Mix Examples - Real Numbers Questions in English

(A) To find the decimal expansion,we express the denominator in the form $2^n \cdot 5^m$.
The given number is $\frac{33}{2^2 \cdot 5^1}$.
To make the powers of $2$ and $5$ equal,we multiply the numerator and denominator by $5^{(2-1)} = 5^1$.
$\frac{33 \cdot 5}{2^2 \cdot 5^1 \cdot 5^1} = \frac{165}{2^2 \cdot 5^2} = \frac{165}{(2 \cdot 5)^2} = \frac{165}{10^2} = \frac{165}{100} = 1.65$.
The decimal expansion terminates after $2$ decimal places.

(B) Euclid's division lemma states that for any two positive integers $a$ and $b,$ there exist unique integers $q$ and $r$ such that $a = bq + r,$ where the remainder $r$ satisfies the condition $0 \leq r < b.$
This means that the remainder $r$ can be zero or greater than zero,but it must always be strictly less than the divisor $b.$

(C) An even integer is defined as any integer that is exactly divisible by $2$.
By definition,an even integer can be expressed as $2m$,where $m$ is any integer $(m \in \mathbb{Z})$.
If we substitute values for $m$:
If $m = 0$,$2m = 0$.
If $m = 1$,$2m = 2$.
If $m = 2$,$2m = 4$.
If $m = -1$,$2m = -2$.
Thus,every even integer is of the form $2m$.

(D) By definition,an integer is even if it is divisible by $2$,which can be expressed as $2q$ for some integer $q$.
An odd integer is an integer that is not divisible by $2$.
When we divide any odd integer by $2$,the remainder is always $1$.
According to Euclid's division lemma,any integer $a$ can be expressed as $a = bq + r$,where $0 \le r < b$.
For $b = 2$,we have $a = 2q + r$,where $r = 0$ or $r = 1$.
If $r = 0$,$a = 2q$ (even integer).
If $r = 1$,$a = 2q + 1$ (odd integer).
Therefore,every odd integer is of the form $2q + 1$ for some integer $q$.

(A) Let $a = n^{2} - 1$.
Case $I$: If $n$ is an even integer,let $n = 2k$ for some integer $k$.
Then $a = (2k)^{2} - 1 = 4k^{2} - 1$. This is always an odd number,so it cannot be divisible by $8$.
Case $II$: If $n$ is an odd integer,let $n = 2k + 1$ for some integer $k$.
Then $a = (2k + 1)^{2} - 1 = (4k^{2} + 4k + 1) - 1 = 4k^{2} + 4k = 4k(k + 1)$.
Since $k$ and $k + 1$ are consecutive integers,one of them must be even. Therefore,the product $k(k + 1)$ is divisible by $2$.
Let $k(k + 1) = 2m$ for some integer $m$.
Then $a = 4(2m) = 8m$.
Since $a$ is a multiple of $8$,$n^{2} - 1$ is divisible by $8$ whenever $n$ is an odd integer.

(B) To find the $HCF$ of $65$ and $117$ using Euclid's division algorithm:
$117 = 65 \times 1 + 52$
$65 = 52 \times 1 + 13$
$52 = 13 \times 4 + 0$
Therefore,the $HCF(65, 117) = 13$.
Given that the $HCF$ is expressible as $65m - 117$,we set up the equation:
$65m - 117 = 13$
$65m = 13 + 117$
$65m = 130$
$m = \frac{130}{65}$
$m = 2$

(C) Since $5$ and $8$ are the remainders when $70$ and $125$ are divided by the required number,respectively.
Subtracting these remainders from the given numbers,we get:
$70 - 5 = 65$
$125 - 8 = 117$
The required number is the $HCF$ of $65$ and $117$.
Using Euclid's division algorithm:
$117 = 65 \times 1 + 52$
$65 = 52 \times 1 + 13$
$52 = 13 \times 4 + 0$
Since the remainder is $0$,the $HCF$ is $13$.
Therefore,$13$ is the largest number that divides $70$ and $125$ leaving remainders $5$ and $8$ respectively.

(D) Given that,$a = x^3 y^2 = x \times x \times x \times y \times y$.
And $b = x y^3 = x \times y \times y \times y$.
The $HCF$ of two numbers is the product of the smallest power of each common prime factor involved in the numbers.
Here,the common prime factors are $x$ and $y$.
The smallest power of $x$ is $x^1$ and the smallest power of $y$ is $y^2$.
Therefore,$HCF(a, b) = x^1 \times y^2 = x y^2$.

(A) Given that,$p = a b^{2}$ and $q = a^{3} b$,where $a$ and $b$ are prime numbers.
To find the $LCM(p, q)$,we take the highest power of each prime factor present in the expressions.
The prime factors involved are $a$ and $b$.
The highest power of $a$ is $a^{3}$ (from $q$).
The highest power of $b$ is $b^{2}$ (from $p$).
Therefore,$LCM(p, q) = a^{3} \times b^{2} = a^{3} b^{2}$.

(B) Let $r$ be a non-zero rational number and $i$ be an irrational number.
Assume the product $p = r \times i$ is rational.
Then $p = \frac{a}{b}$ for some integers $a, b$ where $b \neq 0$.
Since $r$ is a non-zero rational,$r = \frac{p}{q}$ for some integers $p, q$ where $p, q \neq 0$.
Then $i = \frac{p}{r} = \frac{p}{q} \times \frac{b}{a} = \frac{pb}{qa}$.
Since $p, b, q, a$ are integers,$\frac{pb}{qa}$ is a rational number.
This contradicts the fact that $i$ is irrational.
Therefore,the product of a non-zero rational and an irrational number is always irrational.
For example,$\frac{3}{4} \times \sqrt{2} = \frac{3\sqrt{2}}{4}$,which is an irrational number.

(C) To find the least number divisible by all numbers from $1$ to $10$,we need to calculate the Least Common Multiple $(LCM)$ of these numbers.
Prime factorization of each number from $1$ to $10$:
$1 = 1$
$2 = 2^1$
$3 = 3^1$
$4 = 2^2$
$5 = 5^1$
$6 = 2^1 \times 3^1$
$7 = 7^1$
$8 = 2^3$
$9 = 3^2$
$10 = 2^1 \times 5^1$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM = 2^3 \times 3^2 \times 5^1 \times 7^1$
$LCM = 8 \times 9 \times 5 \times 7$
$LCM = 72 \times 35 = 2520$
Thus,the least number divisible by all numbers from $1$ to $10$ is $2520$.

(D) To determine the number of decimal places after which the rational number $\frac{14587}{1250}$ terminates,we express the denominator in its prime factorization form.
$1250 = 2^1 \times 5^4$.
The decimal expansion of a rational number $\frac{p}{q}$ terminates after $n$ decimal places,where $n$ is the maximum of the exponents of $2$ and $5$ in the prime factorization of the denominator $q$ (when the fraction is in its simplest form).
Here,the exponents are $1$ and $4$. The maximum exponent is $4$.
Alternatively,we can write:
$\frac{14587}{1250} = \frac{14587}{2^1 \times 5^4} = \frac{14587 \times 2^3}{2^1 \times 5^4 \times 2^3} = \frac{14587 \times 8}{2^4 \times 5^4} = \frac{116696}{10^4} = \frac{116696}{10000} = 11.6696$.
Since the result is $11.6696$,the decimal expansion terminates after four decimal places.

(B) No.
According to Euclid's division lemma,for any positive integer $a$ and a divisor $b=3$,there exist unique integers $q$ and $r$ such that:
$a = 3q + r$,where $0 \leq r < 3$.
Since $r$ must be an integer satisfying the condition $0 \leq r < 3$,the possible values for the remainder $r$ are $0, 1,$ and $2$.
Therefore,the statement that the remainders are only $0$ and $1$ is incorrect.

(NO) No,the number $6^{n}$ cannot end with the digit $5$.
For any number to end with the digit $5$,its prime factorization must contain the prime factor $5$.
The prime factorization of $6^{n}$ is given by $6^{n} = (2 \times 3)^{n} = 2^{n} \times 3^{n}$.
Since the only prime factors of $6^{n}$ are $2$ and $3$,and $5$ is not a prime factor,$6^{n}$ can never end with the digit $5$ for any natural number $n$.

(B) No,every positive integer cannot be of the form $4q + 2$.
According to Euclid's Division Lemma,for any positive integer $b$ and a positive integer $a = 4$,there exist unique integers $q$ and $r$ such that $b = 4q + r$,where $0 \le r < 4$.
This means the possible values for the remainder $r$ are $0, 1, 2,$ and $3$.
Therefore,any positive integer $b$ can be expressed in one of the following forms: $4q, 4q + 1, 4q + 2,$ or $4q + 3$.
Since a positive integer can also take the forms $4q, 4q + 1,$ or $4q + 3$,it is not restricted to only the form $4q + 2$.

(A) The statement is true.
Let the two consecutive positive integers be $n$ and $(n+1)$.
In any two consecutive integers,one must be even and the other must be odd.
An even number is always divisible by $2$.
Therefore,the product $n(n+1)$ will always be divisible by $2$ because at least one of the factors is an even number.

(A) Let the three consecutive positive integers be $n, (n + 1),$ and $(n + 2).$
In any set of two consecutive integers,one must be even (divisible by $2$). Therefore,in any set of three consecutive integers,at least one must be divisible by $2$.
In any set of three consecutive integers,one must be a multiple of $3$ (divisible by $3$).
Since the product is divisible by both $2$ and $3$,and $2$ and $3$ are coprime (their greatest common divisor is $1$),the product must be divisible by their product,which is $2 \times 3 = 6$.
Thus,the statement is true.

(N/A) No,the square of any positive integer cannot be of the form $3m + 2$.
By Euclid's division lemma,any positive integer $b$ can be expressed as $b = 3q + r$,where $0 \leq r < 3$. Thus,any positive integer is of the form $3k, 3k + 1$,or $3k + 2$.
Case $1$: If $b = 3k$,then $b^2 = (3k)^2 = 9k^2 = 3(3k^2) = 3m$,where $m = 3k^2$.
Case $2$: If $b = 3k + 1$,then $b^2 = (3k + 1)^2 = 9k^2 + 6k + 1 = 3(3k^2 + 2k) + 1 = 3m + 1$,where $m = 3k^2 + 2k$.
Case $3$: If $b = 3k + 2$,then $b^2 = (3k + 2)^2 = 9k^2 + 12k + 4 = 9k^2 + 12k + 3 + 1 = 3(3k^2 + 4k + 1) + 1 = 3m + 1$,where $m = 3k^2 + 4k + 1$.
In all cases,the square of a positive integer is either of the form $3m$ or $3m + 1$. Therefore,it cannot be of the form $3m + 2$.

(B) No. According to Euclid's Division Lemma,any positive integer can be expressed in the form $3q, 3q + 1,$ or $3q + 2$ for some integer $q$.
Let us examine the squares of these forms:
$1.$ If the integer is $3q$,then its square is $(3q)^2 = 9q^2 = 3(3q^2) = 3m$,where $m = 3q^2$.
$2.$ If the integer is $3q + 1$,then its square is $(3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1$,where $m = 3q^2 + 2q$.
$3.$ If the integer is $3q + 2$,then its square is $(3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$,where $m = 3q^2 + 4q + 1$.
Thus,the square of any positive integer is always of the form $3m$ or $3m + 1$. It can never be of the form $3m + 2$.

(A) The $HCF$ (Highest Common Factor) is the largest number that divides both given numbers.
Given that $525$ and $3000$ are divisible by $3, 5, 15, 25,$ and $75$,the largest among these is $75$.
To justify using Euclid's Division Lemma:
$3000 = 525 \times 5 + 375$
$525 = 375 \times 1 + 150$
$375 = 150 \times 2 + 75$
$150 = 75 \times 2 + 0$
Since the remainder is $0$ when the divisor is $75$,the $HCF(525, 3000) = 75$.

(N/A) We have,$3 \times 5 \times 7 + 7 = 105 + 7 = 112$.
Now,we find the prime factorization of $112$:
$112 = 2 \times 56 = 2 \times 2 \times 28 = 2 \times 2 \times 2 \times 14 = 2 \times 2 \times 2 \times 2 \times 7 = 2^4 \times 7$.
Since $112$ can be expressed as a product of prime factors $2$ and $7$,it has more than two factors (specifically,$1, 2, 4, 7, 8, 14, 16, 28, 56, 112$).
By definition,a number that has more than two factors is a composite number. Therefore,$3 \times 5 \times 7 + 7$ is a composite number.

(B) No,two numbers cannot have $18$ as their $HCF$ and $380$ as their $LCM$.
This is because the $HCF$ of any two numbers must always be a factor of their $LCM$.
In this case,we check if $380$ is divisible by $18$:
$380 / 18 = 21.11...$
Since $18$ is not a factor of $380$,such numbers cannot exist.

(A) First,simplify the fraction $\frac{987}{10500}$ by dividing both numerator and denominator by their greatest common divisor,which is $21$.
$\frac{987 \div 21}{10500 \div 21} = \frac{47}{500}$.
Now,find the prime factorization of the denominator $500$:
$500 = 5 \times 100 = 5 \times 10^2 = 5 \times (2 \times 5)^2 = 5 \times 2^2 \times 5^2 = 2^2 \times 5^3$.
According to the theorem,a rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^m \times 5^n$,where $m$ and $n$ are non-negative integers.
Since the denominator $500 = 2^2 \times 5^3$ is in the form $2^m \times 5^n$ (where $m=2$ and $n=3$),the decimal expansion of $\frac{987}{10500}$ is terminating.
To find the decimal value: $\frac{47}{500} = \frac{47 \times 2}{500 \times 2} = \frac{94}{1000} = 0.094$.

(A) The given number $327.7081$ is a terminating decimal number.
Since it is a terminating decimal,it represents a rational number whose denominator $q$,when expressed in the form $\frac{p}{q}$ (where $p$ and $q$ are coprime),must have prime factors of the form $2^{m} \times 5^{n}$,where $m$ and $n$ are non-negative integers.
We can write:
$327.7081 = \frac{3277081}{10000} = \frac{p}{q}$
Here,the denominator $q = 10000 = 10^{4}$.
Expanding the prime factors of $q$:
$q = (2 \times 5)^{4} = 2^{4} \times 5^{4}$.
Thus,the prime factors of $q$ are only $2$ and $5$.

(B) To check if two numbers are co-prime,their $HCF$ must be $1$. We use Euclid's division algorithm:
$396 = 231 \times 1 + 165$
$231 = 165 \times 1 + 66$
$165 = 66 \times 2 + 33$
$66 = 33 \times 2 + 0$
Since the $HCF$ is $33$ (which is not $1$),the numbers $231$ and $396$ are not co-prime.

(B) To find the $HCF$ of $847$ and $2160$ using Euclid's division algorithm,we perform the following steps:
$2160 = 847 \times 2 + 466$
$847 = 466 \times 1 + 381$
$466 = 381 \times 1 + 85$
$381 = 85 \times 4 + 41$
$85 = 41 \times 2 + 3$
$41 = 3 \times 13 + 2$
$3 = 2 \times 1 + 1$
$2 = 1 \times 2 + 0$
Since the remainder is $0$,the $HCF$ is the divisor at this stage,which is $1$.
Two numbers are said to be co-prime if their $HCF$ is $1$. Since the $HCF$ of $847$ and $2160$ is $1$,they are co-prime.

(N/A) Any positive odd integer can be represented in the form $2q+1$,where $q$ is a whole number.
Squaring this integer,we get:
$(2q+1)^2 = 4q^2 + 4q + 1$
$(2q+1)^2 = 4q(q+1) + 1$ $...(1)$
Since $q$ and $q+1$ are consecutive integers,their product $q(q+1)$ is always even. Therefore,we can write $q(q+1) = 2m$ for some whole number $m$.
Substituting this into equation $(1)$:
$(2q+1)^2 = 4(2m) + 1 = 8m + 1$.
Thus,the square of any odd positive integer is of the form $8m+1$ for some whole number $m$.

Let us assume that $\sqrt{2}+\sqrt{3}$ is rational. Let $\sqrt{2}+\sqrt{3} = a$,where $a$ is a rational number.
Then,$\sqrt{2} = a - \sqrt{3}$.
Squaring both sides,we get:
$(\sqrt{2})^2 = (a - \sqrt{3})^2$
$2 = a^2 + 3 - 2a\sqrt{3}$
Rearranging the terms to isolate $\sqrt{3}$:
$2a\sqrt{3} = a^2 + 3 - 2$
$2a\sqrt{3} = a^2 + 1$
$\sqrt{3} = \frac{a^2 + 1}{2a}$
Since $a$ is a rational number,$\frac{a^2 + 1}{2a}$ must also be a rational number. This implies that $\sqrt{3}$ is rational.
However,this contradicts the known fact that $\sqrt{3}$ is an irrational number. This contradiction has arisen because of our initial assumption that $\sqrt{2}+\sqrt{3}$ is rational.
Therefore,we conclude that $\sqrt{2}+\sqrt{3}$ is irrational.

(N/A) Let $a$ be an arbitrary positive integer. By Euclid's division algorithm,for integers $a$ and $4$,there exist non-negative integers $m$ and $r$ such that $a = 4m + r$,where $0 \leq r < 4$.
Squaring both sides,we get $a^2 = (4m + r)^2 = 16m^2 + 8mr + r^2 = 4(4m^2 + 2mr) + r^2$.
Case $I$: If $r = 0$,$a^2 = 4(4m^2) = 4q$,where $q = 4m^2$.
Case $II$: If $r = 1$,$a^2 = 4(4m^2 + 2m) + 1 = 4q + 1$,where $q = 4m^2 + 2m$.
Case $III$: If $r = 2$,$a^2 = 16m^2 + 16m + 4 = 4(4m^2 + 4m + 1) = 4q$,where $q = 4m^2 + 4m + 1$.
Case $IV$: If $r = 3$,$a^2 = 16m^2 + 24m + 9 = 16m^2 + 24m + 8 + 1 = 4(4m^2 + 6m + 2) + 1 = 4q + 1$,where $q = 4m^2 + 6m + 2$.
Thus,the square of any positive integer is of the form $4q$ or $4q + 1$.

(N/A) Let $a$ be any positive integer. By Euclid's division lemma,for $a$ and $b=4$,there exist unique integers $q$ and $r$ such that $a = 4q + r$,where $0 \leq r < 4$.
Taking the cube of both sides:
$a^3 = (4q + r)^3 = 64q^3 + 3(4q)^2r + 3(4q)r^2 + r^3$
$a^3 = 64q^3 + 48q^2r + 12qr^2 + r^3$
$a^3 = 4(16q^3 + 12q^2r + 3qr^2) + r^3$ ... $(i)$
Case $I$: If $r = 0$,then $a^3 = 4(16q^3) = 4m$,where $m = 16q^3$.
Case $II$: If $r = 1$,then $a^3 = 4(16q^3 + 12q^2 + 3q) + 1^3 = 4m + 1$,where $m = 16q^3 + 12q^2 + 3q$.
Case $III$: If $r = 2$,then $a^3 = 4(16q^3 + 12q^2(2) + 3q(2^2)) + 2^3 = 4(16q^3 + 24q^2 + 12q) + 8 = 4(16q^3 + 24q^2 + 12q + 2) = 4m$,where $m = 16q^3 + 24q^2 + 12q + 2$.
Case $IV$: If $r = 3$,then $a^3 = 4(16q^3 + 12q^2(3) + 3q(3^2)) + 3^3 = 4(16q^3 + 36q^2 + 27q) + 27 = 4(16q^3 + 36q^2 + 27q + 6) + 3 = 4m + 3$,where $m = 16q^3 + 36q^2 + 27q + 6$.
Thus,the cube of any positive integer is of the form $4m, 4m+1$ or $4m+3$.

(A) Let $a$ be an arbitrary positive integer.
By Euclid's division algorithm,for integers $a$ and $5$,there exist non-negative integers $m$ and $r$ such that $a = 5m + r$,where $0 \leq r < 5$.
Squaring both sides,we get $a^2 = (5m + r)^2 = 25m^2 + 10mr + r^2 = 5(5m^2 + 2mr) + r^2$.
Let $q' = 5m^2 + 2mr$,then $a^2 = 5q' + r^2$.
We test all possible values of $r \in \{0, 1, 2, 3, 4\}$:
Case $I$: If $r=0$,$a^2 = 5(5m^2) = 5q$,which is of the form $5q$.
Case $II$: If $r=1$,$a^2 = 5(5m^2 + 2m) + 1 = 5q + 1$,which is of the form $5q+1$.
Case $III$: If $r=2$,$a^2 = 5(5m^2 + 4m) + 4 = 5q + 4$,which is of the form $5q+4$.
Case $IV$: If $r=3$,$a^2 = 5(5m^2 + 6m) + 9 = 5(5m^2 + 6m + 1) + 4 = 5q + 4$,which is of the form $5q+4$.
Case $V$: If $r=4$,$a^2 = 5(5m^2 + 8m) + 16 = 5(5m^2 + 8m + 3) + 1 = 5q + 1$,which is of the form $5q+1$.
In all cases,$a^2$ is of the form $5q, 5q+1,$ or $5q+4$. Thus,it cannot be of the form $5q+2$ or $5q+3$.

(N/A) Let $a$ be an arbitrary positive integer. By Euclid's division algorithm,for positive integers $a$ and $6$,there exist non-negative integers $q$ and $r$ such that $a = 6q + r$,where $0 \leq r < 6$.
Squaring both sides,we get $a^2 = (6q + r)^2 = 36q^2 + 12qr + r^2 = 6(6q^2 + 2qr) + r^2$. Let this be equation $(i)$.
Case $I$: If $r = 0$,$a^2 = 6(6q^2) = 6m$,where $m = 6q^2$.
Case $II$: If $r = 1$,$a^2 = 6(6q^2 + 2q) + 1 = 6m + 1$,where $m = 6q^2 + 2q$.
Case $III$: If $r = 2$,$a^2 = 6(6q^2 + 4q) + 4 = 6m + 4$,where $m = 6q^2 + 4q$.
Case $IV$: If $r = 3$,$a^2 = 6(6q^2 + 6q) + 9 = 6(6q^2 + 6q + 1) + 3 = 6m + 3$,where $m = 6q^2 + 6q + 1$.
Case $V$: If $r = 4$,$a^2 = 6(6q^2 + 8q) + 16 = 6(6q^2 + 8q + 2) + 4 = 6m + 4$,where $m = 6q^2 + 8q + 2$.
Case $VI$: If $r = 5$,$a^2 = 6(6q^2 + 10q) + 25 = 6(6q^2 + 10q + 4) + 1 = 6m + 1$,where $m = 6q^2 + 10q + 4$.
Thus,the square of any positive integer can only be of the form $6m, 6m+1, 6m+3,$ or $6m+4$. Therefore,it cannot be of the form $6m+2$ or $6m+5$.

(N/A) Let $a$ be an odd integer. By Euclid's division lemma,for any positive integer $b=4$,we have $a = 4k + r$,where $0 \leq r < 4$.
Since $a$ is odd,$r$ can only be $1$ or $3$.
Case $1$: If $r = 1$,then $a = 4k + 1$.
$a^2 = (4k + 1)^2 = 16k^2 + 8k + 1 = 4(4k^2 + 2k) + 1$.
Let $q = 4k^2 + 2k$,which is an integer. Thus,$a^2 = 4q + 1$.
Case $2$: If $r = 3$,then $a = 4k + 3$.
$a^2 = (4k + 3)^2 = 16k^2 + 24k + 9 = 16k^2 + 24k + 8 + 1 = 4(4k^2 + 6k + 2) + 1$.
Let $q = 4k^2 + 6k + 2$,which is an integer. Thus,$a^2 = 4q + 1$.
In both cases,the square of an odd integer is of the form $4q + 1$ for some integer $q$.

(N/A) Let $n$ be an odd integer. Any odd integer can be represented in the form $n = 2k + 1$ for some integer $k$.
Substituting this into the expression $n^{2} - 1$:
$n^{2} - 1 = (2k + 1)^{2} - 1$
$= (4k^{2} + 4k + 1) - 1$
$= 4k^{2} + 4k$
$= 4k(k + 1)$
Since $k$ and $k + 1$ are consecutive integers,one of them must be even. Therefore,the product $k(k + 1)$ is divisible by $2$.
Let $k(k + 1) = 2m$ for some integer $m$.
Then,$n^{2} - 1 = 4(2m) = 8m$.
Since $n^{2} - 1$ is a multiple of $8$,it is divisible by $8$ for any odd integer $n$.

(N/A) Let $x = 2m + 1$ and $y = 2n + 1$ be any two odd positive integers,where $m$ and $n$ are non-negative integers.
Then,$x^{2} + y^{2} = (2m + 1)^{2} + (2n + 1)^{2}$.
Using the identity $(a + b)^{2} = a^{2} + 2ab + b^{2}$,we get:
$x^{2} + y^{2} = (4m^{2} + 4m + 1) + (4n^{2} + 4n + 1)$.
$x^{2} + y^{2} = 4m^{2} + 4m + 4n^{2} + 4n + 2$.
$x^{2} + y^{2} = 4(m^{2} + m + n^{2} + n) + 2$.
Since $x^{2} + y^{2} = 2[2(m^{2} + m + n^{2} + n) + 1]$,it is clearly an even number.
However,when divided by $4$,it leaves a remainder of $2$. Therefore,$x^{2} + y^{2}$ is not divisible by $4$.

(D) Let $a = 693$,$b = 567$,and $c = 441$.
By Euclid's division algorithm,$a = bq + r$,where $0 \le r < b$.
First,we find the $HCF$ of $693$ and $567$:
$693 = 567 \times 1 + 126$
$567 = 126 \times 4 + 63$
$126 = 63 \times 2 + 0$
Since the remainder is $0$,the $HCF(693, 567) = 63$.
Now,we find the $HCF$ of the result $(63)$ and the remaining number $(441)$:
$441 = 63 \times 7 + 0$
Since the remainder is $0$,the $HCF(63, 441) = 63$.
Therefore,the $HCF$ of $441, 567, 693$ is $63$.

(A) Since $1, 2$ and $3$ are the remainders of $1251, 9377$ and $15628$ respectively,we subtract these remainders from the numbers to get numbers that are exactly divisible by the required number.
$1251 - 1 = 1250$
$9377 - 2 = 9375$
$15628 - 3 = 15625$
The required number is the $HCF$ of $1250, 9375$ and $15625$.
Using Euclid's division algorithm,$a = bq + r$:
First,find $HCF$ of $15625$ and $9375$:
$15625 = 9375 \times 1 + 6250$
$9375 = 6250 \times 1 + 3125$
$6250 = 3125 \times 2 + 0$
So,$HCF(15625, 9375) = 3125$.
Now,find $HCF$ of $3125$ and $1250$:
$3125 = 1250 \times 2 + 625$
$1250 = 625 \times 2 + 0$
So,$HCF(3125, 1250) = 625$.
Thus,the largest number that divides $1251, 9377$ and $15628$ leaving remainders $1, 2$ and $3$ respectively is $625$.

(N/A) Let us assume that $\sqrt{3}+\sqrt{5}$ is a rational number.
Let $\sqrt{3}+\sqrt{5} = a$,where $a$ is a rational number.
Then,$\sqrt{3} = a - \sqrt{5}$.
Squaring both sides,we get:
$(\sqrt{3})^2 = (a - \sqrt{5})^2$
$3 = a^2 + 5 - 2a\sqrt{5}$ (using the identity $(x - y)^2 = x^2 + y^2 - 2xy$).
Rearranging the terms to isolate $\sqrt{5}$:
$2a\sqrt{5} = a^2 + 5 - 3$
$2a\sqrt{5} = a^2 + 2$
$\sqrt{5} = \frac{a^2 + 2}{2a}$.
Since $a$ is a rational number,$\frac{a^2 + 2}{2a}$ must also be a rational number.
This implies that $\sqrt{5}$ is a rational number,which contradicts the fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption is incorrect,and $\sqrt{3}+\sqrt{5}$ is an irrational number.

(N/A) If any number ends with the digit $0$ or $5$,it is always divisible by $5$.
If $12^{n}$ ends with the digit $0$ or $5$,it must be divisible by $5$.
This is possible only if the prime factorization of $12^{n}$ contains the prime number $5$.
Now,the prime factorization of $12$ is $12 = 2^{2} \times 3$.
Therefore,$12^{n} = (2^{2} \times 3)^{n} = 2^{2n} \times 3^{n}$.
By the Fundamental Theorem of Arithmetic,the prime factorization of $12^{n}$ is unique and contains only the prime factors $2$ and $3$.
Since $5$ is not a prime factor of $12^{n}$,$12^{n}$ cannot be divisible by $5$ for any natural number $n$.
Hence,$12^{n}$ cannot end with the digit $0$ or $5$ for any natural number $n$.

(D) To find the minimum distance that each should walk so that each can cover the same distance in complete steps,we need to calculate the Least Common Multiple $(LCM)$ of the step measurements: $40\, cm$,$42\, cm$,and $45\, cm$.
First,we find the prime factorization of each number:
$40 = 2^3 \times 5$
$42 = 2 \times 3 \times 7$
$45 = 3^2 \times 5$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM(40, 42, 45) = 2^3 \times 3^2 \times 5^1 \times 7^1$
$LCM = 8 \times 9 \times 5 \times 7$
$LCM = 72 \times 35$
$LCM = 2520$
Therefore,the minimum distance each should walk is $2520\, cm$.

(D) The denominator of the rational number $\frac{257}{5000}$ is $5000$.
First,we find the prime factorization of $5000$:
$5000 = 5 \times 1000 = 5 \times 10^3 = 5 \times (2 \times 5)^3 = 5 \times 2^3 \times 5^3 = 2^3 \times 5^4$.
This is in the form $2^m \times 5^n$,where $m = 3$ and $n = 4$ are non-negative integers.
To find the decimal expansion without actual division,we make the powers of $2$ and $5$ equal:
$\frac{257}{5000} = \frac{257}{2^3 \times 5^4}$.
To make the powers equal to $4$,we multiply the numerator and denominator by $2^1$:
$\frac{257 \times 2}{2^3 \times 5^4 \times 2^1} = \frac{514}{2^4 \times 5^4} = \frac{514}{(2 \times 5)^4} = \frac{514}{10^4}$.
$\frac{514}{10000} = 0.0514$.
Thus,the decimal expansion is $0.0514$.

(N/A) Let us assume that $\sqrt{p}+\sqrt{q}$ is a rational number.
Let $\sqrt{p}+\sqrt{q} = a$,where $a$ is a non-zero rational number.
Then,$\sqrt{q} = a - \sqrt{p}$.
Squaring both sides,we get:
$(\sqrt{q})^2 = (a - \sqrt{p})^2$
$q = a^2 + p - 2a\sqrt{p}$.
Rearranging the terms to isolate $\sqrt{p}$:
$2a\sqrt{p} = a^2 + p - q$
$\sqrt{p} = \frac{a^2 + p - q}{2a}$.
Since $a$ is a rational number,the expression $\frac{a^2 + p - q}{2a}$ must also be a rational number.
However,we know that $\sqrt{p}$ is an irrational number because $p$ is a prime number.
This leads to a contradiction because an irrational number cannot be equal to a rational number.
Therefore,our initial assumption is false,and $\sqrt{p} + \sqrt{q}$ must be an irrational number.

(N/A) We know that any positive integer can be expressed in the form $6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4,$ or $6m + 5$ for some integer $m$.
An odd positive integer must be of the form $6m + 1, 6m + 3,$ or $6m + 5$.
Now,we calculate the squares of these forms:
$1.$ $(6m + 1)^2 = 36m^2 + 12m + 1 = 6(6m^2 + 2m) + 1 = 6q + 1$,where $q = 6m^2 + 2m$ is an integer.
$2.$ $(6m + 3)^2 = 36m^2 + 36m + 9 = 36m^2 + 36m + 6 + 3 = 6(6m^2 + 6m + 1) + 3 = 6q + 3$,where $q = 6m^2 + 6m + 1$ is an integer.
$3.$ $(6m + 5)^2 = 36m^2 + 60m + 25 = 36m^2 + 60m + 24 + 1 = 6(6m^2 + 10m + 4) + 1 = 6q + 1$,where $q = 6m^2 + 10m + 4$ is an integer.
Thus,the square of any odd positive integer is always of the form $6q + 1$ or $6q + 3$.

(N/A) Let $a$ be an arbitrary positive integer. By Euclid's division algorithm,for any positive integer $a$ and $6$,there exist non-negative integers $q$ and $r$ such that $a = 6q + r$,where $0 \leq r < 6$.
Taking the cube of both sides:
$a^3 = (6q + r)^3 = 216q^3 + r^3 + 3(6q)(r)(6q + r)$
$a^3 = (216q^3 + 108q^2r + 18qr^2) + r^3 \quad \dots(i)$
Case $I$: If $r = 0$,$a^3 = 216q^3 = 6(36q^3) = 6m$,where $m = 36q^3$.
Case $II$: If $r = 1$,$a^3 = (216q^3 + 108q^2 + 18q) + 1 = 6(36q^3 + 18q^2 + 3q) + 1 = 6m + 1$.
Case $III$: If $r = 2$,$a^3 = (216q^3 + 216q^2 + 72q) + 8 = (216q^3 + 216q^2 + 72q + 6) + 2 = 6(36q^3 + 36q^2 + 12q + 1) + 2 = 6m + 2$.
Case $IV$: If $r = 3$,$a^3 = (216q^3 + 324q^2 + 162q) + 27 = (216q^3 + 324q^2 + 162q + 24) + 3 = 6(36q^3 + 54q^2 + 27q + 4) + 3 = 6m + 3$.
Case $V$: If $r = 4$,$a^3 = (216q^3 + 432q^2 + 288q) + 64 = (216q^3 + 432q^2 + 288q + 60) + 4 = 6(36q^3 + 72q^2 + 48q + 10) + 4 = 6m + 4$.
Case $VI$: If $r = 5$,$a^3 = (216q^3 + 540q^2 + 450q) + 125 = (216q^3 + 540q^2 + 450q + 120) + 5 = 6(36q^3 + 90q^2 + 75q + 20) + 5 = 6m + 5$.
Thus,the cube of any positive integer of the form $6q + r$ is of the form $6m + r$.

(N/A) Let $n$ be any positive integer. By Euclid's division lemma,$n$ can be expressed in the form $3q, 3q+1,$ or $3q+2$ for some integer $q \ge 0$.
Case $1$: If $n = 3q$,then $n$ is divisible by $3$. In this case,$n+2 = 3q+2$ and $n+4 = 3q+4 = 3(q+1)+1$,neither of which is divisible by $3$.
Case $2$: If $n = 3q+1$,then $n+2 = 3q+1+2 = 3q+3 = 3(q+1)$,which is divisible by $3$. In this case,$n = 3q+1$ and $n+4 = 3q+1+4 = 3q+5 = 3(q+1)+2$,neither of which is divisible by $3$.
Case $3$: If $n = 3q+2$,then $n+4 = 3q+2+4 = 3q+6 = 3(q+2)$,which is divisible by $3$. In this case,$n = 3q+2$ and $n+2 = 3q+2+2 = 3q+4 = 3(q+1)+1$,neither of which is divisible by $3$.
Thus,in all possible cases,exactly one of $n, n+2,$ or $n+4$ is divisible by $3$.

(N/A) Any three consecutive positive integers can be represented in the form $n, (n+1),$ and $(n+2)$,where $n$ is any positive integer $(n \in \mathbb{N})$.
By the Division Algorithm,any integer $n$ can be expressed in one of the forms $3q, 3q+1,$ or $3q+2$,where $q$ is a non-negative integer.
Case $1$: If $n = 3q$,then $n$ is divisible by $3$.
Case $2$: If $n = 3q + 1$,then $n + 2 = (3q + 1) + 2 = 3q + 3 = 3(q + 1)$,which is divisible by $3$.
Case $3$: If $n = 3q + 2$,then $n + 1 = (3q + 2) + 1 = 3q + 3 = 3(q + 1)$,which is divisible by $3$.
In all possible cases,one of the three consecutive integers $n, n+1,$ or $n+2$ is divisible by $3$.

(N/A) Let $a = n^{3} - n.$
Factorizing the expression,we get $a = n(n^{2} - 1).$
Using the identity $a^{2} - b^{2} = (a - b)(a + b),$ we have $a = (n - 1)n(n + 1).$
This expression represents the product of three consecutive integers.
In any set of three consecutive integers,at least one must be even (divisible by $2$) and exactly one must be divisible by $3.$
Since $2$ and $3$ are coprime,their product $2 \times 3 = 6$ must divide the product of these three consecutive integers.
Therefore,$(n - 1)n(n + 1)$ is always divisible by $6$ for any positive integer $n.$
Hence,$n^{3} - n$ is divisible by $6.$

(N/A) By Euclid's division lemma,any positive integer $n$ can be expressed in the form $5q, 5q+1, 5q+2, 5q+3,$ or $5q+4$,where $q$ is a non-negative integer.
We examine the divisibility of the set ${n, n+4, n+8, n+12, n+16}$ for each case:
$1$. If $n = 5q$,then $n$ is divisible by $5$.
$2$. If $n = 5q+1$,then $n+4 = 5q+1+4 = 5q+5 = 5(q+1)$,which is divisible by $5$.
$3$. If $n = 5q+2$,then $n+8 = 5q+2+8 = 5q+10 = 5(q+2)$,which is divisible by $5$.
$4$. If $n = 5q+3$,then $n+12 = 5q+3+12 = 5q+15 = 5(q+3)$,which is divisible by $5$.
$5$. If $n = 5q+4$,then $n+16 = 5q+4+16 = 5q+20 = 5(q+4)$,which is divisible by $5$.
In every case,exactly one of the five expressions is divisible by $5$.

(N/A) Let the three consecutive positive integers be $n$,$n+1$,and $n+2$.
We know that any positive integer $n$ can be expressed in the form $3m$,$3m+1$,or $3m+2$,where $m$ is a non-negative integer.
Case $1$: If $n = 3m$,then $n$ is divisible by $3$. Here,$n+1 = 3m+1$ and $n+2 = 3m+2$,which leave remainders $1$ and $2$ respectively when divided by $3$. Thus,only $n$ is divisible by $3$.
Case $2$: If $n = 3m+1$,then $n+1 = 3m+2$ and $n+2 = 3m+3 = 3(m+1)$. Here,only $n+2$ is divisible by $3$.
Case $3$: If $n = 3m+2$,then $n+1 = 3m+3 = 3(m+1)$ and $n+2 = 3m+4 = 3(m+1)+1$. Here,only $n+1$ is divisible by $3$.
Conclusion: In all possible cases,exactly one of the three consecutive integers $n$,$n+1$,and $n+2$ is divisible by $3$.

We know that any positive integer $n$ is of the form $2m$ or $2m+1$,where $m$ is a non-negative integer.
Case $1$: If $n = 2m$,then
$n^{2} - n = (2m)^{2} - (2m) = 4m^{2} - 2m = 2(2m^{2} - m)$.
Since $2(2m^{2} - m)$ is a multiple of $2$,$n^{2} - n$ is divisible by $2$.
Case $2$: If $n = 2m + 1$,then
$n^{2} - n = (2m + 1)^{2} - (2m + 1) = (4m^{2} + 4m + 1) - 2m - 1 = 4m^{2} + 2m = 2(2m^{2} + m)$.
Since $2(2m^{2} + m)$ is a multiple of $2$,$n^{2} - n$ is divisible by $2$.
Thus,in both cases,$n^{2} - n$ is divisible by $2$ for every positive integer $n$.