Here are the step-by-step workings: 1) 224-1 can be factorized as (212+1)(212-1) 2) 212-1 can be further factorized as (26+1)(26-1) 3) Therefore, 224-1 = (212+1)(26+1)(26-1) 4) The numbers between 60 and 70 that divide (26+1)(26-1) are 65 and 63. So the two numbers that exactly divide 224-1 and lie between 60 and 70 are 65 and 63.

2. Euclid’s Division Lemma
If there two positive integers a
and b, there exist unique integer q
and r satisfying
a=bq+r
0≤r<b.
Through this lemma the formation
of the fundamental theorem of
arithmetic took place and Euclid’s
division algorithm is based on this
lemma.

3. Fundamental Theorem of Arithmetic
In number theory, the fundamental
theorem of arithmetic, also called
the unique factorization theorem or
the unique-prime-factorization
theorem, states that
every integer greater than 1 either is
prime itself or is the product
of prime numbers, and that this
product is unique, up to the order of
the factors. It can be represented as
a product of primes.

4. Euclid’s Division Lemma:
Given positive integers a and b,there exist unique integers q
and r satisfying a = bq + r, 0 ≤r < b.
Euclid’s Division Algorithm is stated for only positive integers,
it can be extended for all integers except zero.
Euclid’s division Algorithm:- To obtain the HCF of two positive
integers say a and b, with a>b, follow the steps below:
Step I: Apply Euclid’s division lemma, to a and b, so we find
whole numbers, and r such that
a =bq+r, where 0≤ 𝑟 < 𝑏
Step II: If r=0, b is the HCF of a and b.
Step III: Continue the process till the remainder is zero. The
divisor at this stage will be the required HCF

7. FINDING HCF BY LONG DIVISION
METHOD

8. Use Euclid's division algorithm to find the HCF of:
(i) 135 and 225
To use Euclid's division algorithm, we apply Euclid's division
lemma to given numbers c and d, to find whole numbers q
and r such that
a=bq+r,0≤r<b
Here, a=225,b=135
225=135×1+90
Remainder is not equal to 0. Therefore, we apply the same
process again on 135 and 90
135=90×1+45
Remainder is not equal to 0 again. Therefore, we apply
same process again on 90 and 45.
90=45×2+0
Remainder is equal to 0.
Therefore, HCF of 135 and 225 is equal to 45 which is equal

9. Q. Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q
is some integer.
Solution:
Let a be any positive odd integer and b=6.
We can apply Euclid's division algorithm on a and b=6.
⇒a=6q+r
We know that value of 0≤r<6.
Therefore, all possible values of a are:
a=6q
=6q+1
a=6q+2
a=6q+3
a=6q+4
a=6q+5
We can ignore 6q, 6q+2 and 6q+4 because they are divisible by 2 which means they
are not positive odd integers.
Therefore, we are just left with 6q+1, 6q+3 and 6q+5.
Therefore, any positive odd integer is of the form (6q+1) or (6q+3)or (6q+5).

10. Q.Any army contingent of 616 members is to march behind an army band of 32
members in a parade. The two groups are to march in the same number of columns.
What is the maximum number of columns in which they can march?
Solutions:
HCF of 616 and 32 would be equal to maximum number of columns in which they
can march.
To find HCF, we can use Euclid's division algorithm, we apply Euclid's division lemma
to given numbers c and d, to find whole numbers q and r such that
c=dq+r,0≤r<d
Here, c=616,d=32
616=32×19+8
Remainder is not equal to 0. Therefore, we apply the same process again on 32 and
8.
32=8×4+0
Remainder is equal to 0.
Therefore, HCF of 616 and 32 is equal to 8 which is equal to value of d in the last
step.
It means that they can march in maximum of 8 columns.

11. Q.Use Euclid's division lemma to show that the cube of any
positive integer is of the form 9m, 9m+1 or 9m+8.
Let a be any positive integer and b=3.
According to Euclid's division lemma, we can say that
a=3q+r, 0≤r<3
Therefore, possible values of a are: 3q, 3q+1 or 3q+2
Now, we will find cube of each one of them.
(i) a³ = (3q)³=27
= 27q³= 3. 9m³=3q
q=9m

12. (ii)
(3q+1)³=27q³+1+3(3q)(3q+1)
=27q³+1+27q²+9q
=9(3q³ +3q²+3q)+1 let m= 9(3q³ +3q²+3q)
(3q+1)³=9m + 1
(iii)
(3q+2)³=(3q)³+2³+3(3q)(2)(3q+2)=27q³+8+54q²+36q
=27q³+54q²+36q+8=9(3q³+6q²+4q)+8
Let m=(3q³+6q²+4q) be any integer, then we get
(3q+2)³=9m+8

14. Q:Show that any positive odd integer is of the form
6q +1 or 6q + 3 or 6q + 5 where q is some integer
We know, a = bq + r where 0 ≤ r < b
By substituting b= 6 we get: a = 6q + r where[ r= 0,1,2,3,4,5]
If r = 0 , a= 6q , 6q is divisible by 6 ⇒6q is even.
If r = 1, a= 6q +1, 6q + 1 is not divisible by 2
If r = 2, a= 6q +2, 6q + 2 is divisible by 2 ⇒6q is even
If r = 3, a= 6q +3, 6q + 3 is not divisible by 2
If r = 4, a= 6q +4, 6q + 4 is divisible by 2 ⇒6q is even
If r = 5, a= 6q +5, 6q +5 is not divisible by 2
Hence the remaining integers i.e 6q + 1, 6q + 3
and 6q + 5 are odd

15. If n is an odd integer, then show that n2 – 1 is divisible by 8.

16. Q. Finds the H.C.F. of 65 and 117 and express it in the
form of 65m+117n.
The HCF 65 and 117 by Euclid division Lemma is 13
13 = 65 – 52
13 = 65 – (117 – 65 )
⇒13 = 65 × 2 – 117
⇒13 = 65 × 2 + 117 × (–1)
=65m + 117n
In the above relationship the H.C.F. of 65 and 117 is
of the form 65m + 117 n, where m = 2 and n = –1

17. Q. If the H C F of 657 and 963 is expressible in the
form of 657x + 963x - 15 find x.
Ans: Using Euclid’s Division Lemma a= bq+r ,
0 ≤ r < b
963=657×1+306
657=306×2+45
306=45×6+36
45=36×1+9 36=9×4+0 ∴ HCF (657, 963) = 9
Now 9 = 657x + 963× (-15)
657x=9+963×15
=9+14445 657x=14454
x=14454/657

18. Express the HCF of 48 and 18 as a linear combination.
A=bq+r, where o ≤ r < b
48=18x2+12
18=12x1+6
12=6x2+0 ∴ HCF (18,48) = 6
Now 6= 18- (48-36)
6= 18 +36 -48
6= 54+48
6= 18x3-48x1
6= 18x3+48x(-1)
i.e. 6= 18x +48y
6= 18×3 +48×(-1)
=18×3 +48×(-1) + 18×48-18×48
=18(3+48)+48(-1-18)
=18×51+48×(-19)
6=18x+48y
Hence, x and y are not unique.

19. Q. Prove that any three consecutive integers is divisible by 3.
Ans:
n,n+1,n+2 be three consecutive positive integers We know that n is of
the form 3q, 3q +1, 3q + 2
So we have the following cases
Case – I when n = 3q
In the this case, n is divisible by 3 but n + 1 and n + 2 are not divisible by
3
Case - II When n = 3q + 1
Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not
divisible by 3
Case – III When n = 3q +2
Sub n = 2 = 3q +1 +2 = 3(q +1) is divisible by 3. but n and n+1 are not
divisible by 3
Hence one of n, n + 1 and n + 2 is divisible by 3

20. Q.Find the largest possible positive integer that will divide 398, 436, and
542 leaving
remainder 7, 11, 15 respectively.
(Ans: 17)
Ans: The required number is the HCF of the numbers
Find the HCF of 391, 425 and 527 by Euclid’s algorithm
∴ HCF (425, 391) = 17
Now we have to find the HCF of 17 and 527
527 = 17 х 31 +0
∴ HCF (17,527) = 17
∴ HCF (391, 425 and 527) = 17

21. Q. Show that the square of any positive odd integer
is in the form 8k +1.
Let a is an odd positive integer
Therefore a=2m + 1
Squaring both sides we get
a² = 4m² + 4m + 1
=4m(m + 1) + 1
We know product of two consecutive numbers is
always even m(m+1)=2k
a²=4(2k)+1
a² = 8 k + 1
Hence proved

22. Q. Prove that no number of the type 4k+2
can be a perfect square.
Ans: Given the type of Number is 4k+2 =
2(2k+1)
2 is a factor of 4k+2 but 2k+1 is odd and
cannot have factor 2, so 4k+2 is not
divisible by 4, and therefore cannot be a
perfect square.

23. Q.Find the greatest number of 6 digits exactly
divisible by 24, 15 and 36.
Ans: LCM of 24, 15, 36
LCM = 3 × 2 × 2 × 2 × 3 × 5 = 360
Now, the greatest six digit number is 999999
Divide 999999 by 360 ∴ Q = 2777 , R = 279
∴ the required number = 999999 – 279 = 999720

24. Q.Find all positive integral values of n for which n²+96 is perfect
square.
Answer: Let n² + 96 = x²
⇒ x² – n² = 96
⇒ (x – n) (x + n) = 96
⇒ both x and n must be odd or both even
on these condition the cases are
x – n = 2, x + n = 48
x – n = 4, x + n = 24
x – n = 6, x + n = 16
x – n = 8, x + n = 12
and the solution of these equations can be given as
x = 25, n = 23
x = 14, n = 10
x = 11, n = 5
x = 10, n = 2
So, the required values of n are 23, 10, 5, and 2.

27. Q. Let a, b, c, and p be rational numbers such that p is not a perfect cube and
a +b1/3 +cp2/3 =0 then show that a=b=c

28. Q. Check whether 6n can end with the
digit 0 for any natural number n.
If any digit has last digit 0 that means It
is divisible by 10 And the factors of
10 = 2x5 So value 6n should be
divisible by 2 and 5 both 6 is divisible
by 2 but not divisible by 5 So it can not
end with 0.

29. Example
: Consider the numbers 4n, where n is a natural number. Check
whether
there is any value of n for which 4n ends with the digit zero.
Solution : If the number 4n, for any n, were to end with the digit zero,
then it would
be
divisible by 5. That is, the prime factorisation of 4n would contain the
prime 5. This
is
not possible because 4n = (2)2n; so the only prime in the factorisation
of 4n is 2.
So, the
uniqueness of the Fundamental Theorem of Arithmetic guarantees
that there are no
other primes in the factorisation of 4n. So, there is no natural number
n for which 4n
ends with the digit zero.

30. Theorem: Let p be a prime number. If p divides a², then p
divides a, where a is a positive integer.
Proof : Let the prime factorization of a be as follows :
a = p1.P2.P3………Pn
where p1,p2…. . . ., pn
are primes, not necessarily distinct.
a²=( p1.p2…..pn) ( p1.p2…..pn)
Now, it is given that p divides a².
Therefore, from the Fundamental Theorem of Arithmetic, it
follows that p is one of the prime factors of a²
Theorem: Let x be a rational number whose decimal expansion terminates.
Then x can be expressed in the form , p/q where p and q are co -prime, and the prime
factorization of q is of the form 2ⁿ5m , where n, m are non-negative
integers.

31. Show that 571 is a prime number.

32. Q.If d is the HCF of 30, 72, find the value of x & y satisfying d = 30x + 72y.
(Ans:5, -2 (Not unique)

33. Q.Can two numbers have 18 as their HCF and 380 as their LCM? Give
reasons.

34. What are the two numbers lying between 60 and 70, each of which
exactly divides 224-1?
224-1 =( 212+1)( 212-1)
=( 212+1)(26+1)(26-1)
==( 212+1) 65
Hence numbers are 65 ,63