Numbers Questions in English

(D) The relationship between dividend, divisor, quotient, and remainder is given by the formula:
Dividend $= (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$
Given:
Divisor $= 38$
Quotient $= 90$
Remainder $= 19$
Substituting these values into the formula:
Dividend $= (38 \times 90) + 19$
Dividend $= 3420 + 19$
Dividend $= 3439$
Therefore, the number is $3439$.

(A) The given series is an arithmetic progression where the first term $a = 90$,the last term $l = 110$,and the number of terms $n$ is calculated as $n = (110 - 90) + 1 = 21$.
The sum of an arithmetic progression is given by the formula $S_n = \frac{n}{2}(a + l)$.
Substituting the values: $S_{21} = \frac{21}{2}(90 + 110)$.
$S_{21} = \frac{21}{2}(200)$.
$S_{21} = 21 \times 100 = 2100$.

(D) number is divisible by $9$ if the sum of its digits is divisible by $9$.
The given number is $583x437$.
The sum of the digits is $5 + 8 + 3 + x + 4 + 3 + 7 = 30 + x$.
For the number to be divisible by $9$,$(30 + x)$ must be a multiple of $9$.
The multiples of $9$ greater than $30$ are $36, 45, \dots$.
Setting $30 + x = 36$,we get $x = 36 - 30 = 6$.
Thus,the smallest whole number in place of $x$ is $6$.

(A) The smallest $5$-digit number is $10000$.
To find the least $5$-digit number divisible by $88$,we divide $10000$ by $88$.
$10000 \div 88 = 113$ with a remainder of $56$.
To find the next multiple of $88$,we calculate: $88 - 56 = 32$.
Adding this difference to the smallest $5$-digit number: $10000 + 32 = 10032$.
Verification: $10032 \div 88 = 114$. Since $10032$ is exactly divisible by $88$,it is the required number.

(A) number is divisible by $11$ if the difference between the sum of digits at odd places and the sum of digits at even places is either $0$ or a multiple of $11$.
For the number $34N$,the digits are $3$ and $4N$ (where $N$ is the units digit).
Let the number be represented as $34N$. The digits are $3, 4, N$.
The alternating sum is $(3 + N) - 4$.
For the number to be divisible by $11$,this difference must be $0$ or a multiple of $11$.
$(3 + N) - 4 = 0$
$N - 1 = 0$
$N = 1$.
Thus,the number is $341$,which is $11 \times 31$.

(C) To find the number of positive factors of $36$,we first find its prime factorization.
$36 = 6 \times 6 = (2 \times 3) \times (2 \times 3) = 2^{2} \times 3^{2}$.
If a number is expressed as $p^{a} \times q^{b}$,where $p$ and $q$ are prime numbers,the total number of factors is given by $(a+1)(b+1)$.
Here,$a = 2$ and $b = 2$.
Therefore,the number of factors $= (2+1) \times (2+1) = 3 \times 3 = 9$.

(B) The sum of the first $n$ natural numbers is given by the formula $S = \frac{n(n+1)}{2}.$
For $n = 12,$ the sum is $S = \frac{12 \times (12+1)}{2} = \frac{12 \times 13}{2} = 78.$
Let the number added twice be $x.$
The new sum is the original sum plus the extra number: $S_{new} = S + x.$
Given $S_{new} = 80,$ we have $80 = 78 + x.$
Therefore,$x = 80 - 78 = 2.$
The number added twice is $2.$

(C) First,find the count of numbers between $1$ and $200$ that are divisible by $3$. This is given by $lfloor 200/3 floor = 66$.
Next,identify numbers that are divisible by both $3$ and $7$. Since $3$ and $7$ are coprime,these are multiples of $\text{lcm}(3, 7) = 21$.
The count of numbers between $1$ and $200$ divisible by $21$ is $lfloor 200/21 floor = 9$.
To find the numbers divisible by $3$ but not by $7$,subtract the count of multiples of $21$ from the count of multiples of $3$.
Result $= 66 - 9 = 57$.

(C) To find the remainder when $3401$ is divided by $11$,we perform the division:
$3401 \div 11 = 309$ with a remainder.
$3401 = 11 \times 309 + 2$.
Alternatively,using the divisibility rule for $11$: The difference between the sum of digits at odd places and even places is $(3+0) - (4+1) = 3 - 5 = -2$.
Since the result is $-2$,we add $11$ to get the positive remainder: $-2 + 11 = 9$.
Wait,let's re-calculate: $3401 \div 11$: $3401 = 11 \times 309 + 2$.
The remainder is $2$.
Therefore,if we subtract $2$ from $3401$,the result $3399$ will be perfectly divisible by $11$ $(3399 \div 11 = 309)$.

(A) To find the remainder when $2468$ is divided by $37$,we perform long division:
$2468 \div 37$
First,divide $246$ by $37$: $37 \times 6 = 222$.
Subtract $222$ from $246$: $246 - 222 = 24$.
Bring down the $8$ to make it $248$.
Now,divide $248$ by $37$: $37 \times 6 = 222$.
Subtract $222$ from $248$: $248 - 222 = 26$.
Thus,the remainder is $26$.

(B) number is divisible by $9$ if the sum of its digits is divisible by $9$.
For the number $432P1$,the sum of the digits is $4 + 3 + 2 + P + 1 = 10 + P$.
For this sum to be divisible by $9$,the next multiple of $9$ greater than $10$ is $18$.
Therefore,$10 + P = 18$.
Solving for $P$,we get $P = 18 - 10 = 8$.

(C) number is divisible by $12$ if it is divisible by both $3$ and $4$,since $12 = 3 \times 4$.
$(i)$ For divisibility by $3$,the sum of the digits must be divisible by $3$. The sum of the digits of $34M$ is $3 + 4 + M = 7 + M$.
If $M = 2$,sum $= 9$ (divisible by $3$).
If $M = 5$,sum $= 12$ (divisible by $3$).
If $M = 8$,sum $= 15$ (divisible by $3$).
$(ii)$ For divisibility by $4$,the last two digits of the number must be divisible by $4$. The last two digits are $4M$.
Possible values for $4M$ to be divisible by $4$ are $40, 44, 48$ (where $M$ can be $0, 4, 8$).
Comparing the conditions from $(i)$ and $(ii)$,the only common value for $M$ is $8$.
Therefore,$348$ is divisible by $12$ $(348 / 12 = 29)$.

(D) To find the number of positive factors of $160$,we first perform the prime factorization of $160$.
$160 = 16 \times 10$
$160 = 2^{4} \times (2 \times 5)$
$160 = 2^{5} \times 5^{1}$
If a number is expressed as $p^{a} \times q^{b}$,where $p$ and $q$ are prime numbers,the total number of factors is given by $(a+1)(b+1)$.
Here,$a = 5$ and $b = 1$.
Total number of factors $= (5+1) \times (1+1)$
$= 6 \times 2 = 12$.

(B) To compare the fractions $\frac{2}{3}$,$\frac{4}{5}$,and $\frac{7}{11}$,we convert them into decimal form:
$1$. $\frac{2}{3} = 0.666...$
$2$. $\frac{4}{5} = 0.8$
$3$. $\frac{7}{11} = 0.6363...$
Comparing the decimal values: $0.8 > 0.666... > 0.6363...$
Therefore,the largest fraction is $\frac{4}{5}$.

(C) To compare the fractions $\frac{3}{7}, \frac{5}{11}$,and $\frac{6}{13}$,we convert them into decimal form:
$\frac{3}{7} \approx 0.428$
$\frac{5}{11} \approx 0.454$
$\frac{6}{13} \approx 0.461$
Comparing the decimal values,$0.461 > 0.454 > 0.428$.
Therefore,the largest fraction is $\frac{6}{13}$.

(C) prime number is a natural number greater than $1$ that has no positive divisors other than $1$ and itself.
To find the largest two-digit prime number,we check the numbers starting from $99$ downwards:
$1$. $99$ is divisible by $3, 9, 11, 33$,so it is not prime.
$2$. $98$ is an even number,so it is divisible by $2$,thus not prime.
$3$. $97$ has no divisors other than $1$ and $97$. Therefore,$97$ is a prime number.
Thus,the largest two-digit prime number is $97$.

(A) The total time required to build the wall is $70$ hours.
The mason has already worked for $7$ hours.
The remaining time required to complete the wall is $70 - 7 = 63$ hours.
The fraction of the wall yet to be built is calculated by dividing the remaining hours by the total hours:
$\text{Fraction} = \frac{63}{70} = \frac{9}{10} = 0.9$.

(C) Let the number of oranges in the first basket be $x$ and in the second basket be $y$.
Given that $x + y = 640$.
If $\frac{1}{5}$ of the oranges from the first basket are moved to the second,the new count in the first basket is $x - \frac{1}{5}x = \frac{4}{5}x$.
The new count in the second basket is $y + \frac{1}{5}x$.
Since the number of oranges becomes equal,we have $\frac{4}{5}x = y + \frac{1}{5}x$.
This simplifies to $y = \frac{3}{5}x$.
Substituting $y$ in the first equation: $x + \frac{3}{5}x = 640$.
$\frac{8}{5}x = 640$.
$x = 640 \times \frac{5}{8} = 80 \times 5 = 400$.
Thus,the number of oranges in the first basket is $400$.

(A) To add the fractions $\frac{3}{5}$ and $\frac{7}{9}$,we first find the least common multiple $(LCM)$ of the denominators $5$ and $9$.
The $LCM$ of $5$ and $9$ is $45$.
Now,convert both fractions to have a denominator of $45$:
$\frac{3}{5} = \frac{3 \times 9}{5 \times 9} = \frac{27}{45}$
$\frac{7}{9} = \frac{7 \times 5}{9 \times 5} = \frac{35}{45}$
Adding these fractions together:
$\frac{27}{45} + \frac{35}{45} = \frac{27 + 35}{45} = \frac{62}{45}$.

(D) Let the number be $x$.
According to the problem,the sum of the number and twice its reciprocal is $\frac{33}{4}$.
So,$x + 2(\frac{1}{x}) = \frac{33}{4}$.
Multiply the entire equation by $4x$ to clear the denominators:
$4x^2 + 8 = 33x$.
Rearrange the terms to form a quadratic equation:
$4x^2 - 33x + 8 = 0$.
Factor the quadratic equation:
$4x^2 - 32x - x + 8 = 0$.
$4x(x - 8) - 1(x - 8) = 0$.
$(4x - 1)(x - 8) = 0$.
This gives two possible values for $x$: $x = \frac{1}{4}$ or $x = 8$.
Comparing these with the given options,the correct value is $8$.

(NONE) Step $1$: Find the reciprocals of the given fractions.
The reciprocal of $\frac{6}{5}$ is $\frac{5}{6}$.
The reciprocal of $\frac{3}{7}$ is $\frac{7}{3}$.
Step $2$: Find the sum of these reciprocals.
Sum $= \frac{5}{6} + \frac{7}{3} = \frac{5 + 14}{6} = \frac{19}{6}$.
Step $3$: Find the reciprocal of the sum obtained in Step $2$.
The reciprocal of $\frac{19}{6}$ is $\frac{6}{19}$.

(A) Let the number of girls be $x$ and the number of boys be $y$.
Total number of students $= x + y$.
Number of girls who participated $= \frac{1}{5}x$.
Number of boys who participated $= \frac{1}{8}y$.
Total number of students who participated $= \frac{1}{5}x + \frac{1}{8}y = \frac{8x + 5y}{40}$.
The fraction of total students who participated is given by $\frac{\text{Total participants}}{\text{Total students}} = \frac{\frac{8x + 5y}{40}}{x + y} = \frac{8x + 5y}{40(x + y)}$.
Note: Since the ratio of boys to girls is not provided,the exact numerical fraction cannot be determined unless $x = y$. Assuming the question implies an equal number of boys and girls $(x = y)$:
$\frac{8x + 5x}{40(x + x)} = \frac{13x}{40(2x)} = \frac{13}{80}$.

(B) Let the fraction be $x$.
According to the problem,the fraction is greater than twice its reciprocal by $\frac{7}{15}$.
So,the equation is: $x - 2(\frac{1}{x}) = \frac{7}{15}$.
Multiply by $15x$ to clear the denominators: $15x^2 - 30 = 7x$.
Rearrange into a standard quadratic equation: $15x^2 - 7x - 30 = 0$.
Factor the quadratic equation: $15x^2 - 25x + 18x - 30 = 0$.
$5x(3x - 5) + 6(3x - 5) = 0$.
$(5x + 6)(3x - 5) = 0$.
This gives $x = \frac{5}{3}$ or $x = -\frac{6}{5}$.
Comparing with the given options,the correct fraction is $\frac{5}{3}$.

(A) Let the fraction be $x$.
According to the problem,the fraction is greater than its reciprocal by $\frac{9}{20}$,so:
$x - \frac{1}{x} = \frac{9}{20}$
Multiply the entire equation by $20x$ to clear the denominators:
$20x^2 - 20 = 9x$
Rearrange into a standard quadratic equation form:
$20x^2 - 9x - 20 = 0$
Factor the quadratic equation:
$20x^2 - 16x + 25x - 20 = 0$
$4x(5x - 4) + 5(5x - 4) = 0$
$(4x + 5)(5x - 4) = 0$
This gives two possible values for $x$:
$x = -\frac{5}{4}$ or $x = \frac{4}{5}$ is incorrect based on the positive difference. Let's re-evaluate:
If $x = \frac{5}{4}$,then the reciprocal is $\frac{4}{5}$.
Difference: $\frac{5}{4} - \frac{4}{5} = \frac{25 - 16}{20} = \frac{9}{20}$.
Thus,the fraction is $\frac{5}{4}$.

(D) Let the number be $x$.
According to the problem,the number is greater than $58$ times its reciprocal by $\frac{3}{4}$.
So,$x - \frac{58}{x} = \frac{3}{4}$.
Multiplying by $4x$,we get $4x^2 - 232 = 3x$.
Rearranging the terms,we get the quadratic equation $4x^2 - 3x - 232 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a = 4, b = -3, c = -232$:
$x = \frac{3 \pm \sqrt{(-3)^2 - 4(4)(-232)}}{2(4)}$
$x = \frac{3 \pm \sqrt{9 + 3712}}{8}$
$x = \frac{3 \pm \sqrt{3721}}{8}$
$x = \frac{3 \pm 61}{8}$
Taking the positive root: $x = \frac{64}{8} = 8$.
Taking the negative root: $x = \frac{-58}{8} = -7.25$.
Since $8$ is one of the given options,the correct number is $8$.

(B) Let the number of oranges with Natu be $N$ and the number of oranges with Buchku be $B$.
According to the first statement: $N + 10 = 2(B - 10)$.
Simplifying this: $N + 10 = 2B - 20$,which gives $2B - N = 30$ (Equation $1$).
According to the second statement: $B + 10 = N - 10$.
Simplifying this: $N - B = 20$ (Equation $2$).
From Equation $2$,we get $N = B + 20$.
Substituting this into Equation $1$: $2B - (B + 20) = 30$.
$2B - B - 20 = 30$,so $B = 50$.
Now,$N = 50 + 20 = 70$.
Thus,Natu has $70$ oranges and Buchku has $50$ oranges.

(A) Let the two-digit number be represented as $10x + y$,where $x$ is the ten's digit and $y$ is the unit's digit.
According to the problem,the ten's digit is $7$ more than the unit's digit: $x = y + 7$.
The value of the number is $10x + y$.
When the digits are interchanged,the new number becomes $10y + x$.
According to the problem,subtracting $63$ from the original number gives the new number:
$(10x + y) - 63 = 10y + x$
$9x - 9y = 63$
$x - y = 7$
Substitute $x = y + 7$ into the equation:
$(y + 7) - y = 7$,which is always true for any $y$.
However,we also know $x + y$ must be a digit. Since $x = y + 7$ and $x \le 9$,the possible values for $y$ are $0, 1, 2$.
If $y = 1$,then $x = 8$. The number is $81$.
Check: $81 - 63 = 18$,which is the number with interchanged digits. Thus,the number is $81$.

(D) Let the two numbers be $n$ and $n+2$. The greater number is $n+2$.
Given that the greater number is smaller than the denominator $D$,we have $n+2 < D$.
Let the fraction be $f(n) = \frac{n(n+2)}{D}$.
To find the minimum value,we analyze the quadratic expression in the numerator.
For a fixed denominator $D$,the expression $n^2 + 2n$ represents a parabola opening upwards.
The vertex of this parabola is at $n = -b/2a = -2/2 = -1$.
However,since $n$ must be an integer in the context of such problems,we test values near the vertex.
If we assume the denominator $D=5$,the fraction is $f(n) = \frac{n^2+2n}{5}$.
For $n=0$,$f(0) = 0$.
For $n=-1$,$f(-1) = \frac{1-2}{5} = -\frac{1}{5}$.
For $n=-2$,$f(-2) = 0$.
Thus,the minimum value is $-\frac{1}{5}$.

(C) Given that the remainder $r = 46$.
The divisor $d$ is $5$ times the remainder $r$,so $d = 5 \times 46 = 230$.
The divisor $d$ is also $10$ times the quotient $q$,so $10q = 230$,which implies $q = 23$.
Using the division algorithm: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$.
Substituting the values: $\text{Dividend} = (230 \times 23) + 46$.
Calculating the product: $230 \times 23 = 5290$.
Adding the remainder: $5290 + 46 = 5336$.
Therefore,the dividend is $5336$.

(C) The relationship between dividend, divisor, quotient, and remainder is given by the formula: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$.
Given: $\text{Divisor} = 44$, $\text{Quotient} = 432$, $\text{Remainder} = 0$.
Therefore, $\text{Dividend} = (44 \times 432) + 0 = 19008$.
Now, we need to find the remainder when $19008$ is divided by $31$.
Performing the division: $19008 \div 31 = 613$ with a remainder.
$19008 = 31 \times 613 + 5$.
Thus, the remainder is $5$.

(B) Let the number be $N$. According to the division algorithm,$N = 729q + 56$,where $q$ is the quotient.
We want to find the remainder when $N$ is divided by $27$.
We can rewrite $N$ as: $N = (27 \times 27)q + 56$.
Since $729 = 27^2$,$729$ is a multiple of $27$.
Now,divide $56$ by $27$ to find the remainder: $56 = 27 \times 2 + 2$.
Therefore,$N = 27(27q + 2) + 2$.
This shows that when $N$ is divided by $27$,the remainder is $2$.

(D) The smallest $6$-digit number is $100000$.
Divide $100000$ by $108$ to find the remainder:
$100000 \div 108 = 925$ with a remainder of $100$.
To find the smallest $6$-digit number divisible by $108$,we calculate:
$108 - \text{remainder} = 108 - 100 = 8$.
Add this difference to the smallest $6$-digit number:
$100000 + 8 = 100008$.
Thus,$100008$ is the smallest $6$-digit number divisible by $108$.

(C) Let the number be $N.$ According to the problem,when $N$ is divided by $56,$ the remainder is $29.$
This can be expressed as $N = 56k + 29,$ where $k$ is an integer.
We want to find the remainder when $N$ is divided by $8.$
Substitute $N$ in the expression: $N = 8(7k) + 29.$
Now,divide $29$ by $8$ to find the remainder: $29 = 8 \times 3 + 5.$
Therefore,$N = 8(7k) + 8(3) + 5 = 8(7k + 3) + 5.$
When $N$ is divided by $8,$ the remainder is $5.$

(B) Divisor $= 555 + 445 = 1000$
Quotient $= 2 \times (555 - 445) = 2 \times 110 = 220$
Remainder $= 30$
Using the division algorithm: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$
Dividend $= (1000 \times 220) + 30$
Dividend $= 220000 + 30 = 220030$

(B) Let the $2^{\text{nd}}$ number be $x$.
According to the problem:
$1^{\text{st}}$ number $= \frac{x}{2}$
$3^{\text{rd}}$ number $= \frac{x}{4}$
The sum of the three numbers is $2$,so:
$\frac{x}{2} + x + \frac{x}{4} = 2$
To solve for $x$,find a common denominator,which is $4$:
$\frac{2x + 4x + x}{4} = 2$
$\frac{7x}{4} = 2$
$7x = 8$
$x = \frac{8}{7}$
Thus,the $2^{\text{nd}}$ number is $\frac{8}{7}$.

(A) Let the number be $x$.
According to the problem,adding $\frac{1}{2}$ (which is $0.5$) to the number gives $(x + 0.5)$.
Multiplying this sum by $3$ results in $21$,so the equation is $(x + 0.5) \times 3 = 21$.
Dividing both sides by $3$,we get $x + 0.5 = 7$.
Subtracting $0.5$ from both sides,we get $x = 7 - 0.5 = 6.5$.
Therefore,the number is $6.5$.

(C) To find the value of $x$,we add the given decimal numbers:
$x = 0.3 + 0.6 + 0.7 + 0.8$
$x = 2.4$
Now,convert the decimal $2.4$ into a fraction:
$2.4 = \frac{24}{10}$
Simplify the fraction by dividing both the numerator and the denominator by $2$:
$\frac{24 \div 2}{10 \div 2} = \frac{12}{5}$
Convert the improper fraction $\frac{12}{5}$ into a mixed fraction:
$\frac{12}{5} = 2 \frac{2}{5}$
Therefore,the correct option is $C$.

(B) To find the least number that leaves a remainder of $2$ when divided by $6, 9, 12, 15,$ and $18$,we first find the Least Common Multiple $(LCM)$ of these numbers.
Prime factorization:
$6 = 2 \times 3$
$9 = 3^2$
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$18 = 2 \times 3^2$
$LCM = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$.
Since the number must leave a remainder of $2$ in each case,we add the remainder to the $LCM$.
Required number $= LCM + 2 = 180 + 2 = 182$.

(D) Let the number be $x$.
According to the problem,$\frac{3}{4}x = \frac{1}{6}x + 7$.
Subtract $\frac{1}{6}x$ from both sides: $\frac{3}{4}x - \frac{1}{6}x = 7$.
The least common multiple of $4$ and $6$ is $12$.
So,$\frac{9x - 2x}{12} = 7$.
$\frac{7x}{12} = 7$.
Multiply both sides by $12$: $7x = 84$.
Divide by $7$: $x = 12$.
Now,calculate $\frac{5}{3}$ of the number: $\frac{5}{3} \times 12 = 5 \times 4 = 20$.

(D) The sum of the first $n$ odd natural numbers is given by the formula $S_n = n^2$.
The arithmetic mean of a set of numbers is defined as the sum of the numbers divided by the count of the numbers.
For the first $n$ odd natural numbers,the arithmetic mean is $\frac{S_n}{n} = \frac{n^2}{n} = n$.
Given $n = 20$,the arithmetic mean is $20$.

(A) The first $n$ natural numbers are $1, 2, 3, \dots, n$.
The sum of the first $n$ natural numbers is given by the formula $S_n = \frac{n(n+1)}{2}$.
The Arithmetic Mean $(A.M.)$ is defined as the sum of the observations divided by the number of observations.
Therefore,$A.M. = \frac{S_n}{n} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$.

(B) Let the $2$-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
Given that the units digit is twice the tens digit,we have $y = 2x$.
The sum of the digits is $x + y$. According to the problem,if $2$ is subtracted from this sum,the result is $1/6$ of the number:
$x + y - 2 = \frac{1}{6}(10x + y)$
Substitute $y = 2x$ into the equation:
$x + 2x - 2 = \frac{1}{6}(10x + 2x)$
$3x - 2 = \frac{1}{6}(12x)$
$3x - 2 = 2x$
$x = 2$
Since $y = 2x$,we have $y = 2(2) = 4$.
Therefore,the number is $10(2) + 4 = 24$.

(A) To find the smallest number $x$ that must be added to $2000$ to make it divisible by $12, 16, 18$,and $21$,we first find the Least Common Multiple $(LCM)$ of these numbers.
Prime factorization:
$12 = 2^2 \times 3$
$16 = 2^4$
$18 = 2 \times 3^2$
$21 = 3 \times 7$
$LCM = 2^4 \times 3^2 \times 7 = 16 \times 9 \times 7 = 1008$.
Multiples of $1008$ are $1008, 2016, 3024, \dots$
The smallest multiple of $1008$ greater than $2000$ is $2016$.
Therefore,the number to be added is $x = 2016 - 2000 = 16$.
The sum of the digits of $x$ is $1 + 6 = 7$.

(A) To find the unit digit of $(2467)^{153} \times (341)^{72}$,we look at the unit digits of the bases.
$1$. For $(2467)^{153}$,the unit digit is the same as $7^{153}$.
The powers of $7$ follow a cycle of $4$: $7^1=7, 7^2=49, 7^3=343, 7^4=2401$.
Dividing the exponent $153$ by $4$,we get $153 = 4 \times 38 + 1$. Thus,$7^{153}$ has the same unit digit as $7^1$,which is $7$.
$2$. For $(341)^{72}$,the unit digit is the same as $1^{72}$.
Since $1$ raised to any power is $1$,the unit digit is $1$.
$3$. The unit digit of the product is the product of the unit digits: $7 \times 1 = 7$.

(B) The number $x$ is of the form $x = \operatorname{LCM}(5, 6, 7, 8) \times k + 3$,where $k$ is a positive integer.
First,find the $\operatorname{LCM}$ of $5, 6, 7, 8$:
$5 = 5$
$6 = 2 \times 3$
$7 = 7$
$8 = 2^3$
$\operatorname{LCM}(5, 6, 7, 8) = 5 \times 3 \times 7 \times 8 = 840$.
So,$x = 840k + 3$.
We are given that $x$ is divisible by $9$. Thus,$840k + 3 \equiv 0 \pmod{9}$.
Since $840 = 9 \times 93 + 3$,we have $840 \equiv 3 \pmod{9}$.
So,$3k + 3 \equiv 0 \pmod{9}$,which implies $3(k + 1) = 9n$ for some integer $n$,or $k + 1$ must be a multiple of $3$.
The smallest positive integer $k$ is $2$ (since for $k=1$,$843$ is not divisible by $9$).
For $k = 2$,$x = 840(2) + 3 = 1680 + 3 = 1683$.
Check divisibility by $9$: $1 + 6 + 8 + 3 = 18$,which is divisible by $9$.
The sum of the digits of $x = 1683$ is $1 + 6 + 8 + 3 = 18$.

(B) Let the number be $N$. According to the problem,$N = 361q + 47$,where $q$ is the quotient.
We can rewrite $361$ as $19 \times 19$. So,$N = 19(19q) + 47$.
To find the remainder when $N$ is divided by $19$,we divide the expression by $19$.
The term $19(19q)$ is perfectly divisible by $19$.
Thus,the remainder depends on dividing $47$ by $19$.
$47 = 19 \times 2 + 9$.
The remainder is $9$.

(C) To compare the numbers,we express them with a common exponent of $10$:
$3^{50} = (3^5)^{10} = 243^{10}$
$4^{40} = (4^4)^{10} = 256^{10}$
$5^{30} = (5^3)^{10} = 125^{10}$
$6^{20} = (6^2)^{10} = 36^{10}$
Comparing the bases $243, 256, 125,$ and $36$,we see that $256$ is the largest.
Therefore,$4^{40}$ is the greatest number.

(D) Let the number be $x$.
According to the problem,the number exceeds its two-fifths by $75$,which can be written as:
$x - \frac{2}{5}x = 75$
Subtracting the fractions:
$\frac{5x - 2x}{5} = 75$
$\frac{3x}{5} = 75$
Now,solve for $x$:
$3x = 75 \times 5$
$3x = 375$
$x = \frac{375}{3}$
$x = 125$
Therefore,the number is $125$.

(A) Let the first number be $x$.
Then the second number is $\frac{2}{5}x$.
According to the problem,the sum of these two numbers is $50$:
$x + \frac{2}{5}x = 50$
$\frac{5x + 2x}{5} = 50$
$\frac{7x}{5} = 50$
$7x = 250$
$x = \frac{250}{7}$
Now,find the second number:
$\text{Second number} = \frac{2}{5} \times \frac{250}{7} = \frac{2 \times 50}{7} = \frac{100}{7}$
Thus,the two numbers are $\frac{250}{7}$ and $\frac{100}{7}$.

(B) Given the equation: $x[-2\{-4(-a)\}]+5[-2\{-2(-a)\}]=4 a$
First,simplify the terms inside the brackets:
$-4(-a) = 4a$
$-2(4a) = -8a$
So,the first term is $x(-8a) = -8ax$.
Next,simplify the second part:
$-2(-a) = 2a$
$-2(2a) = -4a$
$5(-4a) = -20a$
Substitute these back into the equation:
$-8ax - 20a = 4a$
Add $20a$ to both sides:
$-8ax = 24a$
Divide both sides by $-8a$ (assuming $a \neq 0$):
$x = \frac{24a}{-8a}$
$x = -3$