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