5 Marks Questions

Question 15 Marks

Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisibleby 5, where n is any positive integer.
[Hint: Any positive integer can be written in the form 5q, 5q+1, 5q+2, 5q+3, 5q+4].

Answer

Let a number n is divided by 5 then quotient is q and remainder is r. Then by Euclid's division algorithm,
n = 5q + r, where n, q, r are non-negative integers and $0\leq\text{r}<5$
When r = 0, n = 5q + 0 = 5q
So, n is divisible by 5.
When r = 1, n - 5a + 1
n + 2 = 5q + 1 + 2 = 5q + 3 is not divisible by 5.
n + 4 = (5q + 1) + 4 = 5q + 5 = 5(q + 1) divisible by 5.
So, (n + 4) is divisible by 5.
When r = 2, n = 5q + 2
(n + 8) = (5q + 2) + 8 = 5q + 10 = 5(q + 2) =5m is divisible by 5.
So, (n + 8) is divisible by 5.
When r = 3, n = 5q + 3
n + 12 = (5q + 3) + 12 = 5q + 15 = 5(q + 3) = 5m is divisible by 5.
So, (n + 12) is divisible by 5.
When r = 4, n = 5q + 4
n + 16 = (5q + 4) + 16 = 5q + 20 = 5(q + 4)
(n + 16) = 5m is divisible by 5.
Hence, n, (n + 4), (n + 8), (n + 12) and (n + 16) are divisible by 5.

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Question 25 Marks

Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Answer

Let a be an arbutrary positive integer, then by Euclid's division algorithm, corresponding to the positive integers a and 6 , there exist non-negative integers $q$ and $r$ such that $a=6 q+r$, where $0<r<6 . a=6 q+r$, where $0 \leq r<6$ $\Rightarrow a ^2=(6 q + r )^2=36 q ^2+ r ^2+12 qr \left[\because( a + b )^2= a ^2+2 ab + b ^2\right] \Rightarrow a ^2=6\left(6 q ^2+2 qr \right)+ r ^2 \ldots \ldots$ (i) where, $0 \leq r <6$
Case I:
when $r=0$, then putting $r=0$ in Eq. (i), we get $a^2=6\left(6 q^2\right)=6 m$ where, $m=6 q^2$ ia an integer.
Case II: where $r=1$, then putting $r=1$ in Eq. (i), we get $a^2=6\left(6 q^2+2 q\right)+1=6 m+1$ where, $m=\left(6 q^2+2 q\right)+1=6 m+$ 1
Case III:
When $r=2$, then putting $r=2$ in Eq. (i), we get $a^2=6\left(6 q^2+4 q\right)+4=6 m+4$ where, $m=\left(6 q^2+4 q\right)$ is an integer.
Case IV:
when $r=3$, then Putting $r=4$ in Eq. (i), we get $a^2=6\left(6 q^2+6 q\right)+9=6\left(6 q^2+6 a\right)+9 \Rightarrow a^2=6\left(6 q^2+6 q+1\right)+3=$ $6 m+3$ where, $m=(6 q+6 q+1)$ is an integer.
Case $V$ :
when $r=4$, then putting $r=4$ in Eq. (i), we get $a^2=6\left(6 q^2+8 q\right)+16=6\left(6 q^2+8 q\right)+12+4 \Rightarrow a^2=6\left(6 q^2+8 q+2\right)$ $+4=6 m+4$ where, $m=\left(6 q^2+8 q+2\right)$ is an integer.
Case VI:
When $r=5$, then putting $r=5$ in Eq. (i), we get
$a^2=6\left(6 q^2+10 q\right)+25$
$=6\left(6 q^2+10 q\right)+24+1$
$\Rightarrow a^2=6\left(6 q^2+10 q+4\right)+1=6 m+1$
Where, $m=\left(6 q^2+10 q+1\right)$ is an integer.
Hencre,the square of any positive integer cannot be of the form $6 m+2 m$ or $6 m+5$ for any integer $m$.

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Question 35 Marks

Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3, for some integer m.

Answer

Let a be an arbitrary positive integer. Then, by Euclid's division algorithm, corresponding to the positive integers a and 4, there exist non-negative integers q and r such that a = 4q + r, where $0\leq\text{r}<4$$\text{a}=4\text{q}+\text{r, where}\ 0\leq\text{r}<4$
$\Rightarrow\ \text{a}^3=(4\text{q}+\text{r})^3=64\text{q}^3+\text{r}^3+12\text{qr}^2+48\text{q}^2\text{r}\ \ \big[\because\ (\text{a}+\text{b})^3=\text{a}^3+\text{b}^3+3\text{ab}^2+3\text{a}^2\text{b}\big]$
$\Rightarrow\ \text{a}^3=(64\text{q}^2+48\text{q}^2\text{r}+12\text{qr}^2)+\text{r}^3\ .....\text{(i)}$
$\text{Where}, 0\leq\text{r}<4$
Case I:
When r = 0,
Putting r = 0 in Eq. (i), we get
$a^3 = 64q^3 = 4(16q^3)$
$\Rightarrow a^3 + $ 4m where $m = 16q^3$ is an integer.
Case II:
When r = 1, then Putting r = 1 in Eq. (i), we get
$a^3 = 64q^3 + 48q^2 + 12q + 1$
$= 4(16q^3 + 12q^2 + 3q) + 1$
$= 4m + 1$
Where, $m = (16q^2 + 12q^2 + 3q)$ is an integer.
Case III:
When r = 2, then putting r = 2 in Eq. (i), We get
$a^3 = 64q^3 + 144q^2 + 108q + 27$
$= 64q^3 + 144q^2 + 108q + 24 + 3$
$= 4(16q^3 + 36q^2 + 27q + 6) + 3 = 4m + 3$
Where, $m = (16q^3 + 36q^2 + 27q + 6)$ is an integer.
Hence, the cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

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Question 45 Marks

Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer.

Answer

Consider the given numbers n, n + 2 and n + 4.
When n = 1, number become 1, 1 + 2, 1 + 4 = (1, 3 and 5)
When n = 2, number become 2, 2 + 2, 2 + 4 = (2, 4, 6)
When n = 3, number become = (3, 5, 7)
When n = 4, number become = (4, 6, 8)
When n = 5, number become = (5, 7, 9)
When n = 6, number become = (6, 8, 10)
When n = 7, number become = (7, 9, 11)
From above, we observe that out of 3 numbers one is divisible by 3.
Alternate Answer
Consider that if a number n is divided by 3, then we get a quotient q and remainder r then by Euclid's division algorithm,
$\text{n}=3\text{q}+\text{r where,}\ 0\leq\text{r}<3$

At	$n_1$	Divisible by 3	$n_2 = n_1 + 3$	Divisible by 3	$n_3 = n_3 + 4$	Divisible by 3
r = 0	3q + 0 = 3q	Yes	3q + 2	No	3q + 4 =3q + 3 + 1 = 3(q + 1) +1 = 3m + 1	No
r = 1	3q + 1	No	3q + 1 + 2 = 3q + 3 = 3(q + 1)	Yes	3q + 3 + 2 = 3q + 3 + 2 = 3(q + 1) + 2 = 3m + 2	No
r = 2	3q + 2	No	3q + 2 + 2 = 3q + 3 + 1 = 3q(q + 1) + 1 = 3m + 1	No	3q + 2 + 4 = 3q + 6 = 3(q + 2) = 3m	Yes

From table, out of $n_1, n_2$ or $n_3$ one number is divisible by 3 when r = 0, 1, 2, are taken.

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Question 55 Marks

Show that the cube of a positive integer of the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

Answer

By Euclid's division algorithm,
a = 6q + r .....(i)
where a, q and r non-negative integers $0\leq\text{r}<6\text{ i.e., r}=0,1,2,3,4,5.$
cubing (i) both sides, we get
$(a)^3 = (6q + r)^3$
$\Rightarrow a^3 = (6q)^3 + (r)^3 + 3(6q)^2(r) + 3(6q)(r)^2$
$= 6^3q^3 + r^3 + 3 \times 6^2q^2r + 6 \times 3qr^2$
$\Rightarrow a^3 = 6[36q^3 + 18q^2r + 3qr^2] + r^3 .....(ii)$
When$ r = 0, then a^3 = 6[36q^3 + 18q^2 \times 0 + 3q0^2] + 0^3 [From (ii)]$
$\Rightarrow a^3 = 0[36q^3]$
$\Rightarrow a^3 = 6m$ is perfect cube for some value of m such that $m = 36q^2$
When $r = 1, a^3 =6[36q^3 + 18q^2 \times 1 + 3q1^2] + 1^3 [From (ii)]$
$= 6[36q^3 + 18q^2 + 3q] + 1$
$\Rightarrow a^3 = 6m + 1$ is perfect cube for some value of m such that $m = (36q^3 + 18q^2 + 3q)$
When$ r = 2, q^3 = 6[36q^3 + 18q^2 \times 2 + 3q \times 2^3 [from (ii)]$
$= 6[36q^3 + 36q^2 + 12q] + 6 + 2$
$= 6[36q^3 + 36q^2 + 12q + 1] + 2$
⇒ = 6m + 2 is perfect cube for some values of m such that $m = 36q^3 + 36q^2 + 12q + 1$
When $r = 3, a^3 = 6[36q^3 + 18q^2 \times 3 + 3q \times 3^2] + 3^3 [From (ii)]$
$\Rightarrow a3 = 6[36q^3 + 54q^2 + 27q] + 24 + 3$
$\Rightarrow a3 = 6[36q^3 + 54q^2 + 27q +4] +3$
$\Rightarrow a3 = 6m + 3$
So, (6m + 3) is perfect cube forpsecified value fo m such that $m = 36q^3 + 54q^2 + 27q + 4$
When r = 4, then eq. (ii) become
$a^3 = 6[36q^3 + 18q^2 (4) + 3q 4^2] + 4^3$
$= 6 [36q^3 + 72q^2 + 48q] + 60 + 4$
$= 6 [36q^3 + 72q^2 + 48q + 10] + 4$
$\Rightarrow a^3 = 6m + 4$
So, (6m + 4) is perfect cube for specified value of m such that $m = 36q^3 + 72q^2 + 48q + 10$
When r =5, eq (ii) become as
$a3 = 6[36q^3 + 18q^2(5) + 3q(5)^2] + (5)^3$
$=6 [36q^3 + 90q^2 + 75q] 120 + 5$
$= 6[36q^3 + 90q^2 + 75q + 20] + 5$
$\Rightarrow a^3 = 6m + 5$
⇒ (6m + 5) is perfect cube for specified value of $m = 36q^3 + 90q^2 + 75q + 20$
Hence, cubes of positive integers is of the form (6m + r), where m is a specified integer and r = 0, 1, 2, 3, 4, 5.

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Question 65 Marks

Show that the square of any positive integer is either of the form 4q or 4q + 1 for some integer q.

Answer

Let a be an arbitrary positive integer. Then, by, Euclid's division algorithm, corresponding to the positivde integers a and 4,
there exist non-negative integers m and r, such that
$\text{a}=4\text{m}+\text{r, where }0\leq\text{r}<4$
$\Rightarrow\ \text{r}^2=16\text{m}^2+\text{r}^2+8\text{mr}\ .....(\text{i})$
$\text{Where,}\ 0\leq\text{r}<4\ \big[\because (\text{a}+\text{b})^2=\text{a}^2+2\text{ab}+\text{b}^2\big]$
Case I:
When $r=0$, then putting $r=0$ in Eq. (i), we get $a^2=16 m^2=4\left(4 m^2\right)=4 q$ Where, $q=4 m^2$ is an integer.
Case II:
When $r=1$, then putting $r=1$ in Eq. (i), we get $a^2=16 m^2+1+8 m=4\left(4 m^2+2 m\right)+1=4 q+1$ Where, $q=\left(4 m^2+\right.$ 2 m ) is an integer.
Case III:
When $r=2$, then putting $r=2$ in Eq. (i), we get $a^2=16 m^2+4+16 m=4\left(4 m^2+4 m+1\right)=4 q$ Where, $q=\left(4 m^2+\right.$ $4 m+1)$ is an integer.
Case IV:
When $r=3$, then putting $r=3$ in Eq. (i), we get $a^2=16 m^2+9+24 m=16 m^2+24 m+8+1=4\left(4 m^2+6 m+2\right)+1$ $=4 q+1$ Where, $q=\left(4 m^2+6 m+2\right)$ is an integer. Hence, the square of any positive integer is either of the form $4 q$ or $4 q+1$ for some integer $m$.

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Question 75 Marks

Prove that one of any three consecutive positive integers must be divisible by 3.

Answer

Consider that if a number n, q and r are positive integers. When n is divided by 3 the quotient is a and remainder r. So, by Euclid's dividsion algorithm,
$\text{n}=3\text{q}+\text{r}\ (0\leq\text{r}<3)\ \text{or r}=0, 1, 2, 3$

At	$n_1$	Divisible by 3	$n_2 = n_1 + 3$	Divisible by 3	$n_3 = n_3 + 4$	Divisible by 3
r =0	3q + 0 = 3q	Yes	3q + 1	No	3q + 2	No
r = 1	3q + 1	No	3q + 2	No	3q + 3 =3(q + 1) =3m	Yes
r = 2	3q + 2	No	3q + 3 = 3(q + 1) = 3m	Yes	3q + 4 =3q + 3 + 1 = 3(q + 1) + 1 = 3m + 1	No

So, one of any three consecutive positive integers is divisible by 3.

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Question 85 Marks

Show that the square of any odd integer is of the form 4q + 1, for some integer q.

Answer

By Euclid's division algorithm, we have $a=b q+r$, where $0<r<4$
On putting $b=4$ in Eq. (i), we get
$a =4 q + r$, where $0 \leq r <4$ i.e., $r =0,1.2,3 \ldots \ldots$. (ii)
If $r =0$
$\Rightarrow a=4 q, 4 q$ is divisible by 2
$\Rightarrow 4 q$ is even.
If $r=1$
$\Rightarrow a=4 q+1,(4 q+1)$ is not divisible by 2 .
If $r =2$
$\Rightarrow a=4 q+2,2(4 q+1)$ is divisible by 2
$\Rightarrow 2(2 q+1)$ is even.
If $r =3$
$\Rightarrow a =4 q +3,(4 q+3)$ is not divisible by 2 .
So, for any positive integer $q,(4 q+1)$ and $(4 q+3)$ are odd integers.
Now, $a^2=(4 q+1)^2=16 q^2+1+8 q=4\left(4 q^2+2 q\right)+1$
$\left[\because(a+b)^2=a^2+2 a b+b^2\right]$
is a square which is of the form am $4 m+1$, where $m=\left(4 q^2+2 q\right)$ is an integer. and $a^2=(4 q+3)^2=16 q^2+9+24 q=4\left(4 q^2=6 q+2\right)+1$ is a square.
$\left[\because(a+b)^2=a^2+2 a b+b^3\right]$
Which is of the $4 m+1$, where $m=\left(4 q^2+6 q+2\right)$ is an integer.
Hence, for some integer $m$, the square of any odd integer is of the form $4 m+1$

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Question 95 Marks

Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Answer

Let a be an arbitrrary positive integer. Then, by Euclid's divisions Algorithm, corresponding to the positive integers a and 5,
there exist non-negative integers m and r such that
$\text{a}=5\text{m}+\text{r, where}\ 0\leq\text{r}<5$
$\Rightarrow\ \text{a}^2=(5\text{m}+\text{r}^5)=25\text{m}^2+\text{r}^2+10\text{mr}\ \ \big[\because\ (\text{a}+\text{b})^2=\text{a}^2+2\text{ab}+\text{b}^2\big]$
$\Rightarrow\ \text{a}^2=5(5\text{m}^2+2\text{mr})+\text{r}^2\ .....(\text{i})$
$\text{where,}\ 0\leq\text{r}< 5$
Case I:
When $r=0$, then putting $r=0$ in Eq. (i), we get $a^2=5\left(5 m^2\right)=5 q$ where, $q =5 m^2$ is an integer.
Case II:
When $r=1$, then putting $r=1$ is Eq. (i), we get $a^2=5\left(5 m^2+2 m\right)+1$ $\Rightarrow q=5 q+1$ where, $q=\left(5 m^2+2 m\right)$ is an integer.
Case III:
When $r=2$, then putting $r=2$ in Eq. (i) we get $a^2=5\left(5 m^2+4 m\right)+4=5 q+4$
Where, $q=\left(5 m^2+4 m\right)$ is an integer.
Case IV: When $r=3$, then putting $r=3$ in Eq. (i), we get $a^2=5\left(5 m^2+6 m\right)+9=5\left(5 m^2+6 m\right)+5+4=5\left(5 m^2+6 m+1\right)+4=5 q+4$ where, $q=\left(5 m^2+6 m+1\right)$ is an integer.
Case V:
When $r=4$, then putting $r=4$ in Eq. (i), we get $a^2=5\left(5 m^2+8 m\right)+16=5\left(5 m^2+8 m\right)+15+1$
$\Rightarrow a^2=5\left(5 m^2+8 m+3\right)+1=5 q+1$
where, $q=\left(5 m^2+8 m+3\right)$ is an integer.
Hence, the square of any positive integer cannot be of the from $5 q+2$ of $5 q+3$ for any integer $q$.

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Question 105 Marks

For any positive integer n, prove that $n^3 – n$ is divisible by 6.

Answer

Let $a = n^3 - n$
$\Rightarrow a = n(n^2 - 1)$
$= n (n - 1)(n + 1)$
$(n - 1), n, (n + 1$) are consecutive integers so out of three consecutive numbers at least one will be even. So, a is divisible by 2.
Sum of numbers $= (n - 1) + n + (n + 1)$
$= n - 1 + n + n + 1$
$= 3n$
Clearly, the sum of three consective numbers is divisible by 3, so any one of them must be divisible by 3.
So, out of n, (n - 1), (n + 1), one is divisible by 2 and one is divisible by 3 and
a = (n - 1) × n × (n + 1)
Hence, out of three factors of a, one is divisible by 2 and one is divisible by 3. So, a is divisible by 6 or $n^3 - n$ is divisible by 6.

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