5 Marks Questions - Maths STD 10 Questions - Vidyadip

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19 questions · self-marked practice — reveal the answer and mark yourself.

Question 15 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 arbitrary 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
$a = 6q + r,$ where $0 \leq r < 6$
$\Rightarrow a^2 = (6q + r)^2 = 36q^2 + r^2 + 12qr$
[$\because$ $(a + b)^2 = a^2 + 2ab + b^2]$
$\Rightarrow a^2 = 6(6q^2 + 2qr) + r^2 ...(i)$
Where, $0 \leq r < 6$
Case I: Where $r=0$, then putting $r=0$ in Eq. (i), we get
$a^2=6\left(6 q^2\right)=6 m$
Where, $m=6 q^2$ is an integer.
Case II: When $r=1$, then putting $r=1$ in Eq. (i), we get
$a^2=6\left(6 q^2+2 q\right)+1=6 m+1$
Where, $m=\left(6 q^2+2 q\right)$ is an integer
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 = 3 in
Eq. (i), we get
$a^2 = 6(6q^2 + 6q) + 9$
$= 6(6q^2 + 6q) + 6 + 3$
$\Rightarrow a^2 = 6(6q^2 + 6q + 1) + 3 = 6m + 3$
Where, $m = (6q^2 + 6q + 1)$ is an integer.
Case V: When r = 4, then putting r = 4 in
Eq. (i), we get
$a^2 = 6(6q^2 + 8q) + 16$
$= 6(6q^2 + 8q) + 12 + 4$
$\Rightarrow a^2 = 6(6q^2 + 8q + 2) + 4 = 6m + 4$
Where, $n = (6q^2 + 8q + 2) + 4 = 6m + 4$
Case VI: When r = 5, then putting r = 5 in
Eq. (i), we get
$a^2 = 6(6q^2 + 10q) + 25$
$= 6(6q^2 + 10q) + 24 + 1$
$\Rightarrow a^2 = 6(6q^2 + 10q + 4) + 1 = 6m + 1$
Where,$ m = (6q^2 + 10q + 1)$ is an integer.
Hence, the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

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Question 25 Marks

Determine the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

Answer

To Find The number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
L.C.M of 8, 15 and 21.
$8 = 2^3$
$15 = 3 \times 5$
$21 = 3 \times 7$
L.C.M of 8, 15 and 21 = 23 × 3 × 5 × 7 = 840
When 110000 is divided by 840, the remainder is obtained as 800.
Now, 110000 - 800 = 109200 is divisible by each of 8, 15 and 21.
Also, 110000 + 40 = 110040 is divisible by each of 8, 15 and 21.
109200 and 110040 are greater than 100000.
Hence, 110040 is the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.

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Question 35 Marks

In a morning walk three persons step off together, their steps measure 80cm, 85cm and 90cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?

Answer

Given that three persons step off together, their steps measure 80cm, 85cm and 90cm.
The minimum distance each should walk is equal to
L.C.M of 80, 85 and 90.
Prime factors of 80, 85 and 90 are $80 = 2 \times 2 \times 2 \times 5 = 2^4 \times 5$
$85 = 5 \times 17 = 5 \times 17 90 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$
$L.C.M(80, 85, 90) = 2^4 \times 3^2 \times 5 \times 17 = 16 \times 9 \times 85 = 12240cm $
$$$\because$ $100cm = 1m$
$\therefore$ $12240cm = 122.40m = 122m 40cm$
So, the minimum distance each should walk is 122m 40cm.

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Question 45 Marks

Using Euclid’s division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

Answer

Since, 1,2 and 3 are the remainders of 1251, 9377 and 15628, respectively.
Thus, after subtracting these remainders from the numbers.
We have the numbers, 1251 - 1 = 1250, 9377 - 2 = 9375 and 15628 - 3 = 15625 which is divisible by the required number.
Now, required number = HCF of 1250, 9375 and 15625 [for the largest number]
By Euclid’s division algorithm,
a = bq + r .....(i)
[$\because$ dividend = divisor × quotient + remainder]
For largest number, put a = 15625 and b = 9375
15625 = 9375 × 1 + 6250
⇒ 9375 = 6250 × 1 + 3125
⇒ 6250 = 3125 × 2 + 0
$\therefore$ H.C.F (15625, 9375) = 3125
Now, we take c = 1250 and d = 3125, then again using Euclid's division algorithm,
d = cq + r [from Eq. (i)]
⇒ 3125 = 1250 × 2 + 625
⇒ 1250 = 625 × 2 + 0
$\therefore$ H.C.F (1250, 9375, 15625) = 625
Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.

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Question 55 Marks

Prove that if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.

Answer

To Prove: that if a positive integer is of the form 6q + 5 then it is of the form 3q + 2 for some integer q, but not conversely.
Proof: Let n = 6q + 5
Since any positive integer n is of the form of 3k or 3k + 1, 3k + 2
If q = 3k
Then, n = 6q + 5

⇒ n = 18k + 5(q = 3k)

⇒ n = 3(6k + 1) + 2

⇒ n = 3m + 2(where m = (6k + 1))

If q = 3k + 1
Then, n = (6q + 5)

⇒ n = (6(3k + 1) + 5)(q = 3k + 1)

⇒ n = 18k + 6 + 5

⇒ n = 18k + 11

⇒ n = 3(6k + 3) + 2

⇒ n = 3m + 2(where m = (6k + 3))

If q = 3k + 2
Then, n = (6q + 5)

⇒ n = (6(3k + 2) + 5)(q = 3k + 2)

⇒ n = 18k + 12 + 5

⇒ n = 18k + 17

⇒ n = 3(6k + 5) + 2

⇒ n = 3m + 2(where m = (6k + 5))

Consider here 8 which is the form 3q + 2 i.e. 3 × 2 + 2 but it can’t be written in the form 6q + 5. Hence the converse is not true.

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Question 65 Marks

105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?

Answer

We are given that, 105 goats, 140 donkeys and 175 cows. There is only one boat which will have to make many y trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. We need to tell the number of animals that went in each trip.
Given that
Number of goats = 105
Number of donkeys = 140
Number of cows = 175.
Therefore, the largest number of animals in 1 trip = H.C.F. of 105, 140 and 175.
First we consider 105 and 140.
By applying Euclid’s division lemma
140 = 105 × 1 + 35
105 = 35 × 3 + 0.
Therefore, H.C.F. of 105 and 140 = 35
Now, we consider 35 and 175.
By applying Euclid’s division lemma
175 = 35 × 5 + 0.
Therefore, H.C.F. of 105, 140 and 175 = 35
Hence, the number of animals went in each trip is 35.

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Question 75 Marks

If the HCF of 408 and 1032 is expressible in the form 1032m - 408 × 5, find m.

Answer

Given that two positive integers 408 and 1032 and 1032 > 408.
So, appliying Euclid’s division algorithem
1032 = 408 × 2 + 216 …..(i)
408 = 216 × 1 + 192 …..(ii)
216 = 192 × 1 + 24 …..(iii)
192 = 24 × 6 + 0 …..(iv)
So, HCF of 408 and 1032 is divison of eq. (iv) and remainder of eq. (iii) i.e. 24.
HCF (408, 1032) = 1032m - 408 × 5
⇒ 24 = 1032m - 2040
⇒ 1032m = 2040 + 24
⇒ 1032m = 2064
$\Rightarrow\text{m} = \frac{2064}{1032}$
⇒ m = 2
Thus, m = 2.

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Question 85 Marks

Show that one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Answer

Consider the numbers n, (n + 4), (n + 8), (n + 12) and (n + 16), where n is any positive integer. Suppose n = 5q + r, where 0 ≤ r < 5 n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q + 4 (By Euclid's division algorithm)Case I:
When n = 5q. n = 5q is divisible by 5. n + 4 = 5q + 4 is not divisible by 5. n + 8 = 5q + 5 + 5 + 3 = 5(q + 1) + 3 is not divisible by 5. n + 12 = 5q + 10 + 2 = 5(q + 2) + 2 is not divisible by 5. n + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5.Case II:
When n = 5q + 1. n = 5q + 1 is not divisible by 5. n + 4 = 5q + 1 + 4 = 5(q + 1) is not divisible by 5. n + 8 = 5q + 1 + 5 + 3 = 5(q + 1) + 4 is not divisible by 5. n + 12 = 5q + 1 + 12 = 5(q + 2) + 3 is not divisible by 5. n + 16 = 5q + 1 + 16 = 5(q + 3) + 2 is not divisible by 5.Case III:
When n = 5q + 2. n = 5q + 2 is not divisible by 5. n + 4 = 5q + 2 + 4 = 5(q + 1) + 1 is not divisible by 5. n + 8 = 5q + 2 + 8 = 5(q + 2) is not divisible by 5. n + 12 = 5q + 2 + 12 = 5(q + 2) + 4 is not divisible by 5. n + 16 = 5q + 2 + 16 = 5(q + 3) + 3 is not divisible by 5.Case IV:
When n = 5q + 3. n = 5q + 3 is not divisible by 5. n + 4 = 5q + 3 + 4 = 5(q + 1) + 2 is not divisible by 5. n + 8 = 5q + 3 + 8 = 5(q + 2) + 1 is not divisible by 5. n + 12 = 5q + 3 + 12 = 5(q + 3) is not divisible by 5. n + 16 = 5q + 3 + 16 = 5(q + 3) + 4 is not divisible by 5.Case V:
When n = 5q + 4. n = 5q + 4 is not divisible by 5. n + 4 = 5q + 4 + 4 = 5(q + 1) + 3 is not divisible by 5. n + 8 = 5q + 4 + 8 = 5(q + 2) + 2 is not divisible by 5. n + 12 = 5q + 4 + 12 = 5(q + 3) + 1 is not divisible by 5. n + 16 = 5q + 4 + 16 = 5(q + 4) is not divisible by 5. Hence, in each case, one and only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

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Question 95 Marks

Explain why 3 × 5 × 7 + 7 is a composite number.

Answer

We have, 3 × 5 × 7 + 7 = 105 + 7 = 112
Now, $112 = 2 \times 2 \times 2 \times 2 \times 7 = 2^4 \times 7$
So, it is the product of prime factors 2 and 7. i.e., it has more than two factors.
Hence, it is a composite number.

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Question 105 Marks

If a and b are two odd positive integers such that a > b, then prove that one of the two numbers $\frac{\text{a}+\text{b}}{2}$ and $\frac{\text{a}-\text{b}}{2}$ is odd and the other is even.

Answer

Let a = 2q + 3 and b = 2q + 1 be two positive odd integers such that a > b
Now, $\frac{\text{a}+\text{b}}{2}=\frac{2\text{q}+3+2\text{q}+1}{2}=\frac{4\text{q}+4}{2}=2\text{q}+2=$ an odd number
and $\frac{\text{a}-\text{b}}{2}=\frac{(2\text{q}+3)-(2\text{q}+1)}{2}=\frac{2\text{q}+3-2\text{q}-1}{2}=\frac{2}{2}=1=$ an odd number
Hence one of the two numbers $\frac{\text{a}+\text{b}}{2}$ and $\frac{\text{a}-\text{b}}{2}$ is odd and the other is even for any two positive odd integer.

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Question 115 Marks

Prove that the square of any positive integer is of the form 3m or, 3m + 1 but not of the from 3m + 2.

Answer

Let, n be any positive integer
We know that positive integer n is a in the form of 3q, (3q + 1) and (3q + 2), where q is some integer
Case 1: When n = 3q
Squaring of both sides
$\Rightarrow n^2 = (3q)^2$
$\Rightarrow n^2 = 9q^2$
$\Rightarrow n^2 = 3q(3q)$
$\Rightarrow n^2 = 3m$ [where m = q(3q)]
Case 2: When n = (3q + 1)
Squaring of both sides
$\Rightarrow n^2 = (3q + 1)^2$
$\Rightarrow n^2 = 9q^2 + 6q + 1$
$\Rightarrow n^2 = 3q(3q + 2) + 1$
$\Rightarrow n^2 = 3m + 1$ [Where m = q(3q + 2)]
Case 3: When n = (3q + 2)
Squaring of both sides
$\Rightarrow n^2 = (3q + 2)^2$
$\Rightarrow n2 = 9q^2 + 12q + 4$
$\Rightarrow n^2 = 9q^2 + 12q + 3 + 1$
$\Rightarrow n^2 = 3(3q^2 + 4q + 1) + 1$
$\Rightarrow n^2 = 3m + 1 [Where m = (3q^2 + 4q + 1)]$
Thus, them case 1, case 2 and case 3, it is proved that square of any positive odd integer is of the form 3m, (3m + 1) but not of the form (3m + 2).

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Question 125 Marks

The following table gives the number of boys of a particular age in a class of 40 students. Calculate the mean age of the students

Age (in yeare)	15	16	17	18	19	20
No. of students	3	8	10	10	5	4

Answer

Given;

Age (in years): $x_i$	15	16	17	18	19	20
No.of students: $f_i$	3	8	10	10	5	4

First of all prepare the frequency table in such a way that its first column consist of the values of the variate($x_i$) and the second column the corresponding frequencies($f_i$).
Thereafter multiply the frequency of each row with corresponding values of variable to obtain third column containing($f_ix_i)$.
Then, sum of all entries in the column second and denoted by $\sum\text{f}_\text{i}$ and in the third column to obtain$\sum\text{f}_\text{i}\text{x}_\text{i}$.

Age (in years): $x_i$	No. of students:$ f_i$	$f_ix_i$
15	3	45
16	8	128
17	10	170
18	10	180
19	5	95
20	4	80
	$\sum\text{f}_\text{i}=40$	$\sum\text{f}_\text{i}\text{x}_\text{i}=698$

We know that mean, $\overline{X}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}$ $\overline{X}=\frac{698}{40}$ $=17.45$ Hence, the mean age of the students $=17.45 \ \text{years}$

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Question 135 Marks

Write the denominator of the rational number $\frac{257}{5000}$ in the form $2^m \times 5^n$, where m, n are non-negative integers. Hence, write the decimal expansion, without actual division.

Answer

Denominator of the rational number $\frac{257}{5000}$ is 5000.
Now, factors of $5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = (2)^3 \times (5)^4$,
which is of the type $2^m \times 5^n,$ where m = 3 and n = 4 are non-negative integers.
$\therefore$ Rational number $=\frac{257}{5000}=\frac{257}{2^3\times5^4}\times\frac{2}{2}$
[Since, multiplying numerator and denominater by 2]
$=\frac{514}{2^4\times5^4}=\frac{514}{(10)^4}$
$=\frac{514}{10000}=0.0514$
Hence, which is the required decimal expansion of the rational $\frac{257}{5000}$ and it is also a terminating decimal number.

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Question 145 Marks

Prove that $\sqrt{5}+\sqrt{3}$ is an irrational number.

Answer

Let us assume that $\sqrt{5}+\sqrt{3}$ is a rational number.
$\therefore\ \sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-prime numbers.
$\sqrt{5}+\sqrt{3}=\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \sqrt{5}=\frac{\text{a}}{\text{b}}-\sqrt{3}$
Squaring both sides,
$\Rightarrow(\sqrt{5})^2=\Big(\frac{\text{a}}{\text{b}}-\sqrt{3}\Big)^2$
$\Rightarrow5=\Big(\frac{\text{a}}{\text{b}}\Big)^2+3-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow5-3=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow2=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{3}}{\text{b}}$
$\Rightarrow\frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-2$
$\Rightarrow\ \frac{2\text{a}\sqrt{3}}{\text{b}}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)$
$\Rightarrow\ \sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{\text{b}^2}\Big)\frac{\text{b}}{2\text{a}}$
$\Rightarrow\sqrt{3}=\Big(\frac{\text{a}^2-2\text{b}^2}{2\text{ab}}\Big)$
We know that $\sqrt{3}$ is an irrational number.
This contradicts our assumption that $\sqrt{5}+\sqrt{3}$ is a rational number.

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Question 155 Marks

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

Answer

By Eudid's division Algorithm
$a=b m+r \text {, where } 0 \leq r<b$
Put $b=4$
$a=4 m+r \text {, where } 0 \leq r<4$
If $r=0$, then $a=4 m$
If $r=1$, then $a=4 m+1$
If $r=2$, then $a=4 m+2$
If $r=3$, then $a=4 m+3$
Now, $(4 m)^2=16 m^2$
$=4 \times 4 m^2$
$=4 q$ where q is some integer
$(4m + 1)^2 = (4m)^2 + 2(4m)(1) + (1)^2$
$= 16m^2 + 8m + 1$
$= 4(4m^2 + 2m) + 1$ where $4m^2 + 2m = q$
$= 4q + 1$ where q is some integer
$(4m + 3)^2 = (4m)^2 + 2(4m)(3) + (3)^2$
$= 16m^2 + 24m + 9$
$= 16m^2 + 24m + 8 + 1$
$= 4(4m^2 + 6m + 2) + 1$
$= 4q + 1$, where q is some integre
Hence, the square of any positive in teger is of the form 4q or 4q + 1 for some integer m.

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Question 165 Marks

Show that any positive odd integer is of the form $6q + 1, or 6q + 3$, or $6q + 5$, where q is some integer.

Answer

Let $a$ be any positive integer and $b=6$. Then, by Euclid's algorithm, $a=6 q+r$ for some integer $q \geq 0$, and $r=0,1,2,3,4,5$ because $0 \leq r<6$.
Therefore, $a=6 q$ or $6 q+1$ or $6 q+2$ or $6 q+3$ or $6 q+4$ or $6 q+5$
Also, $6 q +1=2 \times 3 q +1=2 k _1+1$, where $k _1$ is a positive integer
$6 q+3=(6 q+2)+1=2(3 q+1)+1=2 k_2+1$, where $k_2$ is an integer
$6 q+5=(6 q+4)+1=2(3 q+2)+1=2 k_3+1$, where $k_3$ is an integer
Clearly, $6 q+1,6 q+3,6 q+5$ are of the form $2 k+1$, where $k$ is an integer.
Therefore, $6 q+1,6 q+3,6 q+5$ are not exactly divisible by 2 . Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form $6 q+1$, or $6 q+3$, or $6 q+5$

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Question 175 Marks

If p, q are prime positive integers, prove that $\sqrt{\text{p}}+\sqrt{\text{qp + q}}$ is an irrational number.

Answer

Let us asssume that $\sqrt{\text{p}}+\sqrt{\text{q}}$ is rational.
Then there exist positive co-primes a and b such that
$\sqrt{\text{p}}+\sqrt{\text{q}}=\frac{\text{a}}{\text{b}}$
$\sqrt{\text{p}}=\frac{\text{a}}{\text{b}}-\sqrt{\text{q}}$
$(\sqrt{\text{p}})^2=\Big(\frac{\text{a}}{\text{b}}-\sqrt{\text{q}}\Big)^2$
$\text{p}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}+\text{q}$
$\text{p}-\text{q}=\Big(\frac{\text{a}}{\text{b}}\Big)^2-\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\Big(\frac{\text{a}}{\text{b}}\Big)^2-(\text{p}-\text{q})=\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{\text{b}^2}=\frac{2\text{a}\sqrt{\text{q}}}{\text{b}}$
$\bigg(\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{\text{b}^2}\bigg)\Big(\frac{\text{b}}{2\text{a}}\Big)=\sqrt{\text{q}}$
$\sqrt{\text{q}}=\bigg(\frac{\text{a}^2-\text{b}^2(\text{p}-\text{q})}{2\text{ab}}\bigg)$
Here we see that $\sqrt{\text{q}}$ is rational number which is a contradiction as we know that $\sqrt{\text{q}}$ is an irrational number
Hence $\sqrt{\text{q}}+\sqrt{\text{q}}$ is irrational

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Question 185 Marks

Prove that the product of three consecutive positive integer is divisible by 6.

Answer

Let, n be any positive integer. Since any positive is of the form 6q or 6q + 1 or, 6q + 2 or, 6q + 3 or 6q + 4 or 6q + 5.
If n - 6q, then

n(n + 1)(n + 2) = [(6q)(6q + 1)(6q + 2)]

= 6[q(6q + 1)(6q + 2)]

= 6m, which is divisible by 6

If n - 6q + 1, then

n(n + 1)(n + 2) = (6q + 1)(6q + 2)(6q + 3)

= 6[(6q + 1)(3q + 1)(2q + 1)]

= 6m, which is divisible by 6

If n - 6q + 2, then

n(n + 1)(n + 2) = (6q + 2)(6q + 3)(6q + 4)

= 6[(3q + 1)(2q + 1)(6q + 4)]

= 6m, which is divisible by 6

If n - 6q + 3, then

n(n + 1)(n + 2) = (6q + 3)(6q + 4)(6q + 5)

= 6[(3q + 1)(3q + 2)(6q + 5)]

= 6m, which is divisible by 6

If n - 6q + 4, then

n(n + 1)(n + 2) = (6q + 4)(6q + 5)(6q + 6)

= 6[(6q + 4)(6q + 5)(q + 1)]

= 6m, which is divisible by 6

If n - 6q + 5, then

n(n + 1)(n + 2) = (6q + 5)(6q + 6)(6q + 7)

= 6[(6q + 5)(q + 1)(6q + 7)]

= 6m, which is divisible by 6

Hence, the product of three consecutive positive integer is divisible by 6.

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Question 195 Marks

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

Answer

Let a be an arbitrary 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.
$a = 6q + r,$ where, $0 \leq r < 6$
$\Rightarrow a^3 = (6q + r)^3 = 216q^3 + r^3 + 3.6q.r(6q + r)$
[$\because$ $(a + b)^3 = a^3 + b^3 + 3ab(a + b)]$
$\Rightarrow a^3 = (216q^3 + 108q^2r + 18qr^2) + r^3 ...(i)$
Where, 0 ≤ r < 6
Case I: When r = 0, then putting r = 0 in
Eq. (i), we get
$a^3 = 216q^3 = 6(36q^3) = 6m$
Where, $m = 36q^3$ is an integer.
Case II: Where r = 1, then putting r = 1 in
Eq. (i), we get
$a^3 = (216q^3 + 108q^3 + 18q) + 1 $
$= 6(36q^3 + 18q^3 + 3q) + 1$
$a^3 = 6m + 1,$
Where $m = (36q^3 + 18q^3 + 3q)$ is an integer.
Case III: When r = 2, then putting r = 2 in
Eq. (i), we get
$a^3 = (216q^3 + 216q^2 + 72q) + 8$
$a^3 = (216q^3 + 216q^2 + 72q + 6) + 2$
$\Rightarrow a^3 = 6(36q^3 + 36q^2 + 12q + 1) + 2= 6m + 2$
Where, $m = (36q^3 + 36q^2 + 12q + 1)$ is an integer.
Case IV: When r = 3, then putting r = 3 in
Eq. (i), we get
$a^3 = (216q^3 + 324q^2 + 162q) + 27$
$= (216q^3 + 324q^2 + 162q + 24) + 3$
$= 6(36q^3 + 54q^2 + 27q + 4) + 3 = 6m + 3$
Where, $m = (36q^3 + 64q^2 + 27q + 4)$ is an integer.
Case V: When r = 4, then putting r = 4 in
Eq. (i), we get
$a^3 = (216q^3 + 432q^2 + 288q) + 64$
$a^3 = 6(36q^3 + 72q^2 + 48q) + 60 + 4$
$a^3 = 6(36q^3 + 72q^2 + 48q + 10)$ is an integer.
Case VI: When r = 5, then putting r = 5 in
Eq. (i), we get
$a^3 = (216q^3 + 540q^2 + 450q) + 125$
$\Rightarrow a^3 = (216q^3 + 540q^2 + 450q) + 120 + 5$
$\Rightarrow a^3 = 6(36q^3 + 90q^2 + 75q + 20) + 5$
$\Rightarrow a^3 = 6m + 5$
Where, $m = (36q^3 + 90q^2 + 75q + 20)$ is an integer.
Hence, 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 forms 6m, 6m + 1, 6m + 3, 6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.

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