Fundamental Theorem of Arithmetic Questions in English

(N/A) Let $a$ be any positive integer and $b=2$. By Euclid's division lemma,for any positive integer $a$ and divisor $b=2$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \leq r < b$.
Since $b=2$,the possible values for the remainder $r$ are $0$ and $1$ (i.e.,$0 \leq r < 2$).
Case $1$: If $r=0$,then $a = 2q + 0 = 2q$. Since $2q$ is divisible by $2$,$a$ is an even integer.
Case $2$: If $r=1$,then $a = 2q + 1$. Since $2q$ is even,$2q+1$ is not divisible by $2$,so $a$ is an odd integer.
Thus,every positive even integer is of the form $2q$ and every positive odd integer is of the form $2q+1$.

(N/A) Let $a$ be any positive odd integer. According to Euclid's division algorithm,for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \leq r < b$.
Here,we take $b = 4$. Thus,$a = 4q + r$,where $0 \leq r < 4$.
The possible values for the remainder $r$ are $0, 1, 2,$ and $3$.
This means $a$ can be expressed as $4q, 4q+1, 4q+2,$ or $4q+3$.
Since $a$ is an odd integer,it cannot be divisible by $2$.
- $4q = 2(2q)$,which is divisible by $2$ (even).
- $4q+2 = 2(2q+1)$,which is divisible by $2$ (even).
Therefore,$a$ cannot be $4q$ or $4q+2$.
Hence,any positive odd integer must be of the form $4q+1$ or $4q+3$.

(N/A) According to Euclid's Division Lemma,for any two positive integers $a$ and $b$,there exist unique integers $q$ and $r$ such that $a = bq + r$,where $0 \leq r < b$.
Let $a$ be any positive odd integer and $b = 6$.
Substituting $b = 6$ in the lemma,we get $a = 6q + r$,where $0 \leq r < 6$.
This means the possible values for $r$ are $0, 1, 2, 3, 4, 5$.
If $r = 0$,then $a = 6q = 2(3q)$,which is even.
If $r = 1$,then $a = 6q + 1 = 2(3q) + 1$,which is odd.
If $r = 2$,then $a = 6q + 2 = 2(3q + 1)$,which is even.
If $r = 3$,then $a = 6q + 3 = 2(3q + 1) + 1$,which is odd.
If $r = 4$,then $a = 6q + 4 = 2(3q + 2)$,which is even.
If $r = 5$,then $a = 6q + 5 = 2(3q + 2) + 1$,which is odd.
Since $a$ is a positive odd integer,it cannot be of the form $6q, 6q+2,$ or $6q+4$. Thus,any positive odd integer must be of the form $6q+1, 6q+3,$ or $6q+5$.

(N/A) Let $a$ be any positive integer and $b=3$.
By Euclid's division lemma,$a = 3q + r$,where $q \geq 0$ and $r \in \{0, 1, 2\}$.
Case $1$: If $r=0$,then $a = 3q$. Squaring both sides,$a^2 = (3q)^2 = 9q^2 = 3(3q^2) = 3m$,where $m = 3q^2$.
Case $2$: If $r=1$,then $a = 3q+1$. Squaring both sides,$a^2 = (3q+1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1$,where $m = 3q^2 + 2q$.
Case $3$: If $r=2$,then $a = 3q+2$. Squaring both sides,$a^2 = (3q+2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$,where $m = 3q^2 + 4q + 1$.
Thus,the square of any positive integer is always of the form $3m$ or $3m+1$.

(N/A) Let $a$ be any positive integer and $b=3$.
By Euclid's division lemma,$a = 3q + r$,where $q \geq 0$ and $0 \leq r < 3$.
Therefore,$a$ can be $3q, 3q+1$,or $3q+2$.
Case $1$: If $a = 3q$,then $a^3 = (3q)^3 = 27q^3 = 9(3q^3) = 9m$,where $m = 3q^3$.
Case $2$: If $a = 3q+1$,then $a^3 = (3q+1)^3 = 27q^3 + 27q^2 + 9q + 1 = 9(3q^3 + 3q^2 + q) + 1 = 9m + 1$,where $m = 3q^3 + 3q^2 + q$.
Case $3$: If $a = 3q+2$,then $a^3 = (3q+2)^3 = 27q^3 + 54q^2 + 36q + 8 = 9(3q^3 + 6q^2 + 4q) + 8 = 9m + 8$,where $m = 3q^3 + 6q^2 + 4q$.
Thus,the cube of any positive integer is of the form $9m, 9m+1$,or $9m+8$.

(C) If the number $4^{n}$ for any natural number $n$ were to end with the digit zero,it would be divisible by $5$.
This implies that the prime factorization of $4^{n}$ must contain the prime number $5$.
However,we can write $4^{n} = (2^{2})^{n} = 2^{2n}$.
The only prime factor in the prime factorization of $4^{n}$ is $2$.
According to the Fundamental Theorem of Arithmetic,this prime factorization is unique.
Since $5$ is not a factor of $4^{n}$,there is no natural number $n$ for which $4^{n}$ ends with the digit zero.

(A) To express $140$ as a product of its prime factors,we perform prime factorization:
$140 = 2 \times 70$
$70 = 2 \times 35$
$35 = 5 \times 7$
Therefore,$140 = 2 \times 2 \times 5 \times 7 = 2^{2} \times 5 \times 7$.

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

(B) To express $3825$ as a product of its prime factors,we perform prime factorization:
$3825 \div 3 = 1275$
$1275 \div 3 = 425$
$425 \div 5 = 85$
$85 \div 5 = 17$
$17 \div 17 = 1$
Thus,$3825 = 3 \times 3 \times 5 \times 5 \times 17 = 3^{2} \times 5^{2} \times 17$.

(A) To express $5005$ as a product of its prime factors,we perform prime factorization:
$5005$ is divisible by $5$: $5005 = 5 \times 1001$
$1001$ is divisible by $7$: $1001 = 7 \times 143$
$143$ is divisible by $11$: $143 = 11 \times 13$
$13$ is a prime number.
Therefore,the prime factorization of $5005$ is $5 \times 7 \times 11 \times 13$.

(A) To find the prime factors of $7429$,we test divisibility by prime numbers starting from the smallest.
$1$. $7429$ is not divisible by $2, 3, 5, 7, 11,$ or $13$.
$2$. Dividing $7429$ by $17$: $7429 \div 17 = 437$.
$3$. Now,we find the prime factors of $437$. Testing divisibility by $19$: $437 \div 19 = 23$.
$4$. Since $23$ is a prime number,the prime factorization is complete.
Therefore,$7429 = 17 \times 19 \times 23$.

(N/A) If any number ends with the digit $0,$ it must be divisible by $10.$ This implies that it must also be divisible by both $2$ and $5,$ since $10 = 2 \times 5.$
The prime factorisation of $6^{n}$ is $(2 \times 3)^{n} = 2^{n} \times 3^{n}.$
It can be observed that the prime factor $5$ is not present in the prime factorisation of $6^{n}.$
Since $5$ is not a factor of $6^{n},$ $6^{n}$ is not divisible by $5$ for any natural number $n.$
Therefore,$6^{n}$ cannot end with the digit $0$ for any natural number $n.$

(N/A) Numbers are of two types: prime and composite. Prime numbers have only two factors,$1$ and the number itself,whereas composite numbers have more than two factors.
For the first expression:
$7 \times 11 \times 13+13 = 13 \times (7 \times 11 + 1) = 13 \times (77 + 1) = 13 \times 78 = 13 \times 13 \times 6$.
Since this expression has factors other than $1$ and itself (specifically $6, 13, 78$),it is a composite number.
For the second expression:
$7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1+5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$.
Since $1009$ is a prime number,the expression has factors $5$ and $1009$ (besides $1$ and the number itself). Therefore,it is a composite number.

(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 $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,then the decimal expansion is terminating.
Given the fraction $\frac{13}{3125}$,the denominator is $3125$.
Prime factorization of $3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.
This can be written as $2^0 \times 5^5$,which is in the form $2^n \times 5^m$ (where $n=0$ and $m=5$).
Since the denominator is of the form $2^n \times 5^m$,the decimal expansion of $\frac{13}{3125}$ is terminating.

(A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we examine the prime factorization of the denominator $q$.
If $q$ is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers,then the decimal expansion is terminating.
Given the rational number $\frac{17}{8}$.
The denominator is $8$.
Prime factorization of $8 = 2 \times 2 \times 2 = 2^3$.
We can write $8 = 2^3 \times 5^0$.
Since the denominator is of the form $2^n \times 5^m$,the decimal expansion of $\frac{17}{8}$ is terminating.

(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.

(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.

(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 $n$.

(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 $3m + 2$.

(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.

(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 $5$.

(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 $m$.

(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 $n$ 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 $4$.

(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 $n$.

(B) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,the prime factorization of the denominator $q$ must be of the form $2^{m} \times 5^{n}$,where $m$ and $n$ are non-negative integers.
Given the rational number $\frac{64}{455}$.
First,find the prime factorization of the denominator $455$:
$455 = 5 \times 7 \times 13$.
Since the prime factorization of the denominator $455$ contains factors other than $2$ and $5$ (specifically $7$ and $13$),the rational number $\frac{64}{455}$ does not have a terminating decimal expansion.
Therefore,the decimal expansion of $\frac{64}{455}$ is non-terminating repeating.

(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 $q$ is of the form $2^{m} \times 5^{n}$,where $m$ and $n$ are non-negative integers,then the decimal expansion is terminating.
Given fraction: $\frac{15}{1600}$.
First,simplify the fraction: $\frac{15}{1600} = \frac{3}{320}$.
Now,find the prime factorization of the denominator $320$:
$320 = 32 \times 10 = 2^{5} \times 2 \times 5 = 2^{6} \times 5^{1}$.
Since the denominator is in the form $2^{m} \times 5^{n}$ (where $m=6$ and $n=1$),the decimal expansion of $\frac{15}{1600}$ is terminating.

(B) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,the prime factorization of the denominator $q$ must be in the form $2^{m} \times 5^{n}$,where $m$ and $n$ are non-negative integers.
Given the rational number $\frac{29}{343}$:
The denominator is $343$.
Prime factorization of $343 = 7 \times 7 \times 7 = 7^{3}$.
Since the prime factorization of the denominator is $7^{3}$,which is not in the form $2^{m} \times 5^{n}$,the decimal expansion of $\frac{29}{343}$ is non-terminating repeating.

(A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we examine the prime factorization of the denominator $q$.
If the denominator $q$ is of the form $2^{m} \times 5^{n}$,where $m$ and $n$ are non-negative integers,then the rational number has a terminating decimal expansion.
In the given expression $\frac{23}{2^{3} \times 5^{2}}$,the denominator is $2^{3} \times 5^{2}$.
Here,$m = 3$ and $n = 2$,which are non-negative integers.
Since the denominator is in the form $2^{m} \times 5^{n}$,the rational number $\frac{23}{2^{3} \times 5^{2}}$ has a terminating decimal expansion.

(A) First,simplify the fraction to its lowest terms:
$\frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5}$
$A$ 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 $5$,which can be written as $2^0 \times 5^1$.
Since the denominator is in the form $2^n \times 5^m$,the decimal expansion of $\frac{6}{15}$ is terminating.

(A) To determine if a rational number $\frac{p}{q}$ has a terminating decimal expansion,we check 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.
Given the rational number $\frac{35}{50}$,we first simplify it to its lowest form:
$\frac{35}{50} = \frac{7 \times 5}{10 \times 5} = \frac{7}{10}$.
The denominator is $10$.
The prime factorization of $10$ is $2^{1} \times 5^{1}$.
Since the denominator is of the form $2^{m} \times 5^{n}$ (where $m=1$ and $n=1$),the rational number $\frac{35}{50}$ has a terminating decimal expansion.

(0.00416) To find the decimal expansion of $\frac{13}{3125}$,we can perform long division or express the denominator as a power of $10$.
Step $1$: Prime factorization of the denominator $3125 = 5^5$.
Step $2$: To make the denominator a power of $10$,multiply the numerator and denominator by $2^5 = 32$.
$\frac{13}{3125} = \frac{13 \times 2^5}{5^5 \times 2^5} = \frac{13 \times 32}{(5 \times 2)^5} = \frac{416}{10^5} = \frac{416}{100000} = 0.00416$.

(N/A) To find the decimal expansion of $\frac{17}{8}$,we perform long division:
$1$. Divide $17$ by $8$.
$2$. $8 \times 2 = 16$,so $17 - 16 = 1$.
$3$. Place a decimal point in the quotient and add a $0$ to the remainder,making it $10$.
$4$. $8 \times 1 = 8$,so $10 - 8 = 2$.
$5$. Add another $0$ to the remainder,making it $20$.
$6$. $8 \times 2 = 16$,so $20 - 16 = 4$.
$7$. Add another $0$ to the remainder,making it $40$.
$8$. $8 \times 5 = 40$,so $40 - 40 = 0$.
Thus,$\frac{17}{8} = 2.125$.

(B) 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,$q = 455$.
Prime factorization of $455 = 5 \times 7 \times 13$.
Since the prime factorization contains factors other than $2$ and $5$ (specifically $7$ and $13$),the decimal expansion of $\frac{64}{455}$ is non-terminating repeating.

(N/A) To find the decimal expansion of $\frac{15}{1600}$,we perform long division:
$1$. Divide $15$ by $1600$.
$2$. Since $15 < 1600$,we place a decimal point and add zeros to $15$ to make it $15.000000$.
$3$. $1600$ goes into $15000$ nine times $(1600 \times 9 = 14400)$,leaving a remainder of $600$.
$4$. Bring down $0$ to make it $6000$. $1600$ goes into $6000$ three times $(1600 \times 3 = 4800)$,leaving a remainder of $1200$.
$5$. Bring down $0$ to make it $12000$. $1600$ goes into $12000$ seven times $(1600 \times 7 = 11200)$,leaving a remainder of $800$.
$6$. Bring down $0$ to make it $8000$. $1600$ goes into $8000$ five times $(1600 \times 5 = 8000)$,leaving a remainder of $0$.
Thus,the decimal expansion of $\frac{15}{1600}$ is $0.009375$.

(N/A) To find the decimal expansion of the rational number $\frac{23}{2^{3} 5^{2}}$,we first simplify the denominator:
$\frac{23}{2^{3} \times 5^{2}} = \frac{23}{8 \times 25} = \frac{23}{200}$
Now,we perform the division of $23$ by $200$:
$23 \div 200 = 0.115$
Alternatively,we can make the powers of $2$ and $5$ equal:
$\frac{23}{2^{3} \times 5^{2}} = \frac{23 \times 5}{2^{3} \times 5^{2} \times 5} = \frac{115}{2^{3} \times 5^{3}} = \frac{115}{(2 \times 5)^{3}} = \frac{115}{10^{3}} = \frac{115}{1000} = 0.115$

(N/A) To find the decimal expansion of the rational number $\frac{6}{15}$,we first simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor,which is $3$.
$\frac{6}{15} = \frac{2 \times 3}{5 \times 3} = \frac{2}{5}$.
Now,we convert $\frac{2}{5}$ into a decimal by dividing $2$ by $5$:
$2 \div 5 = 0.4$.
Thus,the decimal expansion of $\frac{6}{15}$ is $0.4$.

(N/A) To find the decimal expansion of the rational number $\frac{35}{50}$,we can perform long division or simplify the fraction first.
Method $1$: Long Division
Divide $35$ by $50$:
$35 \div 50 = 0.7$
Method $2$: Converting to a denominator of power of $10$
$\frac{35}{50} = \frac{35 \times 2}{50 \times 2} = \frac{70}{100} = 0.7$
Thus,the decimal expansion of $\frac{35}{50}$ is $0.7$.

(B) First,simplify the fraction $\frac{77}{210}$ by dividing both numerator and denominator by their greatest common divisor,which is $7$.
$\frac{77 \div 7}{210 \div 7} = \frac{11}{30}$.
Next,find the prime factorization of the denominator $30$:
$30 = 2 \times 3 \times 5$.
$A$ rational number has a terminating decimal expansion if and only if the prime factorization of its denominator (in simplest form) is of the form $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
Since the denominator $30$ contains a factor of $3$ (which is not $2$ or $5$),the decimal expansion of $\frac{11}{30}$ is non-terminating and repeating.
Performing the division $11 \div 30$ gives $0.3666...$ or $0.3\overline{6}$.