Mix Examples - Real Numbers Questions in English

$(1)$ For $k=1$,$10^{1}+1=11$. By Euclid's Division Lemma,$11 = 11 \times 1 + 0$. Thus,the quotient is $1$ and the remainder is $0$.
$(2)$ For $k=2$,$10^{2}+1=101$. By Euclid's Division Lemma,$101 = 11 \times 9 + 2$. Thus,the quotient is $9$ and the remainder is $2$.
$(3)$ For $k=3$,$10^{3}+1=1001$. By Euclid's Division Lemma,$1001 = 11 \times 91 + 0$. Thus,the quotient is $91$ and the remainder is $0$.
$(4)$ For $k=4$,$10^{4}+1=10001$. By Euclid's Division Lemma,$10001 = 11 \times 909 + 2$. Thus,the quotient is $909$ and the remainder is $2$.
$(5)$ For $k=5$,$10^{5}+1=100001$. By Euclid's Division Lemma,$100001 = 11 \times 9091 + 0$. Thus,the quotient is $9091$ and the remainder is $0$.

Any positive integer $n$ 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,$n+2 = 3q+2$ (remainder $2$) and $n+4 = 3q+4 = 3(q+1)+1$ (remainder $1$). Thus,only $n$ 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 $3$. In this case,$n = 3q+1$ (remainder $1$) and $n+4 = 3q+5 = 3(q+1)+2$ (remainder $2$). Thus,only $n+2$ 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,$n = 3q+2$ (remainder $2$) and $n+2 = 3q+4 = 3(q+1)+1$ (remainder $1$). Thus,only $n+4$ is divisible by $3$.
Conclusion: In all possible cases,exactly one of the numbers $n, n+2, n+4$ is divisible by $3$.

(N/A) Let the two consecutive positive integers be $n$ and $n+1$,where $n$ is a positive integer.
Their product is $P = n(n+1) = n^2 + n$.
We can consider two cases for $n$:
Case $1$: If $n$ is even,then $n = 2k$ for some integer $k$. Thus,$P = 2k(2k+1) = 2(2k^2 + k)$,which is clearly divisible by $2$.
Case $2$: If $n$ is odd,then $n = 2k+1$ for some integer $k$. Thus,$P = (2k+1)(2k+1+1) = (2k+1)(2k+2) = 2(2k+1)(k+1)$,which is also divisible by $2$.
Since in both cases the product is divisible by $2$,the product of any two consecutive positive integers is always divisible by $2$.

(N/A) Let $a$ be any positive odd integer. By Euclid's division lemma,for any positive integer $a$ and $b=6$,there exist unique integers $q$ and $r$ such that $a = 6q + r$,where $0 \le r < 6$.
Since $r$ can be $0, 1, 2, 3, 4, 5$,the possible forms of $a$ are $6q, 6q+1, 6q+2, 6q+3, 6q+4,$ and $6q+5$.
If $a = 6q, 6q+2,$ or $6q+4$,then $a$ is even because these are divisible by $2$ (e.g.,$6q = 2(3q)$).
Since $a$ is an odd integer,it cannot be of the form $6q, 6q+2,$ or $6q+4$.
Therefore,any positive odd integer must be of the form $6q+1, 6q+3,$ or $6q+5$ for some integer $q \ge 0$. Replacing $q$ with $m$,we get the forms $6m+1, 6m+3,$ or $6m+5$.

(N/A) Let $a$ be any integer. According to Euclid's division lemma,for any integer $a$ and divisor $b=3$,there exist unique integers $q$ and $r$ such that $a = 3q + r$,where $0 \le r < 3$.
Thus,the possible values for $r$ are $0, 1, 2$.
Case $1$: If $r = 0$,then $a = 3q$. Squaring both sides,$a^2 = (3q)^2 = 9q^2 = 3(3q^2)$. Let $m = 3q^2$,then $a^2 = 3m$.
Case $2$: If $r = 1$,then $a = 3q + 1$. Squaring both sides,$a^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1$. Let $m = 3q^2 + 2q$,then $a^2 = 3m + 1$.
Case $3$: If $r = 2$,then $a = 3q + 2$. Squaring both sides,$a^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1$. Let $m = 3q^2 + 4q + 1$,then $a^2 = 3m + 1$.
In all cases,the square of an integer is of the form $3m$ or $3m+1$.

(A) Let $a$ be any positive integer. By Euclid's division lemma,we can express $a$ as $a = 3q + r$,where $r \in \{0, 1, 2\}$ and $q \ge 0$.
Case $1$: If $r = 0$,then $a = 3q$. Cubing both sides,$a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m$,where $m = 3q^3$.
Case $2$: If $r = 1$,then $a = 3q + 1$. Cubing both sides,$a^3 = (3q + 1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 = 9m + 1$,where $m = 3q^3 + 3q^2 + q$.
Case $3$: If $r = 2$,then $a = 3q + 2$. Cubing both sides,$a^3 = (3q + 2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8$,where $m = 3q^3 + 6q^2 + 4q$.
Thus,the cube of any positive integer is of the form $9m$,$9m+1$,or $9m+8$.

Since $a$ and $b$ are odd positive integers,we can represent them as $a = 2m + 1$ and $b = 2n + 1$ for some non-negative integers $m$ and $n$ where $m > n$.
Now,consider the sum: $\frac{a+b}{2} = \frac{(2m+1) + (2n+1)}{2} = \frac{2m + 2n + 2}{2} = m + n + 1$.
Next,consider the difference: $\frac{a-b}{2} = \frac{(2m+1) - (2n+1)}{2} = \frac{2m - 2n}{2} = m - n$.
Let $S = m + n + 1$ and $D = m - n$.
Consider the difference between these two results: $S - D = (m + n + 1) - (m - n) = 2n + 1$.
Since $2n + 1$ is an odd number,the difference between $S$ and $D$ is odd.
If the difference between two integers is odd,one must be even and the other must be odd.
Therefore,one of $\frac{a+b}{2}$ and $\frac{a-b}{2}$ is odd and the other is even.

Let $a$ be an odd positive integer. By Euclid's division lemma,any positive integer $a$ can be expressed as $a = 4q + r$,where $r \in \{0, 1, 2, 3\}$.
Since $a$ is odd,$r$ must be $1$ or $3$.
Case $1$: If $a = 4q + 1$,then $a^2 = (4q + 1)^2 = 16q^2 + 8q + 1 = 8(2q^2 + q) + 1$. Let $m = 2q^2 + q$,then $a^2 = 8m + 1$.
Case $2$: If $a = 4q + 3$,then $a^2 = (4q + 3)^2 = 16q^2 + 24q + 9 = 16q^2 + 24q + 8 + 1 = 8(2q^2 + 3q + 1) + 1$. Let $m = 2q^2 + 3q + 1$,then $a^2 = 8m + 1$.
In both cases,the square of an odd positive integer is of the form $8m + 1$.

(A) Given that the positive integer is of the form $6m + 5$,where $m$ is an integer.
We can rewrite the expression $6m + 5$ as $6m + 3 + 2$.
Factoring out $3$ from the first two terms,we get $3(2m + 1) + 2$.
Let $n = 2m + 1$. Since $m$ is an integer,$n$ is also an integer.
Substituting $n$ into the expression,we get $3n + 2$.
Thus,any integer of the form $6m + 5$ can be expressed in the form $3n + 2$.

(N/A) We can factorize the expression $n^{3}-n$ as follows:
$n^{3}-n = n(n^{2}-1) = n(n-1)(n+1) = (n-1)n(n+1)$.
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 the product contains a factor of $2$ and a factor of $3$,the entire product must be divisible by $2 \times 3 = 6$.
Therefore,$n^{3}-n$ is divisible by $6$ for any positive integer $n$.

(N/A) Let $a$ and $b$ be two odd positive integers. Any odd positive integer can be expressed in the form $2n + 1$ or $2n - 1$ for some integer $n$.
Let $a = 2m + 1$ and $b = 2n + 1$ for some integers $m$ and $n$.
Then,$a^{2} = (2m + 1)^{2} = 4m^{2} + 4m + 1 = 4(m^{2} + m) + 1$.
Similarly,$b^{2} = (2n + 1)^{2} = 4n^{2} + 4n + 1 = 4(n^{2} + n) + 1$.
Now,$a^{2} + b^{2} = [4(m^{2} + m) + 1] + [4(n^{2} + n) + 1] = 4(m^{2} + m + n^{2} + n) + 2$.
Let $k = m^{2} + m + n^{2} + n$. Then $a^{2} + b^{2} = 4k + 2$.
Since $a^{2} + b^{2} = 2(2k + 1)$,it is clearly an even number.
However,when $4k + 2$ is divided by $4$,the remainder is $2$. Therefore,$a^{2} + b^{2}$ is not divisible by $4$.

(B) Here,$210 > 55$.
Applying Euclid's division lemma,$a = bq + r$:
$210 = 55 \times 3 + 45$
$55 = 45 \times 1 + 10$
$45 = 10 \times 4 + 5$
$10 = 5 \times 2 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.(210, 55) = 5$.

(A) Let $d$ be the number of items in each stack.
Since each stack must contain the same number of items,$d$ must be a common divisor of $420$ and $130$.
To occupy the least area of the tray,the number of stacks must be minimized,which implies that the number of items in each stack $(d)$ must be the greatest common divisor $(GCD)$ of $420$ and $130$.
Using Euclid's division algorithm:
$420 = 130 \times 3 + 30$
$130 = 30 \times 4 + 10$
$30 = 10 \times 3 + 0$
The last non-zero divisor is $10$.
Therefore,$GCD(420, 130) = 10$.
Thus,the maximum number of burfis that can be placed in each stack is $10$.

(A) To ensure that each stack has the same height and the number of stacks is minimized (which minimizes the surface area),we need to find the Greatest Common Divisor $(GCD)$ of the number of English books $(56)$ and Gujarati books $(72)$.
Step $1$: Find the prime factorization of $56$ and $72$.
$56 = 2^3 \times 7$
$72 = 2^3 \times 3^2$
Step $2$: The $GCD$ is the product of the lowest powers of common prime factors.
$GCD(56, 72) = 2^3 = 8$.
Therefore,the maximum number of books that can be placed in each stack is $8$.

(A) To find the $g.c.d.$ of $12576$ and $4052$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $12576 > 4052$,we apply Euclid's division lemma to $12576$ and $4052$:
$12576 = 4052 \times 3 + 420$
Step $2$: Since the remainder $420 \neq 0$,we apply the lemma to $4052$ and $420$:
$4052 = 420 \times 9 + 272$
Step $3$: Since the remainder $272 \neq 0$,we apply the lemma to $420$ and $272$:
$420 = 272 \times 1 + 148$
Step $4$: Since the remainder $148 \neq 0$,we apply the lemma to $272$ and $148$:
$272 = 148 \times 1 + 124$
Step $5$: Since the remainder $124 \neq 0$,we apply the lemma to $148$ and $124$:
$148 = 124 \times 1 + 24$
Step $6$: Since the remainder $24 \neq 0$,we apply the lemma to $124$ and $24$:
$124 = 24 \times 5 + 4$
Step $7$: Since the remainder $4 \neq 0$,we apply the lemma to $24$ and $4$:
$24 = 4 \times 6 + 0$
Since the remainder is now $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $12576$ and $4052$ is $4$.

(B) To find the $g.c.d.$ of $81$ and $237$ using Euclid's division algorithm,we follow these steps:
Step $1$: Apply Euclid's division lemma to $237$ and $81$ $(237 > 81)$:
$237 = 81 \times 2 + 75$
Step $2$: Apply the lemma to $81$ and $75$:
$81 = 75 \times 1 + 6$
Step $3$: Apply the lemma to $75$ and $6$:
$75 = 6 \times 12 + 3$
Step $4$: Apply the lemma to $6$ and $3$:
$6 = 3 \times 2 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $81$ and $237$ is $3$.

(C) To find the $g.c.d.$ of $117$ and $65$ using Euclid's division algorithm,we follow these steps:
Step $1$: Since $117 > 65$,we apply Euclid's division lemma to $117$ and $65$:
$117 = 65 \times 1 + 52$
Step $2$: Since the remainder $52 \neq 0$,we apply the lemma to $65$ and $52$:
$65 = 52 \times 1 + 13$
Step $3$: Since the remainder $13 \neq 0$,we apply the lemma to $52$ and $13$:
$52 = 13 \times 4 + 0$
Since the remainder is now $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $117$ and $65$ is $13$.

(D) To find the $g.c.d.$ of $240$ and $6552$ using Euclid's division algorithm,we follow these steps:
Step $1$: Apply Euclid's division lemma to $6552$ and $240$:
$6552 = 240 \times 27 + 72$
Step $2$: Apply the lemma to $240$ and $72$:
$240 = 72 \times 3 + 24$
Step $3$: Apply the lemma to $72$ and $24$:
$72 = 24 \times 3 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $240$ and $6552$ is $24$.

(A) To find the $g.c.d.$ of $155$ and $1385$ using Euclid's division algorithm,we follow these steps:
Step $1$: Apply Euclid's division lemma to $1385$ and $155$:
$1385 = 155 \times 8 + 145$
Step $2$: Apply the lemma to $155$ and $145$:
$155 = 145 \times 1 + 10$
Step $3$: Apply the lemma to $145$ and $10$:
$145 = 10 \times 14 + 5$
Step $4$: Apply the lemma to $10$ and $5$:
$10 = 5 \times 2 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $155$ and $1385$ is $5$.

(B) To find the $g.c.d.$ of $75$ and $243$ using Euclid's division algorithm,we apply the lemma $a = bq + r$ where $0 \le r < b$.
Step $1$: Divide $243$ by $75$.
$243 = 75 \times 3 + 18$
Step $2$: Divide $75$ by $18$.
$75 = 18 \times 4 + 3$
Step $3$: Divide $18$ by $3$.
$18 = 3 \times 6 + 0$
Since the remainder is $0$,the divisor at this stage is the $g.c.d.$
Therefore,the $g.c.d.$ of $75$ and $243$ is $3$.

(C) To find the largest number that divides $245$ and $1029$ leaving a remainder of $5$ in each case,we first subtract the remainder from both numbers:
$245 - 5 = 240$
$1029 - 5 = 1024$
Now,we need to find the Highest Common Factor $(HCF)$ of $240$ and $1024$.
Prime factorization of $240 = 2^4 \times 3 \times 5$
Prime factorization of $1024 = 2^{10}$
The $HCF$ is the product of the lowest power of common prime factors:
$HCF(240, 1024) = 2^4 = 16$.
Therefore,the largest number is $16$.

(D) To find the largest number that divides $2053$ and $967$ leaving remainders $5$ and $7$ respectively,we first subtract the remainders from the given numbers.
$2053 - 5 = 2048$
$967 - 7 = 960$
Now,we need to find the Highest Common Factor $(HCF)$ of $2048$ and $960$.
Using the prime factorization method:
$2048 = 2^{11}$
$960 = 2^6 \times 3 \times 5$
The $HCF$ is the product of the lowest powers of common prime factors:
$HCF = 2^6 = 64$
Therefore,the largest number is $64$.

(A) Given numbers are $a = 96$ and $b = 404$.
First,find the prime factorization of $96$ and $404$:
$96 = 2^5 \times 3$
$404 = 2^2 \times 101$
The $g.c.d.(96, 404)$ is the product of the smallest power of each common prime factor:
$g.c.d.(96, 404) = 2^2 = 4$.
Using the formula $g.c.d.(a, b) \times l.c.m.(a, b) = a \times b$:
$4 \times l.c.m.(96, 404) = 96 \times 404$
$l.c.m.(96, 404) = \frac{96 \times 404}{4}$
$l.c.m.(96, 404) = 96 \times 101 = 9696$.

(B) Given the property: $g.c.d. (a, b) \times l.c.m. (a, b) = a \times b$.
First,find the prime factorization of $26$ and $91$:
$26 = 2 \times 13$
$91 = 7 \times 13$
Therefore,$g.c.d. (26, 91) = 13$.
Using the formula: $13 \times l.c.m. (26, 91) = 26 \times 91$.
$l.c.m. (26, 91) = \frac{26 \times 91}{13}$.
$l.c.m. (26, 91) = 2 \times 91 = 182$.

(B) Given expression: $7 \times 11 \times 13 + 13$
Taking $13$ as a common factor:
$= 13 \times (7 \times 11 + 1)$
$= 13 \times (77 + 1)$
$= 13 \times 78$
$= 13 \times (13 \times 6)$
$= 2 \times 3 \times 13^2 = 1014$
$A$ composite number is a positive integer that has at least one divisor other than $1$ and itself.
Since the expression $1014$ can be expressed as a product of prime factors $(2, 3, 13)$,it is a composite number.

(N/A) If $6^{n}$ ends with the digit $0$,then it must be divisible by $10$,which means it must be divisible by both $2$ and $5$.
The prime factorization of $6^{n}$ is $(2 \times 3)^{n} = 2^{n} \times 3^{n}$.
According to the Fundamental Theorem of Arithmetic,the prime factorization of any number is unique. The prime factors of $6^{n}$ are only $2$ and $3$.
Since $5$ is not a prime factor of $6^{n}$,$6^{n}$ is not divisible by $5$.
Therefore,$6^{n}$ cannot be divisible by $10$,which implies that $6^{n}$ cannot end with the digit $0$ for any natural number $n \in N$.

(N/A) Assume,to the contrary,that $\sqrt{3}$ is a rational number.
Then,there exist coprime positive integers $a$ and $b$ such that $\sqrt{3} = \frac{a}{b}$,where $gcd(a, b) = 1$.
Squaring both sides,we get $3 = \frac{a^2}{b^2}$,which implies $a^2 = 3b^2$ .......... $(1)$.
Since $3$ divides $a^2$,it follows that $3$ must also divide $a$ (by the Fundamental Theorem of Arithmetic).
Let $a = 3k$ for some integer $k$.
Substituting this into equation $(1)$,we get $(3k)^2 = 3b^2$,which simplifies to $9k^2 = 3b^2$,or $b^2 = 3k^2$.
This implies that $3$ divides $b^2$,and consequently,$3$ must divide $b$.
Thus,$3$ is a common factor of both $a$ and $b$,which contradicts our initial assumption that $gcd(a, b) = 1$.
Therefore,our assumption that $\sqrt{3}$ is rational is false.
Hence,$\sqrt{3}$ is an irrational number.

(N/A) Assume,for the sake of contradiction,that $\sqrt[3]{6}$ is a rational number.
Let $\sqrt[3]{6} = \frac{a}{b}$,where $a, b \in N$ and $g.c.d.(a, b) = 1$.
Since $1 < 6 < 8$,we have $\sqrt[3]{1} < \sqrt[3]{6} < \sqrt[3]{8}$,which implies $1 < \frac{a}{b} < 2$.
This means $b > 1$,because if $b = 1$,then $\frac{a}{b}$ would be an integer,but there is no integer between $1$ and $2$.
Cubing both sides,we get $6 = \frac{a^3}{b^3}$,which implies $6b^3 = a^3$.
Since $g.c.d.(a, b) = 1$,it follows that $g.c.d.(a^3, b^3) = 1$.
From $6b^3 = a^3$,we see that $b^3$ must divide $a^3$. Since $g.c.d.(a^3, b^3) = 1$,this is only possible if $b^3 = 1$,which means $b = 1$.
However,we already established that $b > 1$. This is a contradiction.
Therefore,our assumption that $\sqrt[3]{6}$ is rational is false.
Hence,$\sqrt[3]{6}$ is an irrational number.

(N/A) First,rationalize the denominator: $\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.
Assume,for the sake of contradiction,that $\frac{\sqrt{2}}{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $\frac{\sqrt{2}}{2} = \frac{a}{b}$.
This implies $\sqrt{2} = \frac{2a}{b}$.
Since $a$ and $b$ are integers,$\frac{2a}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the well-known fact that $\sqrt{2}$ is an irrational number.
Therefore,our assumption that $\frac{1}{\sqrt{2}}$ is rational must be false.
Hence,$\frac{1}{\sqrt{2}}$ is an irrational number.

(N/A) Assume that $3+2 \sqrt{5}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3+2 \sqrt{5} = \frac{a}{b}$.
Subtracting $3$ from both sides,we get $2 \sqrt{5} = \frac{a}{b} - 3 = \frac{a-3b}{b}$.
Dividing by $2$,we get $\sqrt{5} = \frac{a-3b}{2b}$.
Since $a$ and $b$ are integers,$\frac{a-3b}{2b}$ is a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the fact that $\sqrt{5}$ is an irrational number.
Therefore,our assumption that $3+2 \sqrt{5}$ is rational is incorrect.
Hence,$3+2 \sqrt{5}$ is an irrational number.

(A) Let $d$ be the number of books in each stack.
Since all stacks must have the same height and books are of the same thickness,$d$ must be a common divisor of $96$,$240$,and $336$.
To minimize the number of stacks,we need to maximize the number of books per stack,$d$. Thus,$d = \text{HCF}(96, 240, 336)$.
Prime factorization:
$96 = 2^5 \times 3$
$240 = 2^4 \times 3 \times 5$
$336 = 2^4 \times 3 \times 7$
$\text{HCF}(96, 240, 336) = 2^4 \times 3 = 16 \times 3 = 48$.
So,each stack contains $48$ books.
Number of stacks of English books $= \frac{96}{48} = 2$.
Number of stacks of Hindi books $= \frac{240}{48} = 5$.
Number of stacks of Mathematics books $= \frac{336}{48} = 7$.
Therefore,the number of stacks are $2, 5,$ and $7$ respectively.

(A) To express $156$ as a product of primes,we perform prime factorization:
$156 = 2 \times 78$
$78 = 2 \times 39$
$39 = 3 \times 13$
Combining these factors,we get: $156 = 2 \times 2 \times 3 \times 13 = 2^{2} \times 3 \times 13$.
Therefore,the correct option is $A$.

(B) To express $96$ as a product of prime factors,we perform prime factorization:
$96 = 2 \times 48$
$96 = 2 \times 2 \times 24$
$96 = 2 \times 2 \times 2 \times 12$
$96 = 2 \times 2 \times 2 \times 2 \times 6$
$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3$
Thus,$96 = 2^{5} \times 3$.

(C) To express $404$ as a product of primes,we perform prime factorization:
$404 = 2 \times 202$
$202 = 2 \times 101$
Since $101$ is a prime number,the prime factorization of $404$ is $2 \times 2 \times 101$,which can be written as $2^{2} \times 101$.

(D) To express $5005$ as a product of prime factors,we perform prime factorization:
$5005$ is divisible by $5$: $5005 \div 5 = 1001$.
Now,we factorize $1001$. It is not divisible by $2, 3,$ or $5$. Checking for $7$: $1001 \div 7 = 143$.
Next,we factorize $143$. It is not divisible by $7$. Checking for $11$: $143 \div 11 = 13$.
Since $13$ is a prime number,the prime factorization is $5 \times 7 \times 11 \times 13$.

(A) To express $7429$ as a product of prime factors,we perform prime factorization:
Step $1$: Check for divisibility by small primes. $7429$ is not divisible by $2, 3, 5, 7, 11,$ or $13$.
Step $2$: Test $17$: $7429 \div 17 = 437$.
Step $3$: Test $437$ for divisibility by $19$: $437 \div 19 = 23$.
Step $4$: Since $23$ is a prime number,the prime factorization is $17 \times 19 \times 23$.

(N/A) Step $1$: Prime factorization of $96$:
$96 = 2^5 \times 3^1$
Step $2$: Prime factorization of $404$:
$404 = 2^2 \times 101^1$
Step $3$: The $\text{g.c.d.}$ is the product of the smallest power of each common prime factor:
$\text{g.c.d.}(96, 404) = 2^2 = 4$
Step $4$: The $\text{l.c.m.}$ is the product of the greatest power of each prime factor involved:
$\text{l.c.m.}(96, 404) = 2^5 \times 3^1 \times 101^1 = 32 \times 3 \times 101 = 9696$
Therefore,the $\text{g.c.d.}$ is $4$ and the $\text{l.c.m.}$ is $9696$.

(N/A) Step $1$: Prime factorization of the numbers:
$144 = 2^4 \times 3^2$
$180 = 2^2 \times 3^2 \times 5^1$
$192 = 2^6 \times 3^1$
Step $2$: To find the $\text{g.c.d.}$,take the product of the smallest power of each common prime factor:
$\text{g.c.d.} = 2^2 \times 3^1 = 4 \times 3 = 12$
Step $3$: To find the $\text{l.c.m.}$,take the product of the highest power of each prime factor involved:
$\text{l.c.m.} = 2^6 \times 3^2 \times 5^1 = 64 \times 9 \times 5 = 2880$

(N/A) To find the $\text{l.c.m.}$ and $\text{g.c.d.}$ using the fundamental theorem of arithmetic,we first find the prime factorization of each number:
$84 = 2^2 \times 3^1 \times 7^1$
$90 = 2^1 \times 3^2 \times 5^1$
$120 = 2^3 \times 3^1 \times 5^1$
$\text{g.c.d.}$ is the product of the smallest power of each common prime factor:
$\text{g.c.d.}(84, 90, 120) = 2^1 \times 3^1 = 6$
$\text{l.c.m.}$ is the product of the highest power of each prime factor present in the numbers:
$\text{l.c.m.}(84, 90, 120) = 2^3 \times 3^2 \times 5^1 \times 7^1 = 8 \times 9 \times 5 \times 7 = 2520$

(A) The given numbers are $17$,$23$,and $29$.
Since $17$,$23$,and $29$ are all prime numbers,their prime factorizations are:
$17 = 17^1 \times 1^1$
$23 = 23^1 \times 1^1$
$29 = 29^1 \times 1^1$
$\text{GCD}$ (Greatest Common Divisor) is the product of the smallest power of each common prime factor. Here,the only common factor is $1$.
$\text{GCD}(17, 23, 29) = 1$
$\text{LCM}$ (Least Common Multiple) is the product of the highest power of each prime factor involved in the numbers.
$\text{LCM}(17, 23, 29) = 17 \times 23 \times 29$
$17 \times 23 = 391$
$391 \times 29 = 11339$
Therefore,$\text{LCM} = 11339$ and $\text{GCD} = 1$.

(N/A) composite number is a positive integer that has at least one divisor other than $1$ and itself.
Given expression: $7 \times 11 \times 17 + 17$.
We can factor out $17$ from the expression:
$7 \times 11 \times 17 + 17 = 17 \times (7 \times 11 + 1)$.
Now,simplify the expression inside the parentheses:
$17 \times (77 + 1) = 17 \times 78$.
Since $78$ can be further factored as $2 \times 39$ or $2 \times 3 \times 13$,the expression becomes:
$17 \times 2 \times 3 \times 13$.
Because the number can be expressed as a product of prime factors other than $1$ and itself,it is a composite number.

(B) composite number is a positive integer that has at least one divisor other than $1$ and itself.
Given expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$.
We can factor out $5$ from the expression:
$= 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1)$
$= 5 \times (1008 + 1)$
$= 5 \times 1009$
Since the number can be expressed as a product of two factors ($5$ and $1009$),both of which are greater than $1$,it has factors other than $1$ and itself.
Therefore,the given number is a composite number.

(N/A) For a number to end with the digit $0$,its prime factorization must contain the factors $2$ and $5$.
The prime factorization of $8$ is $2^3$.
Therefore,$8^n = (2^3)^n = 2^{3n}$.
The only prime factor in the prime factorization of $8^n$ is $2$.
Since the factor $5$ is not present in the prime factorization of $8^n$,it is impossible for $8^n$ to end with the digit $0$ for any natural number $n \in N$.

(N/A) For a number to end in the digit $0$,it must be divisible by $10$.
This implies that the prime factorization of the number must contain at least one pair of the prime factors $2$ and $5$.
The prime factorization of $21$ is $3 \times 7$.
Therefore,the prime factorization of $21^{n}$ is $(3 \times 7)^{n} = 3^{n} \times 7^{n}$.
Since the prime factors of $21^{n}$ are only $3$ and $7$,it does not contain the prime factor $2$ or $5$.
Because the prime factor $2$ and $5$ are missing,$21^{n}$ is not divisible by $10$.
Hence,$21^{n}$ cannot end in the digit $0$ for any natural number $n \in N$.

(N/A) To determine if $5^{n} \times 6^{n}$ ends in zero,we must check if it is divisible by $10$. $A$ number is divisible by $10$ if its prime factorization contains at least one pair of $2$ and $5$.
Given expression: $5^{n} \times 6^{n}$.
We can factorize $6^{n}$ as $(2 \times 3)^{n} = 2^{n} \times 3^{n}$.
Substituting this into the expression: $5^{n} \times 2^{n} \times 3^{n} = (2 \times 5)^{n} \times 3^{n} = 10^{n} \times 3^{n}$.
Since the expression can be written as $10^{n} \times 3^{n}$,it is clearly divisible by $10$ for all $n \in N$.
Therefore,$5^{n} \times 6^{n}$ always ends in zero for any natural number $n$.

(B) Assume,to the contrary,that $\sqrt{7}$ is a rational number.
Then,there exist coprime positive integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{7} = \frac{a}{b}$.
Squaring both sides,we get $7 = \frac{a^2}{b^2}$,which implies $a^2 = 7b^2$.
This means that $a^2$ is divisible by $7$,and by the Fundamental Theorem of Arithmetic,$a$ must also be divisible by $7$.
So,we can write $a = 7k$ for some integer $k$.
Substituting this into the equation $a^2 = 7b^2$,we get $(7k)^2 = 7b^2$,which simplifies to $49k^2 = 7b^2$,or $b^2 = 7k^2$.
This implies that $b^2$ is divisible by $7$,and therefore $b$ must also be divisible by $7$.
Since both $a$ and $b$ are divisible by $7$,they have a common factor of $7$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption that $\sqrt{7}$ is rational is false,and $\sqrt{7}$ is an irrational number.

(N/A) Assume,to the contrary,that $\sqrt{21}$ is a rational number.
Then,there exist two coprime positive integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{21} = \frac{a}{b}$.
Squaring both sides,we get $21 = \frac{a^2}{b^2}$,which implies $a^2 = 21b^2$.
This means $a^2$ is divisible by $21$,so $a$ must also be divisible by $21$ (since $21$ is a product of two distinct primes $3$ and $7$).
Let $a = 21k$ for some integer $k$.
Substituting this into the equation,we get $(21k)^2 = 21b^2$,which simplifies to $441k^2 = 21b^2$,or $b^2 = 21k^2$.
This implies $b^2$ is divisible by $21$,so $b$ must also be divisible by $21$.
Thus,$a$ and $b$ have a common factor of $21$,which contradicts our assumption that $a$ and $b$ are coprime.
Therefore,our assumption is false,and $\sqrt{21}$ is an irrational number.

(B) Assume that $\sqrt{5}+1$ is a rational number.
Then,we can write $\sqrt{5}+1 = \frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
Rearranging the equation,we get $\sqrt{5} = \frac{p}{q} - 1$.
This simplifies to $\sqrt{5} = \frac{p-q}{q}$.
Since $p$ and $q$ are integers,$\frac{p-q}{q}$ must be a rational number.
This implies that $\sqrt{5}$ is a rational number.
However,this contradicts the known fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption is false,and $\sqrt{5}+1$ must be an irrational number.

(N/A) Let us assume,to the contrary,that $\sqrt{3}+\sqrt{7}$ is a rational number. Let $\sqrt{3}+\sqrt{7} = r$,where $r$ is a rational number.
Squaring both sides,we get: $(\sqrt{3}+\sqrt{7})^2 = r^2$.
$3 + 7 + 2\sqrt{21} = r^2$.
$10 + 2\sqrt{21} = r^2$.
$2\sqrt{21} = r^2 - 10$.
$\sqrt{21} = \frac{r^2 - 10}{2}$.
Since $r$ is a rational number,$r^2$ is also a rational number. Therefore,$\frac{r^2 - 10}{2}$ is a rational number.
This implies that $\sqrt{21}$ is a rational number.
However,we know that $\sqrt{21}$ is an irrational number because $21$ is not a perfect square.
This is a contradiction to our assumption that $\sqrt{3}+\sqrt{7}$ is rational.
Therefore,our assumption is false,and $\sqrt{3}+\sqrt{7}$ must be an irrational number.

(N/A) Let us assume,to the contrary,that $3\sqrt{2}$ is a rational number.
Then,there exist coprime integers $a$ and $b$ $(b \neq 0)$ such that $3\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{3b}$.
Since $a$ and $b$ are integers,$\frac{a}{3b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the fact that $\sqrt{2}$ is an irrational number.
Our assumption that $3\sqrt{2}$ is rational is incorrect.
Therefore,$3\sqrt{2}$ is an irrational number.