67 + x \(\equiv \) 1 (mod 4)
67 + x  - 1  = 4n, for some integer n,
66 + x  = 4n
66 + x is a multiple of 4
Therefore, the least positive value of x must be 2, since 68 is the nearest multiple of 4 more than 66.

15 \(\equiv \) 3 (mod d) means 15 - 3 = kd, for some integer k,
12 = kd
gives d divides 12.
The divisors of 12 are 1,2,3,4,6,12. But d should be larger than 3 and so the possible values for d are 4, 6, 12.

The number 13824 can be factorized as
2^a x 3^b = 13824 = 2⁹ x 3³
a = 9 and b = 3.

Consider 2ⁿ \(\times\) 5^m
Since the product has 2 as a factor
2ⁿ \(\times\) 5^m is even
But if a number ends with the digit 5, then the number is an odd number
It is impossible.
For no value of m, 2ⁿ \(\times\) 5^m ends in 5.

for some natural number n,
4ⁿ = (2)ⁿ 
So 2 is a factor of 4ⁿ
By fundamental theorem of arithmetic, we know the factorization of 4n is unique.
Only factor of 4ⁿ is 2, even number of times.
But 4ⁿ always end with 4 or 6.
If n is odd then 4ⁿ end with 4.
If n is even, then 4ⁿ end with the digit 6.

The number 800 can be factorized as
800 = 2 x 2 x 2 x 2 x 2 x 5  x 5 = 2⁵ x 5²
Hence a^b x b^a = 2⁵ x 5²
This implies that a = 2 and b = 5  or a = 5 and b = 2

Yes, the given number is a composite number, because
7 x 5 x 3 x 2 + 3 = 3 (7 x 5 x 2 + 1) = 3 x 71
Since the given number can be factorized in terms of two primes, it is a composite number.

Since 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
2 is a factor of 6ⁿ. So, 6 n is always even.
But any number whose last digit is 5 is always odd.
Hence, 6ⁿ cannot end with the digit 5 

Value of the first box from bottom = 5 x 2 = 10
Value of n = 5 x 10 = 50
Value of the second box from bottom = 3 x 50 = 150
Value of m = 2 x 150 = 300
Thus, the required numbers are
m = 300, n = 50 

Let the two consecutive positive integers be n and n + 1
Here n + 1 > n
By Euclid's Division Algorithm, we have
n + 1 = n x q + 1 for some integer
Remainder = 1
Again applying Euclid' s Division Algorithm
n = 1 x (n) + 0
Here the remainder = 0
H.C.F. is the last divisor 1.
H.C.F.(n, n + 1) = 1
i.e., H.C.F. of two consecutive positive integers = 1
Two consecutive positive integers are always coprime.

Let the positive integer be 'n'
So n = 88 (p) + 61, where p be an integer
n = 88 (p) + (5 x 11 + 6)
n = 8 x 11 x p + 5 x 11 + 6
n = 11 (8p + 5) + 6
Dividing both the sides by 11, we get
\(\frac{n}{11}=(8 p+5)+\frac{6}{11}\)
When the same number n is divided by 11 the remainder will be 6.

All the integers 'a' must be either even or odd.
If it is even then a = 2q.
If it is odd then a = 2q + 1
Case 1:
lf a = 2q
a² = (2q)²
a² = 4q²,  remainder 0 when divided by 4.
Case 2:
If a = 2q+ 1
a² = (2q + 2)²
= 4q² + 4q + 1
= 4q ( q + 1) + 1
a² = 4m + 1 Where m = q (q + 1) is an integer
It is of the form bq + 1 where 1 is the remainder when divided by 4.
The square of any integer leaves the remainder either 0 or 1 when divided by 4.

Given the positive integers a, b and c divided by 13 leaves remainders 9, 7, 10 respectively
By Euclid's division Lemma we have
a = 13q₁+ 9
b = 13q₂ + 7
c = 13q₃ + 10
a + b + c = 13q₁ + 9 + 13q₂ + 7 + 13q₃ + 10
= 39[q₁ + q₂ + q₃] =13m
where m = q₁ + q₂ + q₃ +2 and m is positive integer. which is divisible by 13
a + b + c is divisible by 13.

Given number of flower pots = 532
By Euclid's Division Lemma
a = bq + r where 0 < r < lbl
532 = 21 x 25 + 7
where 0 < r < l21l
Number of completed rows = 25
Number of flower pots left over = 7

Let the required positive integer be a.
Given a is divided by 3 leaves remainder 2.
By Euclid's Division Lemma a and b are any positive integers, then there exist unique integer q and r such that
a = bq + r where 0 < r < lbl
Here a = 3q + r where 0 < r < l3l
r = 0, 1, 2
Considering r = 2
a = 3q + 2,where q > 0
i.e., a = 3q + 12, q = 0, 1, 2,3,...
a = 3 (0) + 2,3 (1) + 2, 3 (2) + 2,3 (3) +2, ...
a = 2, 5, 8, 11, ...
The required numbers are 2,5, 8, 11, ...........

Let x be any odd integer. Since any odd integer is one more than an even integer, we have x = 2k + 1 for some integer k.
x² = (2k + 1)²
=  4k² + 4k + 1
= 4k(k + 1) + 1
= 4q + 1, where q = k (k + 1) is some integer

a = -12,  b = 5
By Euclid’s division lemma
a = bq + r , Where 0 \(\le\) r < |b|
-12 = 5 x (-3) + 3 0 \(\le\) r  < |5|
Therefore , Quotient q = -3, Remainder r = 3

We see that 6 boxes are required to pack 30 cakes with 4 cakes left over. This distribution of cakes can be understood as follows:

34	=	5	x	6	+	4
Total number of cakes	=	Number of cakes in each box	x	Number of boxes	+	Number of cakes left over
\(\downarrow\)		\(\downarrow\)		\(\downarrow\)		\(\downarrow\)
Dividend      \(\alpha\)	=	Divisor    \(\beta\)	x	Quotient     q	+	Remainder      r