Fundamental Theorem of Arithmetic Questions in English

(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 $n$.

$(1)$ 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.

(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) 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 $3$.

(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 $n \in N$.

(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) 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 $n$.

(N/A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$.
If $q = 2^n \times 5^m$,where $n$ and $m$ are non-negative integers,the decimal expansion is terminating.
Here,the fraction is $\frac{64}{455}$.
First,we check if the fraction is in its simplest form. The $g.c.d.$ of $64$ and $455$ is $1$,so it is in its simplest form.
Next,we find the prime factorization of the denominator $455$:
$455 = 5 \times 91 = 5 \times 7 \times 13$.
Since the prime factorization of the denominator contains factors other than $2$ and $5$ (specifically $7$ and $13$),the rational number $\frac{64}{455}$ does not have a terminating decimal expansion. It has a non-terminating repeating decimal expansion.

(D) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Given the fraction $\frac{13}{125}$:
$1$. Prime factorization of the denominator: $125 = 5^3 = 2^0 \times 5^3$.
$2$. Since the denominator is in the form $2^n \times 5^m$ (where $n=0, m=3$),the rational number has a terminating decimal expansion.
$3$. To find the decimal expansion,we make the denominator a power of $10$:
$\frac{13}{125} = \frac{13 \times 2^3}{5^3 \times 2^3} = \frac{13 \times 8}{10^3} = \frac{104}{1000} = 0.104$.

(N/A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$. If $q = 2^n \times 5^m$ (where $n, m \ge 0$),the decimal expansion is terminating.
Here,the denominator is $343$.
The prime factorization of $343$ is $7^3$.
Since the denominator is not of the form $2^n \times 5^m$,the rational number $\frac{29}{343}$ does not have a terminating decimal expansion; it has a non-terminating repeating decimal expansion.

(D) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$. If $q = 2^n \times 5^m$ (where $n, m \ge 0$),the expansion is terminating.
First,simplify the fraction: $\frac{77}{210} = \frac{7 \times 11}{7 \times 30} = \frac{11}{30}$.
The denominator is $30 = 2^1 \times 3^1 \times 5^1$.
Since the prime factorization of the denominator contains a factor of $3$ (other than $2$ and $5$),the decimal expansion is non-terminating and repeating.

(A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $8 = 2^3 = 2^3 \times 5^0$.
Since the denominator is in the form $2^n \times 5^m$,the rational number $\frac{17}{8}$ has a terminating decimal expansion.
To find the decimal expansion: $\frac{17}{8} = \frac{17 \times 125}{8 \times 125} = \frac{2125}{1000} = 2.125$.

(C) To find the square root of $24+2 \sqrt{119}$,we assume that $\sqrt{24+2 \sqrt{119}} = \sqrt{x} + \sqrt{y}$.
Squaring both sides,we get $24+2 \sqrt{119} = x + y + 2 \sqrt{xy}$.
Comparing the rational and irrational parts,we have $x + y = 24$ and $xy = 119$.
We need two numbers whose sum is $24$ and product is $119$.
These numbers are $17$ and $7$,since $17+7=24$ and $17 \times 7 = 119$.
Therefore,$\sqrt{24+2 \sqrt{119}} = \sqrt{17} + \sqrt{7}$.

(A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,the denominator is $8 = 2^3 = 2^3 \times 5^0$.
Since the denominator is in the form $2^n \times 5^m$,the rational number $\frac{19}{8}$ has a terminating decimal expansion.
To find the decimal expansion,we can divide $19$ by $8$:
$19 \div 8 = 2.375$.

(N/A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$.
If $q = 2^n \times 5^m$,where $n$ and $m$ are non-negative integers,the decimal expansion is terminating.
Here,the denominator is $343 = 7^3$.
Since the prime factorization of the denominator contains a factor other than $2$ or $5$ (specifically,$7$),the rational number $\frac{17}{343}$ does not have a terminating decimal expansion; it has a non-terminating repeating decimal expansion.

(A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check the prime factorization of the denominator $q$ in its simplest form.
First,simplify the fraction: $\frac{9}{30} = \frac{3 \times 3}{3 \times 10} = \frac{3}{10}$.
The denominator is $10 = 2^1 \times 5^1$.
Since the prime factorization of the denominator is of the form $2^n \times 5^m$ (where $n, m$ are non-negative integers),the rational number has a terminating decimal expansion.
To find the decimal expansion: $\frac{3}{10} = 0.3$.

(N/A) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Here,$q = 256 = 2^8$.
Since the denominator is of the form $2^n \times 5^m$ (where $n=8, m=0$),the rational number $\frac{19}{256}$ has a terminating decimal expansion.
To find the decimal expansion,we multiply the numerator and denominator by $5^8$:
$\frac{19}{2^8} \times \frac{5^8}{5^8} = \frac{19 \times 390625}{10^8} = \frac{7421875}{100000000} = 0.07421875$.

(B) To find the number of decimal places after which the expansion of a rational number $\frac{p}{q}$ terminates,we express the denominator in the form $2^n \times 5^m$.
Given the expression: $\frac{43}{2^{4} \times 5^{3}}$.
To make the powers of $2$ and $5$ equal,we multiply the numerator and denominator by $5^1$:
$\frac{43 \times 5^1}{2^{4} \times 5^{3} \times 5^1} = \frac{43 \times 5}{2^{4} \times 5^{4}} = \frac{215}{(2 \times 5)^{4}} = \frac{215}{10^4} = \frac{215}{10000} = 0.0215$.
The decimal expansion terminates after $4$ digits.

(D) To find the remainder when $(3k+2)^2$ is divided by $3$,we expand the expression:
$(3k+2)^2 = (3k)^2 + 2(3k)(2) + 2^2$
$= 9k^2 + 12k + 4$
We can rewrite this as:
$= 9k^2 + 12k + 3 + 1$
$= 3(3k^2 + 4k + 1) + 1$
Since $3(3k^2 + 4k + 1)$ is clearly divisible by $3$,the remainder when the expression is divided by $3$ is $1$.

(A) Bezout's identity states that for any two integers $a$ and $b$ (not both zero),there exist integers $x$ and $y$ such that $ax + by = \text{gcd}(a, b)$,where $\text{gcd}(a, b)$ is the greatest common divisor of $a$ and $b$.

(B) An integer is defined as odd if it is not divisible by $2$.
By Euclid's division lemma,any integer $a$ can be expressed as $a = bq + r$,where $0 \le r < b$.
For divisibility by $2$,we take $b = 2$.
Thus,$a = 2k + r$,where $r$ can be $0$ or $1$.
If $r = 0$,then $a = 2k$,which is an even integer.
If $r = 1$,then $a = 2k + 1$,which is an odd integer.
Therefore,every odd integer is of the form $2k + 1$ for some integer $k$.

(A) According to the Fundamental Theorem of Arithmetic and the properties of rational numbers,a rational number $\frac{p}{q}$ (where $p$ and $q$ are coprime) has a terminating decimal expansion if and only if the prime factorization of the denominator $q$ is of the form $2^{m} 5^{n}$,where $m$ and $n$ are non-negative integers.
Therefore,the decimal form of $\frac{p}{q}$ is a terminating decimal expansion.

(A) To find the number of decimal places after which the expansion of $\frac{9}{1600}$ terminates,we first express the denominator in prime factors.
$1600 = 16 \times 100 = 2^4 \times 10^2 = 2^4 \times (2 \times 5)^2 = 2^4 \times 2^2 \times 5^2 = 2^6 \times 5^2$.
Now,the fraction is $\frac{9}{2^6 \times 5^2}$.
To make the powers of $2$ and $5$ equal,we multiply the numerator and denominator by $5^4$:
$\frac{9 \times 5^4}{2^6 \times 5^2 \times 5^4} = \frac{9 \times 625}{2^6 \times 5^6} = \frac{5625}{(2 \times 5)^6} = \frac{5625}{10^6}$.
Since the denominator is $10^6$,the decimal expansion will terminate after $6$ digits.

(C) Two surds are called like surds if their radicands are the same after simplifying them to their simplest form.
First,simplify $\sqrt{12}$:
$\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}$.
Now,check the options:
$A) \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$
$B) \sqrt{36} = 6$ (This is a rational number,not a surd)
$C) \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$
$D) \sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$
Since $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{48} = 4\sqrt{3}$,they share the same irrational factor $\sqrt{3}$.
Therefore,$\sqrt{12}$ and $\sqrt{48}$ are like surds.

(B) Two surds are called 'like surds' if they have the same radicand after simplification.
$A) \sqrt{8} = 2\sqrt{2}$ and $\sqrt{16} = 4$. These are not like surds.
$B) \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2}$ and $\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$. Both have the same surd part $\sqrt{2}$,so they are like surds.
$C) \sqrt{72} = 6\sqrt{2}$ and $\sqrt{6}$. These are not like surds.
$D) \sqrt{24} = 2\sqrt{6}$ and $\sqrt{48} = 4\sqrt{3}$. These are not like surds.
Therefore,the correct pair is $\sqrt{18}$ and $\sqrt{50}$.

(D) Any odd integer $n$ can be expressed in the form $2k+1$ where $k$ is an integer.
Alternatively,any odd integer $n$ can be represented as $n = 4k \pm 1$ or simply by testing values.
Let $n = 1$,then $n^{2}-1 = 1^{2}-1 = 0$ (divisible by $8$).
Let $n = 3$,then $n^{2}-1 = 3^{2}-1 = 9-1 = 8$ (divisible by $8$).
Let $n = 5$,then $n^{2}-1 = 5^{2}-1 = 25-1 = 24$ (divisible by $8$).
General proof: Since $n$ is odd,$n = 2k+1$.
$n^{2}-1 = (2k+1)^{2}-1 = 4k^{2}+4k+1-1 = 4k(k+1)$.
Since $k(k+1)$ is the product of two consecutive integers,it is always even,i.e.,$k(k+1) = 2m$.
Therefore,$n^{2}-1 = 4(2m) = 8m$.
Thus,$n^{2}-1$ is always divisible by $8$.