Demo Questions in English

(B) The fundamental property of $HCF$ and $LCM$ of two numbers $a$ and $b$ is that the $LCM$ must be perfectly divisible by the $HCF$.
In other words, $\frac{LCM(a, b)}{HCF(a, b)} = \text{integer}$.
Given $HCF(a, b) = 16$.
We check each option:
$1$. For $32$: $\frac{32}{16} = 2$ (Possible).
$2$. For $72$: $\frac{72}{16} = 4.5$ (Not an integer, hence impossible).
$3$. For $64$: $\frac{64}{16} = 4$ (Possible).
Therefore, $72$ cannot be the $LCM$.

(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 the remainder $r$ can be zero or greater than zero,but it must always be strictly less than the divisor $b$.

(A) We are given the expression $2^{m} \cdot 5^{n}$ where $m, n \in N$.
We can rewrite this as $2^{m-n} \cdot (2 \cdot 5)^{n}$ if $m \ge n$,or $5^{n-m} \cdot (2 \cdot 5)^{m}$ if $n > m$.
In either case,the expression contains a factor of $(2 \cdot 5) = 10$.
Any positive integer multiplied by $10$ will result in a number ending in $0$.
For example,if $m=1, n=1$,then $2^1 \cdot 5^1 = 10$ (ends in $0$).
If $m=2, n=1$,then $2^2 \cdot 5^1 = 4 \cdot 5 = 20$ (ends in $0$).
If $m=1, n=2$,then $2^1 \cdot 5^2 = 2 \cdot 25 = 50$ (ends in $0$).
Therefore,the last digit is always $0$.

(C) To convert the fraction $\frac{1}{32}$ into decimal form,we perform long division.
Divide $1$ by $32$:
$1 \div 32 = 0.03125$.
Alternatively,we can write $\frac{1}{32} = \frac{1}{2^5}$.
To convert to a decimal,multiply the numerator and denominator by $5^5$:
$\frac{1 \times 5^5}{2^5 \times 5^5} = \frac{3125}{10^5} = \frac{3125}{100000} = 0.03125$.
Thus,the correct option is $C$.

(B) rational number $\frac{p}{q}$ has a terminating decimal expansion if and only 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.
Non-negative integers include all natural numbers $(N)$ and zero $(0)$.
Therefore,$m, n \in N \cup \{0\}$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM(2, 4)$:
Prime factorization of $2 = 2^1$.
Prime factorization of $4 = 2^2$.
$LCM$ is the product of the highest power of each prime factor involved.
$LCM(2, 4) = 2^2 = 4$.
Therefore,the $LCM$ of the smallest prime number and the smallest composite number is $4$.

(C) To convert the fraction $\frac{337}{125}$ into a terminating decimal,we can multiply both the numerator and the denominator by $8$ to make the denominator a power of $10$.
$\frac{337 \times 8}{125 \times 8} = \frac{2696}{1000}$.
Dividing by $1000$ shifts the decimal point three places to the left.
$\frac{2696}{1000} = 2.696$.

(C) To find the number of decimal places after which the decimal expansion of $\frac{2517}{6250}$ terminates,we need to express the denominator in the form $2^n \times 5^m$.
First,find the prime factorization of the denominator $6250$:
$6250 = 625 \times 10 = 5^4 \times (2 \times 5) = 2^1 \times 5^5$.
The decimal expansion of a rational number $\frac{p}{q}$ terminates after $k$ decimal places,where $k = \max(n, m)$ in the prime factorization $q = 2^n \times 5^m$.
Here,$n = 1$ and $m = 5$.
Therefore,$k = \max(1, 5) = 5$.
The decimal expansion will terminate after $5$ decimal places.

(C) Let the four consecutive positive integers be $n, (n+1), (n+2),$ and $(n+3)$.
The product is $P = n(n+1)(n+2)(n+3)$.
We know that the product of $k$ consecutive integers is always divisible by $k!$ (k factorial).
Here,$k = 4$,so the product is divisible by $4!$.
$4! = 4 \times 3 \times 2 \times 1 = 24$.
Therefore,the product of any four consecutive positive integers is always divisible by $24$.

(B) 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.
$1$. For $\frac{17}{32}$: $32 = 2^5$. Since it is in the form $2^n \times 5^m$,it has a terminating decimal expansion.
$2$. For $\frac{17}{248}$: $248 = 8 \times 31 = 2^3 \times 31^1$. Since there is a factor other than $2$ or $5$ (i.e.,$31$),it has a non-terminating repeating decimal expansion.
$3$. For $\frac{17}{160}$: $160 = 16 \times 10 = 2^4 \times 2 \times 5 = 2^5 \times 5^1$. Since it is in the form $2^n \times 5^m$,it has a terminating decimal expansion.
Therefore,$\frac{17}{248}$ has a non-terminating repeating decimal expansion.

(A) To find the Least Common Multiple $(LCM)$ of $23, 35,$ and $46$:
Step $1$: Find the prime factorization of each number.
$23 = 23^1$ (since $23$ is a prime number).
$35 = 5^1 \times 7^1$.
$46 = 2^1 \times 23^1$.
Step $2$: Identify the highest power of each prime factor present in the numbers.
The prime factors involved are $2, 5, 7,$ and $23$.
The highest powers are $2^1, 5^1, 7^1,$ and $23^1$.
Step $3$: Multiply these highest powers to find the $LCM$.
$LCM = 2^1 \times 5^1 \times 7^1 \times 23^1$
$LCM = 10 \times 7 \times 23$
$LCM = 70 \times 23 = 1610$.

(D) To find the number of decimal places in the decimal representation of a rational number $\frac{p}{q}$ (where $q \neq 0$),we first simplify the fraction to its lowest terms.
$\frac{7}{80}$ is already in its simplest form.
Next,we find the prime factorization of the denominator $q = 80$.
$80 = 2^4 \times 5^1$.
The number of decimal places is given by the maximum exponent of $2$ or $5$ in the prime factorization of the denominator.
Here,the exponents are $4$ and $1$. The maximum exponent is $4$.
Therefore,the decimal representation of $\frac{7}{80}$ will have $4$ digits after the decimal point.
Verification: $\frac{7}{80} = \frac{7 \times 125}{80 \times 125} = \frac{875}{10000} = 0.0875$.

(A) To find the largest positive integer that divides $70$ and $125$ leaving remainders $5$ and $8$ respectively,we subtract the remainders from the numbers first.
$70 - 5 = 65$
$125 - 8 = 117$
Now,we need to find the Greatest Common Divisor $(GCD)$ of $65$ and $117$.
Prime factorization of $65 = 5 \times 13$
Prime factorization of $117 = 3^2 \times 13$
The common factor is $13$.
Therefore,the largest positive integer is $13$.

(B) First,find the $HCF$ of $65$ and $117$ using prime factorization:
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
Therefore,$HCF(65, 117) = 13$.
Given the equation: $65m - 117 = HCF(65, 117)$.
Substitute the $HCF$ value: $65m - 117 = 13$.
Add $117$ to both sides: $65m = 13 + 117$.
$65m = 130$.
Divide by $65$: $m = 130 / 65$.
$m = 2$.

(D) In a factor tree, a number is represented as the product of its factors.
Given the structure of a standard factor tree where a number at the top branches into two factors, we apply the property: $\text{Number} = \text{Factor}_1 \times \text{Factor}_2$.
Assuming the standard structure where $x$ branches into $2$ and $5$, then $x = 2 \times 5 = 10$.
Assuming $y$ branches into $x$ and $2$, then $y = x \times 2 = 10 \times 2 = 20$.
Therefore, the values are $x = 10$ and $y = 20$.

(A) We know that for any two positive integers $a$ and $b$,the product of the numbers is equal to the product of their $\text{HCF}$ and $\text{LCM}$.
Mathematically,$a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$.
Given that $\text{HCF}(a, b) = 7$ and $\text{LCM}(a, b) = 385$.
Therefore,$a \times b = 7 \times 385$.
Calculating the product: $7 \times 385 = 2695$.
Thus,$a \times b = 2695$.

(B) The smallest prime number is $2$.
The smallest composite number is $4$.
To find the $LCM$ of $2$ and $4$:
$2 = 2^1$
$4 = 2^2$
The $LCM$ is the product of the highest power of each prime factor involved,which is $2^2 = 4$.
Therefore,the $LCM$ of $2$ and $4$ is $4$.

(B) By Euclid's division lemma,any positive integer $a$ can be expressed in the form $a = bq + r$,where $0 \le r < b$.
For $b = 2$,the possible values of $r$ are $0$ and $1$.
Thus,$a = 2m + 0 = 2m$ (which is an even integer) or $a = 2m + 1$ (which is an odd integer).
Therefore,any odd positive integer is of the form $2m + 1$.

(A) The largest $4$-digit number is $9999$.
To find the largest $4$-digit number divisible by $95$,we divide $9999$ by $95$.
$9999 \div 95 = 105.2526...$
The integer part is $105$.
Now,multiply $105$ by $95$ to find the largest multiple:
$105 \times 95 = 9975$.
Thus,the largest $4$-digit integer divisible by $95$ is $9975$.

(C) 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 $0 \le r < b$.
In this problem,$b = 5$.
Therefore,the condition for $r$ is $0 \le r < 5$.
This means the possible values for $r$ are ${0, 1, 2, 3, 4}$.
Comparing this with the given options,$5$ is not a possible value for $r$ because $r$ must be strictly less than the divisor $5$.

(A) We are given the expression $2^{3} \times 5^{125}$.
First,calculate $2^{3} = 8$.
Now,the expression becomes $8 \times 5^{125}$.
We can rewrite this as $8 \times 5 \times 5^{124} = 40 \times 5^{124}$.
Since any number multiplied by $40$ ends in $0$,the last digit of the expression is $0$.

(A) Let $x = 1. \overline{23} = 1.232323...$
Multiplying by $100$,we get $100x = 123.232323...$
Subtracting $x$ from $100x$,we get $99x = 122$,so $x = \frac{122}{99}$.
Now,$3.2 = \frac{32}{10} = \frac{16}{5}$.
The sum is $\frac{122}{99} + \frac{16}{5} = \frac{122 \times 5 + 16 \times 99}{99 \times 5} = \frac{610 + 1584}{495} = \frac{2194}{495}$.
Since the sum can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$,it is a rational number.

(A) The smallest two-digit prime number is $11$.
The smallest two-digit composite number is $10$.
To find the $GCD$ of $11$ and $10$:
Prime factorization of $11 = 11^1$.
Prime factorization of $10 = 2^1 \times 5^1$.
Since there are no common prime factors,the $GCD(11, 10) = 1$.

(B) To determine the nature of the number,we simplify the expression:
$3 \times 15 \times 11 + 11$
$= 11 \times (3 \times 15 + 1)$
$= 11 \times (45 + 1)$
$= 11 \times 46$
$= 11 \times 2 \times 23$
Since the number can be expressed as a product of prime factors other than $1$ and itself (i.e.,$2, 11, 23$),it is a composite number.

(C) By definition,a prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
Since $a$ and $b$ are distinct prime numbers,they have no common factors other than $1$.
The Least Common Multiple $(LCM)$ of two numbers is the smallest positive integer that is divisible by both numbers.
For any two numbers $a$ and $b$,the relationship between their product,$HCF$,and $LCM$ is given by: $a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b)$.
Since $a$ and $b$ are distinct prime numbers,their Highest Common Factor $(HCF)$ is $1$.
Therefore,$a \times b = 1 \times \text{LCM}(a, b)$.
This implies that $\text{LCM}(a, b) = ab$.

(B) We know that for any two positive integers $p$ and $q$,the product of the numbers is equal to the product of their $HCF$ and $LCM$.
$p \times q = HCF(p, q) \times LCM(p, q)$
Given that $HCF(p, q) = p$.
Substituting this into the formula:
$p \times q = p \times LCM(p, q)$
Dividing both sides by $p$ (assuming $p \neq 0$):
$LCM(p, q) = q$
Therefore,the correct option is $B$.

(C) For any two prime numbers $p_1$ and $p_2$,the Highest Common Factor $(HCF)$ is always $1$ because they have no common factors other than $1$.
The Least Common Multiple $(LCM)$ of two prime numbers is their product,i.e.,$p_1 \times p_2$.
Given prime numbers are $7$ and $11$.
$HCF(7, 11) = 1$.
$LCM(7, 11) = 7 \times 11 = 77$.
Difference = $LCM - HCF = 77 - 1 = 76$.

(A) To find the $LCM$ of $12, 15,$ and $21$,we first find their prime factorizations:
$12 = 2^2 \times 3^1$
$15 = 3^1 \times 5^1$
$21 = 3^1 \times 7^1$
The $LCM$ is the product of the highest powers of all prime factors involved:
$LCM(12, 15, 21) = 2^2 \times 3^1 \times 5^1 \times 7^1 = 4 \times 3 \times 5 \times 7 = 420$
Given the equation: $LCM(12, 15, 21) = 500 + 2x$
$420 = 500 + 2x$
$2x = 420 - 500$
$2x = -80$
$x = -40$

(B) To find the Greatest Common Divisor $(GCD)$ of $20 a^{2} b$ and $30 a b^{2}$:
$1$. Prime factorization of $20 a^{2} b = 2^{2} \times 5 \times a^{2} \times b$.
$2$. Prime factorization of $30 a b^{2} = 2 \times 3 \times 5 \times a \times b^{2}$.
$3$. The $GCD$ is the product of the lowest power of each common prime factor:
- Common factors are $2, 5, a, b$.
- Lowest powers are $2^{1}, 5^{1}, a^{1}, b^{1}$.
- $GCD$ $= 2 \times 5 \times a \times b = 10 ab$.
Since $10 ab
eq 10 a^{2} b^{2}$,the statement is False.

(B) The factors of $5$ are $1, 5$.
The factors of $15$ are $1, 3, 5, 15$.
The common factors are $1$ and $5$.
The greatest common factor $(GCD)$ is $5$.
Since the statement claims the $GCD$ is $10$,the statement is False.

(B) Given expression: $(\sqrt{2} - \sqrt{3})(\sqrt{3} + \sqrt{2})$
This is in the form of $(a - b)(a + b) = a^2 - b^2$.
Here,$a = \sqrt{2}$ and $b = \sqrt{3}$ is not quite right,let us rewrite it as $(\sqrt{2} - \sqrt{3})(\sqrt{2} + \sqrt{3})$.
Using the identity $(a - b)(a + b) = a^2 - b^2$:
$(\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$.
Since $-1$ is an integer,it is a rational number.
Therefore,the expression is not an irrational number.

(A) Let the positive integer be $x = 3q + 1$.
To find the square of this integer,we calculate $x^2 = (3q + 1)^2$.
Using the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$,we get:
$x^2 = (3q)^2 + 2(3q)(1) + (1)^2$
$x^2 = 9q^2 + 6q + 1$
We can factor out $3$ from the first two terms:
$x^2 = 3(3q^2 + 2q) + 1$
Let $m = 3q^2 + 2q$,where $m$ is an integer.
Thus,$x^2 = 3m + 1$.
Therefore,the statement is true.

(B) Let any positive integer be $n$. According to Euclid's division lemma,any positive integer can be expressed in the form $3q$,$3q + 1$,or $3q + 2$ for some integer $q \ge 0$.
Case $1$: If $n = 3q$,then $n^2 = (3q)^2 = 9q^2 = 3(3q^2) = 3m$,where $m = 3q^2$.
Case $2$: If $n = 3q + 1$,then $n^2 = (3q + 1)^2 = 9q^2 + 6q + 1 = 3(3q^2 + 2q) + 1 = 3m + 1$,where $m = 3q^2 + 2q$.
Case $3$: If $n = 3q + 2$,then $n^2 = (3q + 2)^2 = 9q^2 + 12q + 4 = 9q^2 + 12q + 3 + 1 = 3(3q^2 + 4q + 1) + 1 = 3m + 1$,where $m = 3q^2 + 4q + 1$.
Thus,the square of any positive integer is always of the form $3m$ or $3m + 1$. It can never be of the form $3m + 2$. Therefore,the given statement is False.

(B) The statement is $False$.
Every positive even integer is of the form $2n$,where $n$ is an integer.
While it is true that some even integers can be written as $4q + 2$ (where $q$ is an integer),not all even integers follow this form.
For example,the integer $2$ can be written as $4(0) + 2$,but the integer $4$ cannot be expressed in the form $4q + 2$ for any integer $q$,as $4q + 2 = 4 \implies 4q = 2 \implies q = 0.5$,which is not an integer.

(A) Let the three consecutive positive integers be $n, (n+1)$,and $(n+2)$.
In any set of three consecutive integers,at least one must be even (divisible by $2$) and exactly one must be a multiple of $3$.
Since the product contains at least one factor of $2$ and at least one factor of $3$,the product must be divisible by $2 \times 3 = 6$.
For example,if $n=1$,the product is $1 \times 2 \times 3 = 6$,which is divisible by $6$.
If $n=2$,the product is $2 \times 3 \times 4 = 24$,which is divisible by $6$.
Thus,the statement is true.

(A) Let the two consecutive positive integers be $n$ and $n+1$.
Their product is $P = n(n+1)$.
Since $n$ and $n+1$ are consecutive integers,one of them must be even and the other must be odd.
An even number is always divisible by $2$.
Therefore,the product of an even number and an odd number is always even,meaning it is divisible by $2$.
Thus,the statement is true.

(A) For any two positive integers $a$ and $b$,the relationship between their Highest Common Factor $(HCF)$ and Least Common Multiple $(LCM)$ is given by the formula: $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$.
Since $a$ and $b$ are consecutive positive integers,they are coprime,meaning their only common factor is $1$. Thus,$\text{HCF}(a, b) = 1$.
Substituting this into the formula,we get $1 \times \text{LCM}(a, b) = a \times b$,which simplifies to $\text{LCM}(a, b) = a \times b$.
Therefore,$\text{HCF}(a, b) \times \text{LCM}(a, b) = 1 \times (a \times b) = a \times b$.

(B) rational number is defined as a number that can be expressed in the form $\frac{p}{q}$,where $p$ and $q$ are integers and $q \neq 0$.
$\sqrt{2}$ is a non-terminating and non-repeating decimal $(1.41421356...)$.
Since it cannot be expressed as a ratio of two integers,it is an irrational number.
Therefore,$\sqrt{2}$ is not a rational number.

(B) To find the number of digits after the decimal point for a rational number of the form $\frac{p}{2^n \cdot 5^m}$,we look for the maximum value between the exponents $n$ and $m$.
Given the expression $\frac{7}{2^3 \cdot 5^4}$,we have $n = 3$ and $m = 4$.
The number of decimal places is determined by $\max(n, m) = \max(3, 4) = 4$.
Thus,the decimal representation will have $4$ digits after the decimal point.

(A) To find the $HCF$ of $25$ and $52$,we find their prime factors:
$25 = 5^2$
$52 = 2^2 \times 13$
Since there are no common prime factors other than $1$,the Highest Common Factor $(HCF)$ is $1$.
Therefore,the statement is True.

(A) To determine if the decimal expansion of a rational number $\frac{p}{q}$ is terminating,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,the decimal expansion is terminating.
Here,$q = 3125$.
$3125 = 5 \times 5 \times 5 \times 5 \times 5 = 5^5$.
Since $3125$ can be written as $2^0 \times 5^5$,it satisfies the condition $2^n \times 5^m$.
Therefore,the decimal expansion of $\frac{13}{3125}$ is terminating.

(A) To find the smallest integer divisible by both $28$ and $63$,we need to calculate the Least Common Multiple $(LCM)$ of $28$ and $63$.
First,find the prime factorization of each number:
$28 = 2^2 \times 7^1$
$63 = 3^2 \times 7^1$
The $LCM$ is found by taking the highest power of each prime factor present in the numbers:
$LCM = 2^2 \times 3^2 \times 7^1$
$LCM = 4 \times 9 \times 7$
$LCM = 36 \times 7 = 252$
Therefore,the smallest integer divisible by both $28$ and $63$ is $252$.

(A) To find the $LCM$ of $36$ and $100$,we first find their prime factorizations:
$36 = 2^2 \times 3^2$
$100 = 2^2 \times 5^2$
The $LCM$ is the product of the highest powers of all prime factors involved:
$LCM = 2^2 \times 3^2 \times 5^2$
$LCM = 4 \times 9 \times 25$
$LCM = 36 \times 25 = 900$
Therefore,the $LCM$ of $36$ and $100$ is $900$.

(A) To find the $HCF$ of $65$ and $117$,we first find their prime factorizations:
$65 = 5 \times 13$
$117 = 9 \times 13 = 3^2 \times 13$
The common prime factor with the lowest power is $13$.
Therefore,the $HCF$ of $65$ and $117$ is $13$.

(A) To find the $LCM$ of $220$ and $60$,we first find the prime factorization of each number:
$220 = 2^2 \times 5^1 \times 11^1$
$60 = 2^2 \times 3^1 \times 5^1$
The $LCM$ is the product of the highest power of each prime factor present in the numbers:
$LCM(220, 60) = 2^2 \times 3^1 \times 5^1 \times 11^1$
$LCM(220, 60) = 4 \times 3 \times 5 \times 11$
$LCM(220, 60) = 12 \times 55 = 660$
Therefore,the $LCM$ of $220$ and $60$ is $660$.

(A) To find the $HCF(156, 455)$,we use the prime factorization method:
$156 = 2^2 \times 3 \times 13$
$455 = 5 \times 7 \times 13$
The common prime factor is $13$.
Therefore,the $HCF(156, 455) = 13$.

(B) Let the two consecutive positive integers be $n$ and $n+1$.
Their product is $P = n(n+1) = n^2 + n$.
Since one of the two consecutive integers must be even,the product $n(n+1)$ is always an even number.
An even number is always divisible by $2$.
Therefore,the product of two consecutive positive integers is always divisible by $2$.

(Q.1-B, Q.2-A) Step $1$: Find the $\text{LCM}(7, 49)$.
Since $49$ is a multiple of $7$ $(7 \times 7 = 49)$,the $\text{LCM}$ of $7$ and $49$ is $49$.
Step $2$: Find the $\text{LCM}(10, 25)$.
Prime factorization of $10 = 2 \times 5$.
Prime factorization of $25 = 5 \times 5$.
$\text{LCM}(10, 25) = 2 \times 5 \times 5 = 50$.
Therefore,$Q.1$ matches with $B$ and $Q.2$ matches with $A$ (or $D$ as they are identical).

(A-B) $1$. To find the $HCF$ of $25$ and $10$:
Prime factorization of $25 = 5^2$.
Prime factorization of $10 = 2 \times 5$.
The common factor is $5$. Therefore, $HCF$ $(25, 10)$ = $5$.
$2$. To find the $LCM$ of $17$ and $11$:
Since $17$ and $11$ are both prime numbers, their $LCM$ is their product.
$LCM$ $(17, 11)$ = $17 \times 11 = 187$.
Thus, the correct matching is $Q.1-A$ and $Q.2-B$.

(A) To find the Highest Common Factor $(HCF)$ of $30$ and $72$:
Prime factorization of $30 = 2 \times 3 \times 5$.
Prime factorization of $72 = 2^3 \times 3^2$.
The common factors are $2^1$ and $3^1$.
$HCF(30, 72) = 2 \times 3 = 6$.
To find the $HCF$ of $19$ and $23$:
Since both $19$ and $23$ are prime numbers,their only common factor is $1$.
$HCF(19, 23) = 1$.
Thus,the $HCF$ values are $6$ and $1$ respectively. The correct option is $A$.