Textbook - Real Numbers Questions in English

(N/A) Since $12576 > 4052,$ we apply the division lemma to $12576$ and $4052$ to get:
$12576 = 4052 \times 3 + 420$
Since the remainder $420 \neq 0,$ we apply the division lemma to $4052$ and $420$ to get:
$4052 = 420 \times 9 + 272$
Since the remainder $272 \neq 0,$ we apply the division lemma to $420$ and $272$ to get:
$420 = 272 \times 1 + 148$
Since the remainder $148 \neq 0,$ we apply the division lemma to $272$ and $148$ to get:
$272 = 148 \times 1 + 124$
Since the remainder $124 \neq 0,$ we apply the division lemma to $148$ and $124$ to get:
$148 = 124 \times 1 + 24$
Since the remainder $24 \neq 0,$ we apply the division lemma to $124$ and $24$ to get:
$124 = 24 \times 5 + 4$
Since the remainder $4 \neq 0,$ we apply the division lemma to $24$ and $4$ to get:
$24 = 4 \times 6 + 0$
The remainder is now $0,$ so the procedure stops. The divisor at this stage is $4.$ Therefore,the $HCF$ of $12576$ and $4052$ is $4.$

(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 $4q+3$.

(D) To minimize the area of the tray,we need to maximize the number of barfis in each stack. This is equivalent to finding the Highest Common Factor $(HCF)$ 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 $HCF$ of $420$ and $130$ is $10$.
Therefore,the sweetseller can place $10$ barfis in each stack to minimize the area occupied on the tray.

(A) To find the $HCF$ of $135$ and $225$ using Euclid's division algorithm:
Since $225 > 135$,we apply the division lemma to $225$ and $135$ to obtain:
$225 = 135 \times 1 + 90$
Since the remainder $90 \neq 0$,we apply the division lemma to $135$ and $90$ to obtain:
$135 = 90 \times 1 + 45$
Now,we consider the new divisor $90$ and the new remainder $45$,and apply the division lemma to obtain:
$90 = 45 \times 2 + 0$
Since the remainder is $0$,the process stops.
Since the divisor at this stage is $45$,the $HCF$ of $135$ and $225$ is $45$.

(B) To find the $HCF$ of $196$ and $38220$ using Euclid's division algorithm:
Since $38220 > 196$,we apply Euclid's division lemma to $38220$ and $196$:
$38220 = 196 \times 195 + 0$
Since the remainder is $0$,the process stops.
The divisor at this stage is $196$.
Therefore,the $HCF$ of $196$ and $38220$ is $196$.

(C) Given numbers are $867$ and $255$.
Since $867 > 255$,we apply Euclid's division lemma to $867$ and $255$ to obtain:
$867 = 255 \times 3 + 102$
Since the remainder $102 \neq 0$,we apply the division lemma to $255$ and $102$ to obtain:
$255 = 102 \times 2 + 51$
Since the remainder $51 \neq 0$,we apply the division lemma to $102$ and $51$ to obtain:
$102 = 51 \times 2 + 0$
Since the remainder is $0$,the process stops.
The divisor at this stage is $51$.
Therefore,the $HCF$ of $867$ and $255$ is $51$.

(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 $6q+5$.

(A) The maximum number of columns in which they can march is given by the $HCF$ of $616$ and $32$.
We use Euclid's division algorithm to find the $HCF$:
$616 = 32 \times 19 + 8$
$32 = 8 \times 4 + 0$
Since the remainder is $0$,the $HCF$ of $616$ and $32$ is $8$.
Therefore,the maximum number of columns in which they can march is $8$.

(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.

(N/A) First,we find the prime factorisation of the given numbers:
$6 = 2^1 \times 3^1$
$20 = 2^2 \times 5^1$
To find the $HCF$,we take the product of the smallest power of each common prime factor:
$HCF(6, 20) = 2^1 = 2$
To find the $LCM$,we take the product of the greatest power of each prime factor involved in the numbers:
$LCM(6, 20) = 2^2 \times 3^1 \times 5^1 = 4 \times 3 \times 5 = 60$
Thus,the $HCF$ is $2$ and the $LCM$ is $60$.

(B) The prime factorisation of $96$ and $404$ is as follows:
$96 = 2^5 \times 3$
$404 = 2^2 \times 101$
The $HCF$ is the product of the smallest power of each common prime factor in the numbers:
$HCF(96, 404) = 2^2 = 4$
Using the relationship $LCM(a, b) \times HCF(a, b) = a \times b$:
$LCM(96, 404) = \frac{96 \times 404}{HCF(96, 404)}$
$LCM(96, 404) = \frac{96 \times 404}{4}$
$LCM(96, 404) = 96 \times 101 = 9696$

(C) We have the prime factorisation of the numbers:
$6 = 2^1 \times 3^1$
$72 = 2^3 \times 3^2$
$120 = 2^3 \times 3^1 \times 5^1$
To find the $HCF$,we take the product of the smallest power of each common prime factor:
$HCF(6, 72, 120) = 2^1 \times 3^1 = 6$
To find the $LCM$,we take the product of the greatest power of each prime factor involved in the numbers:
$LCM(6, 72, 120) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360$
Thus,the $HCF$ is $6$ and the $LCM$ is $360$.

(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,$7429 = 17 \times 19 \times 23$.

(A) First,find the prime factorization of the given numbers:
$26 = 2 \times 13$
$91 = 7 \times 13$
The $HCF$ is the product of the smallest power of each common prime factor:
$HCF = 13$
The $LCM$ is the product of the greatest power of each prime factor involved:
$LCM = 2 \times 7 \times 13 = 182$
Now,verify the relationship:
Product of the two numbers $= 26 \times 91 = 2366$
$HCF \times LCM = 13 \times 182 = 2366$
Since $2366 = 2366$,the relationship $LCM \times HCF =$ product of the two numbers is verified.

(B) First,find the prime factorization of the given numbers:
$510 = 2 \times 3 \times 5 \times 17$
$92 = 2^2 \times 23 = 2 \times 2 \times 23$
To find the $HCF$,take the product of the smallest power of each common prime factor:
$HCF = 2^1 = 2$
To find the $LCM$,take the product of the highest power of each prime factor involved:
$LCM = 2^2 \times 3 \times 5 \times 17 \times 23 = 4 \times 3 \times 5 \times 17 \times 23 = 23460$
Now,verify the relationship:
Product of the two numbers $= 510 \times 92 = 46920$
$HCF \times LCM = 2 \times 23460 = 46920$
Since $46920 = 46920$,the relationship $LCM \times HCF =$ product of the two numbers is verified.

(C) First,find the prime factorization of the numbers:
$336 = 2^4 \times 3 \times 7$
$54 = 2 \times 3^3$
$HCF$ is the product of the smallest power of each common prime factor:
$HCF = 2^1 \times 3^1 = 6$
$LCM$ is the product of the greatest power of each prime factor involved:
$LCM = 2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024$
Now,verify the relationship:
Product of the two numbers $= 336 \times 54 = 18144$
$LCM \times HCF = 3024 \times 6 = 18144$
Since $18144 = 18144$,the relationship is verified.

(D) To find the $LCM$ and $HCF$ of $12, 15$ and $21$ using the prime factorisation method:
Step $1$: Find the prime factors of each number.
$12 = 2^{2} \times 3$
$15 = 3 \times 5$
$21 = 3 \times 7$
Step $2$: The $HCF$ is the product of the smallest power of each common prime factor.
The only common prime factor is $3$,and its smallest power is $3^{1}$.
Therefore,$HCF = 3$.
Step $3$: The $LCM$ is the product of the greatest power of each prime factor involved.
The prime factors involved are $2, 3, 5$ and $7$.
The greatest powers are $2^{2}, 3^{1}, 5^{1}$ and $7^{1}$.
$LCM = 2^{2} \times 3^{1} \times 5^{1} \times 7^{1} = 4 \times 3 \times 5 \times 7 = 420$.

(A) To find the $LCM$ and $HCF$ of $17, 23$ and $29$ using the prime factorisation method:
Step $1$: Write the prime factorisation of each number.
$17 = 1 \times 17$
$23 = 1 \times 23$
$29 = 1 \times 29$
Step $2$: Find the $HCF$.
The $HCF$ is the product of the smallest power of each common prime factor in the numbers.
Since there are no common prime factors other than $1$,$HCF = 1$.
Step $3$: Find the $LCM$.
The $LCM$ is the product of the greatest power of each prime factor involved in the numbers.
$LCM = 17 \times 23 \times 29 = 11339$.
Thus,the $HCF$ is $1$ and the $LCM$ is $11339$.

(B) To find the $LCM$ and $HCF$ of $8, 9$ and $25$ using the prime factorisation method:
Step $1$: Write the prime factorisation of each number:
$8 = 2 \times 2 \times 2 = 2^3$
$9 = 3 \times 3 = 3^2$
$25 = 5 \times 5 = 5^2$
Step $2$: Find the $HCF$ (Highest Common Factor):
The $HCF$ is the product of the smallest power of each common prime factor. Since there are no common prime factors other than $1$,the $HCF = 1$.
Step $3$: Find the $LCM$ (Least Common Multiple):
The $LCM$ is the product of the greatest power of each prime factor involved:
$LCM = 2^3 \times 3^2 \times 5^2$
$LCM = 8 \times 9 \times 25$
$LCM = 72 \times 25 = 1800$
Thus,the $LCM$ is $1800$ and the $HCF$ is $1$.

(C) We are given that $HCF (306, 657) = 9$.
We know the fundamental relationship between two numbers,their $HCF$,and their $LCM$:
$LCM (a, b) \times HCF (a, b) = a \times b$
Substituting the given values:
$LCM (306, 657) \times 9 = 306 \times 657$
To find the $LCM$,divide the product of the numbers by their $HCF$:
$LCM (306, 657) = \frac{306 \times 657}{9}$
First,divide $306$ by $9$:
$306 \div 9 = 34$
Now,multiply the result by $657$:
$LCM (306, 657) = 34 \times 657 = 22338$
Therefore,the $LCM (306, 657)$ is $22338$.

(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.

(B) To find the time when Sonia and Ravi will meet again at the starting point,we need to find the Least Common Multiple $(LCM)$ of the time taken by them to complete one round.
Sonia takes $18$ minutes and Ravi takes $12$ minutes.
Prime factorization of $18 = 2 \times 3^2$.
Prime factorization of $12 = 2^2 \times 3$.
$LCM(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$.
Therefore,they will meet again at the starting point after $36$ minutes.

(N/A) Let us assume,to the contrary,that $\sqrt{3}$ is rational.
That is,we can find integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{3} = \frac{a}{b}$.
Suppose $a$ and $b$ have a common factor other than $1$. Then we can divide by the common factor and assume that $a$ and $b$ are coprime.
So,$b\sqrt{3} = a$.
Squaring on both sides,we get $3b^2 = a^2$.
This implies that $a^2$ is divisible by $3$,and by the fundamental theorem of arithmetic,$a$ is also divisible by $3$.
So,we can write $a = 3c$ for some integer $c$.
Substituting $a = 3c$ in $3b^2 = a^2$,we get $3b^2 = (3c)^2 = 9c^2$,which simplifies to $b^2 = 3c^2$.
This means that $b^2$ is divisible by $3$,and consequently,$b$ is also divisible by $3$.
Therefore,$a$ and $b$ have at least $3$ as a common factor.
But this contradicts the fact that $a$ and $b$ are coprime.
This contradiction has arisen because of our incorrect assumption that $\sqrt{3}$ is rational. Hence,we conclude that $\sqrt{3}$ is irrational.

(N/A) Let us assume,to the contrary,that $5-\sqrt{3}$ is rational.
That is,we can find coprime integers $a$ and $b$ $(b \neq 0)$ such that $5-\sqrt{3} = \frac{a}{b}$.
Rearranging the equation,we get $5 - \frac{a}{b} = \sqrt{3}$,which simplifies to $\sqrt{3} = \frac{5b - a}{b}$.
Since $a$ and $b$ are integers,$\frac{5b - a}{b}$ is a rational number. This implies that $\sqrt{3}$ is rational.
However,this contradicts the established fact that $\sqrt{3}$ is an irrational number.
This contradiction has arisen because of our incorrect assumption that $5-\sqrt{3}$ is rational.
Therefore,we conclude that $5-\sqrt{3}$ is irrational.

(N/A) Let us assume,to the contrary,that $3 \sqrt{2}$ is rational.
That is,we can find 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 $3$,$a$,and $b$ are integers,$\frac{a}{3b}$ must be a rational number,which implies that $\sqrt{2}$ is rational.
However,this contradicts the established fact that $\sqrt{2}$ is irrational.
Therefore,our initial assumption is false,and we conclude that $3 \sqrt{2}$ is irrational.

(N/A) Assume that $\sqrt{5}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{5} = \frac{a}{b}$,where $a$ and $b$ are co-prime (i.e.,their only common factor is $1$).
Squaring both sides,we get $a^2 = 5b^2$.
This implies that $a^2$ is divisible by $5$,and by the Fundamental Theorem of Arithmetic,$a$ is also divisible by $5$.
Let $a = 5k$ for some integer $k$.
Substituting this into the equation $a^2 = 5b^2$,we get $(5k)^2 = 5b^2$,which simplifies to $25k^2 = 5b^2$,or $b^2 = 5k^2$.
This implies that $b^2$ is divisible by $5$,and consequently,$b$ is also divisible by $5$.
Since both $a$ and $b$ are divisible by $5$,they have a common factor of $5$,which contradicts our initial assumption that $a$ and $b$ are co-prime.
Therefore,our assumption that $\sqrt{5}$ is rational is false,and we conclude that $\sqrt{5}$ is irrational.

(N/A) Assume that $3+2 \sqrt{5}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $3+2 \sqrt{5} = \frac{a}{b}$.
Subtracting $3$ from both sides,we get $2 \sqrt{5} = \frac{a}{b} - 3$.
Simplifying the right side,$2 \sqrt{5} = \frac{a-3b}{b}$.
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 rational.
However,this contradicts the established fact that $\sqrt{5}$ is an irrational number.
This contradiction has arisen because of our incorrect assumption that $3+2 \sqrt{5}$ is rational.
Therefore,we conclude that $3+2 \sqrt{5}$ is irrational.

(B) Assume that $\frac{1}{\sqrt{2}}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $\frac{1}{\sqrt{2}} = \frac{a}{b}$.
By rearranging the terms,we get $\sqrt{2} = \frac{b}{a}$.
Since $a$ and $b$ are integers,$\frac{b}{a}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the established fact that $\sqrt{2}$ is an irrational number.
Therefore,our initial assumption is false,and $\frac{1}{\sqrt{2}}$ must be an irrational number.

(N/A) Assume that $7 \sqrt{5}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $7 \sqrt{5} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{5} = \frac{a}{7b}$.
Since $a$ and $b$ are integers,$\frac{a}{7b}$ is a rational number.
This implies that $\sqrt{5}$ must also be a rational number.
However,this contradicts the established fact that $\sqrt{5}$ is an irrational number.
Therefore,our initial assumption that $7 \sqrt{5}$ is rational is incorrect.
Hence,$7 \sqrt{5}$ is an irrational number.

(N/A) Assume that $6+\sqrt{2}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $6+\sqrt{2} = \frac{a}{b}$.
Rearranging the equation,we get $\sqrt{2} = \frac{a}{b} - 6$.
Since $a$ and $b$ are integers,$\frac{a}{b} - 6 = \frac{a-6b}{b}$ is a rational number.
This implies that $\sqrt{2}$ is a rational number.
However,this contradicts the established fact that $\sqrt{2}$ is an irrational number.
This contradiction has arisen due to our incorrect assumption that $6+\sqrt{2}$ is rational.
Therefore,we conclude that $6+\sqrt{2}$ is an irrational 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) 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.

(B) rational number $\frac{p}{q}$ has a terminating decimal expansion if the prime factorization of the denominator $q$ is of the form $2^{m} \cdot 5^{n}$,where $m$ and $n$ are non-negative integers.
In the given rational number $\frac{129}{2^{2} \cdot 5^{7} \cdot 7^{5}}$,the denominator is $q = 2^{2} \cdot 5^{7} \cdot 7^{5}$.
Since the prime factorization of the denominator contains a factor other than $2$ or $5$ (which is $7^{5}$),the decimal expansion is non-terminating repeating.

(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.

(B) 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 $2^n \times 5^m$,where $n$ and $m$ are non-negative integers.
First,simplify the fraction $\frac{77}{210}$ by dividing both numerator and denominator by their greatest common divisor,which is $7$:
$\frac{77 \div 7}{210 \div 7} = \frac{11}{30}$.
Now,find the prime factorization of the denominator $30$:
$30 = 2 \times 3 \times 5$.
Since the prime factorization of the denominator contains a factor $3$ (other than $2$ and $5$),the rational number $\frac{11}{30}$ has a non-terminating repeating 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,$\frac{17}{8} = 2.125$.