Hence, we can say that every integer which is a multiple of 2 must be an even integer, while 1 added to every integer which is multiplied by 2 is an odd integer.

Therefore, let us conclude that,

for an integer ‘q’, every odd integer must be of the form

(2 × q)+1 = 2q+1.

Hence, option (D) is the correct answer.

Question - 3 : -
n² – 1 is divisible by 8, if n is
(A) an integer (B) anatural number
(C) an odd integer(D) an even integer

Answer - 3 : -

Solution:

Explanation:

Let x = n² – 1

In the above equation, n can be either even orodd.

Let us assume that n= even.

So, when n = even i.e., n = 2k, where k is aninteger,

We get,

⇒ x = (2k)²-1

⇒ x = 4k² – 1

At k = -1, x = 4(-1)² – 1 = 4– 1 = 3, is not divisible by 8.

At k = 0, x = 4(0)² – 1 = 0 –1 = -1, is not divisible by 8

Let us assume that n= odd:

So, when n = odd i.e., n = 2k + 1, where k isan integer,

We get,

⇒ x = 2k + 1

⇒ x = (2k+1)² – 1

⇒ x = 4k² + 4k + 1 – 1

⇒ x = 4k² + 4k

⇒ x = 4k(k+1)

At k = -1, x = 4(-1)(-1+1) = 0 which isdivisible by 8.

At k = 0, x = 4(0)(0+1) = 0 which is divisibleby 8 .

At k = 1, x = 4(1)(1+1) = 8 which is divisibleby 8.

From the above two observation, we canconclude that, if n is odd, n²-1 is divisible by 8.

Hence, option (C) is thecorrect answer.

Question - 4 : -
If the HCF of 65 and 117 is expressible in theform 65m – 117, then the value of m is
(A) 4 (B) 2
(C) 1 (D) 3

Answer - 4 : -

Solution:

(B) 2

Explanation:

Let us find the HCF of 65 and 117,

117 = 1×65 + 52

65 = 1× 52 + 13

52 = 4 ×13 + 0

Hence, we get the HCF of 65 and 117 = 13.

According to the question,

65m – 117 = 13

65m = 117+13 = 130

∴ m =130/65 = 2

Hence, option (B) is thecorrect answer.

Question - 5 : -
The largest numberwhich divides 70 and 125, leaving remainders 5 and 8, respectively, is
(A) 13 (B) 65
(C) 875 (D) 1750

Answer - 5 : -

Solution:

(A) 13

Explanation:

According to the question,

We have to find the largest number whichdivides 70 and 125, leaving remainders 5 and 8.

This can be also written as,

To find the largest number which exactlydivides (70 – 5), and (125 – 8)

The largest number that divides 65 and 117 isalso the Highest Common Factor of 65 and 117

Therefore, the required number is the HCF of65 and 117

Factors of 65 = 1, 5, 13, 65

Factors of 117 = 1, 3, 9, 13, 39, 117

Common Factors = 1, 13

Highest Common factor (HCF) = 13

i.e., the largest number which divides 70 and125, leaving remainders 5 and 8, respectively = 13

Hence, option (A) is thecorrect answer.

Question - 6 : - Use Euclid’s division algorithm to find the HCF of:

Answer - 6 : -

i. 135 and 225

ii. 196 and 38220

iii. 867 and 225

Solutions:

Q i. 135 and 225

i. 135 and 225

As you can see, from the question 225 isgreater than 135. Therefore, by Euclid’s division algorithm, we have,

225 = 135 × 1 + 90

Now, remainder 90 ≠ 0, thus again usingdivision lemma for 90, we get,

135 = 90 × 1 + 45

Again, 45 ≠ 0, repeating the above step for45, we get,

90 = 45 × 2 + 0

The remainder is now zero, so our method stopshere. Since, in the last step, the divisor is 45, therefore, HCF (225,135) =HCF (135, 90) = HCF (90, 45) = 45.

Hence, the HCF of 225 and 135 is 45.

Q ii. 196 and 38220

ii. 196 and 38220

In this given question, 38220>196,therefore the by applying Euclid’s division algorithm and taking 38220 asdivisor, we get,

38220 = 196 × 195 + 0

We have already got the remainder as 0 here.Therefore, HCF(196, 38220) = 196.

Hence, the HCF of 196 and 38220 is 196.

Q iii. 867 and 225

iii. 867 and 225

As we know, 867 is greater than 225. Let usapply now Euclid’s division algorithm on 867, to get,

867 = 225 × 3 + 102

Remainder 102 ≠ 0, therefore taking 225 asdivisor and applying the division lemma method, we get,

225 = 102 × 2 + 51

Again, 51 ≠ 0. Now 102 is the new divisor, sorepeating the same step we get,

102 = 51 × 2 + 0

The remainder is now zero, so our procedurestops here. Since, in the last step, the divisor is 51, therefore, HCF(867,225) = HCF(225,102) = HCF(102,51) = 51.

Hence, the HCF of 867 and 225 is 51.

Question - 7 : -
Show that any positive odd integer is of theform 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Answer - 7 : -

Question - 8 : -
An army contingent of 616 members is to marchbehind an army band of 32 members in a parade. The two groups are to march inthe same number of columns. What is the maximum number of columns in which theycan march?

Answer - 8 : -

Solution:

Given,

Number of army contingent members=616

Number of army band members = 32

By Using Euclid’s algorithm to find their HCF,we get,

Since, 616>32, therefore,

616 = 32 × 19 + 8

Since, 8 ≠ 0, therefore, taking 32 as newdivisor, we have,

32 = 8 × 4 + 0

Now we have got remainder as 0, therefore, HCF(616, 32) = 8.

Hence, the maximum number of columns in whichthey can march is 8.

Question - 9 : -
Use Euclid’s division lemma to show that thesquare of any positive integer is either of the form 3m or 3m + 1 for someinteger m.

Answer - 9 : -

Question - 10 : -
Use Euclid’s division lemma to show that thecube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Answer - 10 : -

Solution:

Let x be any positive integer and y = 3.

By Euclid’s division algorithm

Now

x = 3q+r, where q≥0 and r = 0, 1, 2, as r ≥ 0and r < 3.

Then, putting the value of r

We get,

x = 3q or x = 3q + 1 or x = 3q + 2

Now, by taking the cube of all the three aboveexpressions.

Case (i): Whenr = 0

Then,

x²= (3q)³ = 27q³=9(3q³)= 9m; where m = 3q³

Case (ii): Whenr = 1

Then,

x³ = (3q+1)³ =(3q)³+1³+3×3q×1(3q+1) = 27q³+1+27q²+9q

Taking 9 as common above factor

We get,

x³= 9(3q³+3q²+q)+1

Putting = m

We get,

Putting (3q³+3q²⁺q) = m,we get ,

x³ = 9m+1

Case (iii): When r = 2

Then,

x³ = (3q+2)³

=(3q)³+2³+3×3q×2(3q+2)

=27q³+54q²+36q+8

Taking 9 as common above factor

We get,

x³=9(3q³+6q²+4q)+8

Putting (3q³+6q²+4q) = m

We get ,

x³ = 9m+8

Therefore, from all the three cases explainedabove, it is proved that the cube of any positive integer is of the form 9m, 9m+ 1 or 9m + 8.

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