by }5.
\displaystyle \text{Case 4: If }r=3,\text{ then }n=5q+3.
\displaystyle \therefore n+12=5q+15=5(q+3)
\displaystyle \therefore n+12\text{ is divisible by }5.
\displaystyle \text{Case 5: If }r=4,\text{ then }n=5q+4.
\displaystyle \therefore n+16=5q+20=5(q+4)
\displaystyle \therefore n+16\text{ is divisible by }5.
\displaystyle \therefore \text{One and only one out of }n,\ n+4,\ n+8,\ n+12\text{ and }n+16\text{ is divisible by }5.
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\displaystyle \textbf{Question 11. }\text{Show that the square of an odd positive integer can be of the form }
\displaystyle 6q+1 \text{ or }6q+3\text{ for some integer }q.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be any odd positive integer.}
\displaystyle \text{By Euclid's division lemma, }a=6m+r,\text{ where }0\le r<6.
\displaystyle \therefore r=0,1,2,3,4,5
\displaystyle \text{Since }a\text{ is odd, }a\text{ can be of the form }6m+1,\ 6m+3,\text{ or }6m+5.
\displaystyle \text{If }a=6m+1,\text{ then }a^{2}=36m^{2}+12m+1
\displaystyle =6(6m^{2}+2m)+1
\displaystyle \text{If }a=6m+3,\text{ then }a^{2}=36m^{2}+36m+9
\displaystyle =6(6m^{2}+6m+1)+3
\displaystyle \text{If }a=6m+5,\text{ then }a^{2}=36m^{2}+60m+25
\displaystyle =6(6m^{2}+10m+4)+1
\displaystyle \therefore \text{The square of an odd positive integer can be of the form }6q+1\text{ or }6q+3.
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\displaystyle \textbf{Question 12. }\text{A positive integer is of the form }3q+1,\ q\text{ being a natural number.}
\displaystyle \text{Can you write its square in any form other than }3m+1,\ 3m\text{ or }3m+2
\displaystyle \text{for some integer }m?\text{ Justify your answer.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the positive integer be }a=3q+1.
\displaystyle \therefore a^{2}=(3q+1)^{2}
\displaystyle =9q^{2}+6q+1
\displaystyle =3(3q^{2}+2q)+1
\displaystyle \text{Let }m=3q^{2}+2q.
\displaystyle \therefore a^{2}=3m+1
\displaystyle \text{No, its square cannot be written in any form other than }3m,\ 3m+1\text{ or }3m+2.
\displaystyle \text{By Euclid's division lemma, every positive integer is of one of these three forms.}
\displaystyle \text{Here, the square is specifically of the form }3m+1.
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\displaystyle \textbf{Question 13. }\text{Show that the square of any positive integer cannot be of the form }
\displaystyle 3m+2, \text{ where }m\text{ is a natural number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }a\text{ be any positive integer.}
\displaystyle \text{By Euclid's division lemma, }a=3q+r,\text{ where }0\le r<3.
\displaystyle \therefore r=0,1,2
\displaystyle \text{If }a=3q,\text{ then }a^{2}=9q^{2}=3(3q^{2})
\displaystyle \text{If }a=3q+1,\text{ then }a^{2}=9q^{2}+6q+1=3(3q^{2}+2q)+1
\displaystyle \text{If }a=3q+2,\text{ then }a^{2}=9q^{2}+12q+4=3(3q^{2}+4q+1)+1
\displaystyle \therefore a^{2}\text{ is of the form }3m\text{ or }3m+1.
\displaystyle \therefore a^{2}\text{ cannot be of the form }3m+2.
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\displaystyle \textbf{Question 14. }\text{If }a\text{ and }b\text{ are two odd positive integers such that }a>b,\text{ then prove that one}
\displaystyle \text{of the two numbers }\frac{a+b}{2}\text{ and }\frac{a-b}{2}\text{ is odd and the other is even.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }a\text{ and }b\text{ are odd positive integers, let}
\displaystyle a=2m+1\text{ and }b=2n+1,\text{ where }m>n.
\displaystyle \frac{a+b}{2}=\frac{(2m+1)+(2n+1)}{2}
\displaystyle =\frac{2m+2n+2}{2}
\displaystyle =m+n+1
\displaystyle \frac{a-b}{2}=\frac{(2m+1)-(2n+1)}{2}
\displaystyle =\frac{2m-2n}{2}
\displaystyle =m-n
\displaystyle \text{Now, }(m+n+1)+(m-n)=2m+1,
\displaystyle \text{which is odd.}
\displaystyle \text{If the sum of two integers is odd, then one integer is odd and the other is even.}
\displaystyle \therefore \text{One of }\frac{a+b}{2}\text{ and }\frac{a-b}{2}\text{ is odd and the other is even.}
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