Numbers Questions in English

(C) Given, remainder $= 46$.
According to the problem, the divisor is $5$ times the remainder.
Divisor $= 5 \times 46 = 230$.
Also, the divisor is $10$ times the quotient.
$10 \times \text{quotient} = 230$.
Quotient $= 230 / 10 = 23$.
The formula for the dividend is: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$.
Dividend $= (230 \times 23) + 46$.
Dividend $= 5290 + 46 = 5336$.

(B) The smallest four-digit number is $1000$.
To find the smallest four-digit number divisible by $71$,we first divide $1000$ by $71$.
$1000 \div 71 = 14$ with a remainder of $6$.
This means $71 \times 14 = 994$,which is a three-digit number.
The next multiple of $71$ will be $71 \times 15$.
$71 \times 15 = 1065$.
Alternatively,the smallest four-digit number divisible by $71$ is calculated as $1000 + (71 - 6) = 1000 + 65 = 1065$.

(A) To find the number of integers up to $100$ that are divisible by $7$,we perform the division $100 \div 7$.
The quotient obtained is $14$ with a remainder of $2$ $(100 = 7 \times 14 + 2)$.
Therefore,there are $14$ multiples of $7$ between $1$ and $100$ (i.e.,$7, 14, 21, \dots, 98$).

(C) To find the number of integers up to $500$ that are divisible by $23$,we perform the division $500 \div 23$.
$500 = 23 \times 21 + 17$.
The quotient obtained is $21$.
Therefore,there are $21$ numbers up to $500$ that are divisible by $23$.

(B) number is divisible by both $2$ and $3$ if and only if it is divisible by their least common multiple $(L.C.M.)$.
The $L.C.M.$ of $2$ and $3$ is $6$.
To find the number of integers up to $200$ that are divisible by $6$,we divide $200$ by $6$ and take the quotient.
$200 \div 6 = 33.33...$
The integer part of the quotient is $33$.
Therefore,there are $33$ numbers up to $200$ that are divisible by both $2$ and $3$.

(A) To find the number of integers between $100$ and $300$ divisible by $11$,we identify the first and last multiples of $11$ in this range.
The first multiple of $11$ greater than $100$ is $110$ $(11 \times 10)$.
The last multiple of $11$ less than $300$ is $286$ $(11 \times 26)$.
This forms an arithmetic progression: $110, 121, 132, \dots, 286$.
Using the formula for the $n^{th}$ term of an arithmetic progression: $a_n = a + (n - 1)d$,where $a = 110$,$d = 11$,and $a_n = 286$.
$286 = 110 + (n - 1)11$
$176 = (n - 1)11$
$n - 1 = 16$
$n = 17$.
Alternatively,the number of multiples of $k$ up to $N$ is given by $\lfloor N/k \rfloor$.
The number of multiples of $11$ up to $300$ is $\lfloor 300/11 \rfloor = 27$.
The number of multiples of $11$ up to $100$ is $\lfloor 100/11 \rfloor = 9$.
The number of multiples between $100$ and $300$ is $27 - 9 = 18$.

(B) number divisible by $2, 3,$ and $7$ must be divisible by their $L.C.M.$
Since $2, 3,$ and $7$ are prime numbers,their $L.C.M. = 2 \times 3 \times 7 = 42.$
We need to find the number of multiples of $42$ between $150$ and $500$.
First,find the smallest multiple of $42$ greater than $150$: $42 \times 4 = 168$.
Next,find the largest multiple of $42$ less than $500$: $500 \div 42 \approx 11.9$,so $42 \times 11 = 462$.
The multiples are $42 \times 4, 42 \times 5, \dots, 42 \times 11$.
The number of such terms is $11 - 4 + 1 = 8$.

(A) The least number of six figures is $100000$.
First,we divide $100000$ by $357$ to find the remainder:
$100000 \div 357 = 280$ with a remainder of $40$.
To find the least six-figure number divisible by $357$,we add the difference between the divisor and the remainder to the dividend:
$100000 + (357 - 40) = 100000 + 317 = 100317$.
Now,we need to find the number to be added to $5555$ to get $100317$:
$x + 5555 = 100317$
$x = 100317 - 5555 = 94762$.
Thus,the required number is $94762$.

(A) To find the nearest number to $58701$ divisible by $567$,we first divide $58701$ by $567$.
$58701 \div 567 = 103$ with a remainder of $300$.
Since the remainder $300$ is greater than half of the divisor $(567 / 2 = 283.5)$,
we add the difference $(567 - 300) = 267$ to the original number.
Nearest number $= 58701 + 267 = 58968$.

(A) Let the number be $N = 3422213xy$,where $x$ and $y$ are the missing digits.
For $N$ to be divisible by $99$,it must be divisible by both $9$ and $11$.
For divisibility by $9$,the sum of the digits must be a multiple of $9$:
$3+4+2+2+2+1+3+x+y = 17+x+y$. For this to be divisible by $9$,$x+y$ must be $1$ or $10$ or $19$. Since $x, y$ are digits,$x+y=10$ (as $17+10=27$,which is divisible by $9$).
For divisibility by $11$,the difference between the sum of digits at odd places and even places must be $0$ or a multiple of $11$:
Sum of digits at odd places (from right): $y+1+2+4 = y+7$.
Sum of digits at even places (from right): $x+3+2+2+3 = x+10$.
Difference: $(y+7) - (x+10) = y-x-3$.
For divisibility by $11$,$y-x-3 = 0$ or $y-x-3 = -11$.
Case $1$: $y-x = 3$. We have $x+y=10$ and $y-x=3$. Adding these,$2y=13$ (no integer solution).
Case $2$: $y-x-3 = -11 \implies y-x = -8$. We have $x+y=10$ and $y-x=-8$. Adding these,$2y=2 \implies y=1$. Then $x=9$.
Thus,the digits are $9$ and $1$.

(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 $11$.
Let the missing digit be $x$.
The number is $5x3457$.
Sum of digits at odd places (from right) $= 7 + 4 + x = 11 + x$.
Sum of digits at even places (from right) $= 5 + 3 + 5 = 13$.
For the number to be divisible by $11$,the difference must be $0$ or a multiple of $11$.
$(11 + x) - 13 = 0$ or $(11 + x) - 13 = 11$.
Case $1$: $x - 2 = 0 \implies x = 2$.
Case $2$: $x - 2 = 11 \implies x = 13$ (not possible as $x$ must be a single digit).
Therefore,the least value is $2$.

(B) To find the nearest number to $1,000,000$ divisible by $537$,we first divide $1,000,000$ by $537$.
$1,000,000 \div 537 = 1862$ with a remainder of $106$.
Since the remainder $106$ is less than half of the divisor $(537 / 2 = 268.5)$,the nearest multiple is obtained by subtracting the remainder from $1,000,000$.
$1,000,000 - 106 = 999,894$.
Thus,the nearest whole number is $999,894$.

(B) To find the smallest number between $400$ and $500$ divisible by $9$,we first divide $400$ by $9$.
$400 \div 9 = 44$ with a remainder of $4$.
To find the next multiple of $9$ greater than $400$,we subtract the remainder from the divisor and add it to the number:
$9 - 4 = 5$.
Adding this to $400$,we get $400 + 5 = 405$.
Since $405$ is between $400$ and $500$ and is divisible by $9$,it is the smallest such number.

(A) The greatest number of five digits is $99,999$.
To find the greatest five-digit number divisible by $231$,we divide $99,999$ by $231$.
$99,999 \div 231 = 432$ with a remainder.
$99,999 = 231 \times 432 + 207$.
The remainder is $207$.
To find the required number,we subtract the remainder from the greatest five-digit number:
$99,999 - 207 = 99,792$.
Thus,$99,792$ is the greatest five-digit number divisible by $231$.

(D) To find the number nearest to $16386$ that is divisible by $425$,we first divide $16386$ by $425$.
$16386 \div 425 = 38$ with a remainder of $361$.
This means $16386 = (425 \times 38) + 361$.
The two multiples of $425$ closest to $16386$ are $(425 \times 38) = 16150$ and $(425 \times 39) = 16575$.
Alternatively,we can calculate the distance to the nearest multiple:
The remainder is $361$. The distance to the next multiple is $425 - 361 = 64$.
The distance to the previous multiple is $361$.
Since $64 < 361$,the nearest multiple is $16386 + 64 = 16450$.

(C) To find the number to be subtracted,we divide $9269$ by $73$.
$9269 \div 73 = 126$ with a remainder.
$9269 = 73 \times 126 + 71$.
Since the remainder is $71$,subtracting $71$ from $9269$ will make the resulting number exactly divisible by $73$.
Therefore,the required least number is $71$.

(A) To find the number to be added,first divide $15463$ by $107$.
$15463 \div 107 = 144$ with a remainder of $55$.
Since the remainder is $55$,the number that must be added to $15463$ to make it exactly divisible by $107$ is the difference between the divisor and the remainder.
Required number $= 107 - 55 = 52$.

(C) To find the number just greater than $5000$ that is exactly divisible by $73$,we first divide $5000$ by $73$.
$5000 \div 73 = 68$ with a remainder of $36$.
To find the next multiple of $73$ greater than $5000$,we calculate the difference between the divisor and the remainder: $73 - 36 = 37$.
Adding this difference to the original number: $5000 + 37 = 5037$.
Therefore,the required number is $5037$.

(D) Let the two numbers be $a$ and $b$.
Given that the sum of the numbers is $a + b = 100$.
Given that the difference of the numbers is $a - b = 37$.
We need to find the difference of their squares,which is $a^2 - b^2$.
Using the algebraic identity $a^2 - b^2 = (a + b)(a - b)$,
Substitute the given values: $a^2 - b^2 = 100 \times 37$.
Therefore,$a^2 - b^2 = 3700$.

(B) Let the number of times $79$ is subtracted be $n$.
According to the problem,we have the equation:
$50000 - (n \times 79) = 43759$
Rearranging the terms to solve for $n$:
$n \times 79 = 50000 - 43759$
$n \times 79 = 6241$
$n = \frac{6241}{79}$
$n = 79$
Therefore,$79$ must be subtracted $79$ times.

(C) Let the two numbers be $3x$ and $4x$,where $x$ is a common multiplier.
According to the problem,the sum of these two numbers is $420$.
So,$3x + 4x = 420$.
$7x = 420$.
$x = 420 / 7 = 60$.
The two numbers are $3 \times 60 = 180$ and $4 \times 60 = 240$.
Therefore,the greater of the two numbers is $240$.

(C) Let the two consecutive numbers be $x$ and $(x+1)$.
According to the problem,the difference between their squares is $35$.
So,$(x+1)^{2} - x^{2} = 35$.
Expanding the square: $(x^{2} + 2x + 1) - x^{2} = 35$.
Simplifying the equation: $2x + 1 = 35$.
Subtracting $1$ from both sides: $2x = 34$.
Dividing by $2$: $x = 17$.
Therefore,the two consecutive numbers are $17$ and $17+1 = 18$.

(B) Let the number be $x$.
According to the problem,three-fourths of one-fifth of $x$ is $60$.
This can be written as: $\frac{3}{4} \times \frac{1}{5} \times x = 60$.
Simplifying the equation: $\frac{3}{20} \times x = 60$.
Multiplying both sides by $20$: $3x = 60 \times 20$.
$3x = 1200$.
Dividing by $3$: $x = \frac{1200}{3} = 400$.
Therefore,the number is $400$.

(A) Let the two numbers be $a$ and $b$.
Given that the sum of their squares is $a^2 + b^2 = 80$.
Also,the square of their difference is $(a - b)^2 = 36$.
We know the algebraic identity $(a - b)^2 = a^2 + b^2 - 2ab$.
Substituting the given values into the identity:
$36 = 80 - 2ab$.
Rearranging the terms to solve for $2ab$:
$2ab = 80 - 36 = 44$.
Dividing by $2$ to find the product $ab$:
$ab = 22$.
Therefore,the product of the two numbers is $22$.

(A) Let the number be $N$ and the quotient be $K.$
According to the division algorithm,$N = 357 \times K + 37.$
We need to find the remainder when $N$ is divided by $17.$
$N = (17 \times 21) \times K + (17 \times 2 + 3).$
$N = 17 \times (21K + 2) + 3.$
Since $N$ is expressed in the form $17 \times Q' + R,$ where $Q' = 21K + 2$ and $R = 3,$ the remainder is $3.$

(B) Let the two numbers be $a$ and $b$.
Given that the product of the numbers is $ab = 120$.
Given that the sum of their squares is $a^2 + b^2 = 289$.
We know the algebraic identity: $(a + b)^2 = a^2 + b^2 + 2ab$.
Substituting the given values into the identity:
$(a + b)^2 = 289 + 2(120)$.
$(a + b)^2 = 289 + 240$.
$(a + b)^2 = 529$.
Taking the square root on both sides,we get $a + b = \sqrt{529} = 23$.
Thus,the sum of the two numbers is $23$.

(C) Let the three numbers be $4x$,$5x$,and $6x$.
The average of these numbers is given by the formula: $\frac{\text{Sum of numbers}}{\text{Count of numbers}} = \text{Average}$.
Substituting the given values: $\frac{4x + 5x + 6x}{3} = 25$.
Simplifying the equation: $\frac{15x}{3} = 25$,which gives $5x = 25$.
Solving for $x$: $x = \frac{25}{5} = 5$.
The largest number is $6x$. Substituting the value of $x$: $6 \times 5 = 30$.

(A) Let the required number be $x$.
According to the problem,the number exceeds $20 \%$ of itself by $40$.
This can be written as the equation: $x - 0.20x = 40$.
Simplifying the equation: $0.80x = 40$.
Solving for $x$: $x = \frac{40}{0.80} = \frac{4000}{80} = 50$.
Therefore,the number is $50$.

(C) Let the number be $x$.
According to the problem,$16 \%$ of $40 \%$ of $x$ is $8$.
This can be written as: $\frac{16}{100} \times \frac{40}{100} \times x = 8$.
Simplify the expression: $\frac{16 \times 40}{10000} \times x = 8$.
$\frac{640}{10000} \times x = 8$.
$0.064 \times x = 8$.
$x = \frac{8}{0.064}$.
$x = \frac{8000}{64}$.
$x = 125$.

(B) Let the number be $N = d_2 d_1 d_0 341$,where $d_2, d_1, d_0$ are the missing digits.
Since $4767$ divides $N$ exactly,we can write $N = 4767 \times Q$.
The last digit of $N$ is $1$ and the last digit of $4767$ is $7$.
For the product to end in $1$,the last digit of the quotient $Q$ must be $3$ (since $7 \times 3 = 21$).
Let us estimate the quotient $Q$ by dividing $xxx,341$ by $4767$.
$4767 \times 100 = 476700$.
$N - 476700 = xxx,341 - 476,700 = yyy,641$.
Now,$4767 \times 20 = 95340$.
$N - 476700 - 95340 = 586341 - 572040 = 14301$.
Finally,$4767 \times 3 = 14301$.
Thus,$Q = 100 + 20 + 3 = 123$.
$N = 4767 \times 123 = 586341$.
The missing digits are $5, 8, 6$.

(A) Let the number be $N$ and the divisor be $D$.
According to the division algorithm,$N = D \times q_1 + 241$,where $q_1$ is the quotient.
This implies $N > 241$,so $D > 241$.
When $2N$ is divided by $D$,the remainder is $112$.
So,$2N = D \times q_2 + 112$.
Substituting $N$ from the first equation: $2(D \times q_1 + 241) = D \times q_2 + 112$.
$2D \times q_1 + 482 = D \times q_2 + 112$.
$482 - 112 = D \times q_2 - 2D \times q_1$.
$370 = D(q_2 - 2q_1)$.
This shows that $D$ must be a factor of $370$.
Since $D > 241$,the only factor of $370$ greater than $241$ is $370$ itself.
Therefore,the divisor is $370$.

(B) Let the two numbers be $N_1$ and $N_2$,and the divisor be $d$.
According to the problem,$N_1 = q_1d + 43$ and $N_2 = q_2d + 37$,where $q_1$ and $q_2$ are quotients.
The sum of the numbers is $N_1 + N_2 = (q_1 + q_2)d + (43 + 37) = (q_1 + q_2)d + 80$.
When the sum is divided by $d$,the remainder is $13$. Thus,$80$ divided by $d$ must leave a remainder of $13$.
This implies $80 = kd + 13$ for some integer $k$.
$kd = 80 - 13 = 67$.
Since $67$ is a prime number,the divisor $d$ must be $67$ (as the remainder $13$ must be less than the divisor $d$,and $13 < 67$ holds true).

(D) Let the two numbers be $3x$ and $5x$.
According to the problem,if each number is increased by $10$,the new ratio becomes $5:7$.
So,$\frac{3x + 10}{5x + 10} = \frac{5}{7}$.
Cross-multiplying,we get $7(3x + 10) = 5(5x + 10)$.
$21x + 70 = 25x + 50$.
$25x - 21x = 70 - 50$.
$4x = 20$.
$x = 5$.
Therefore,the numbers are $3(5) = 15$ and $5(5) = 25$.

(A) Let the two parts be $x$ and $(50 - x)$.
According to the problem,the sum of their reciprocals is $1/12$:
$\frac{1}{x} + \frac{1}{50 - x} = \frac{1}{12}$
Taking the common denominator on the left side:
$\frac{(50 - x) + x}{x(50 - x)} = \frac{1}{12}$
$\frac{50}{50x - x^2} = \frac{1}{12}$
Cross-multiplying gives:
$50 \times 12 = 50x - x^2$
$600 = 50x - x^2$
Rearranging into a standard quadratic equation:
$x^2 - 50x + 600 = 0$
Factoring the quadratic equation:
$(x - 30)(x - 20) = 0$
Thus,$x = 30$ or $x = 20$.
If $x = 30$,the other part is $50 - 30 = 20$. If $x = 20$,the other part is $50 - 20 = 30$.
Therefore,the two parts are $20$ and $30$.

(A) Let the seven numbers be $n_1, n_2, n_3, n_4, n_5, n_6, n_7$.
The sum of the seven numbers is given as $n_1 + n_2 + n_3 + n_4 + n_5 + n_6 + n_7 = 235$.
The average of the first three numbers is $23$,so their sum is $n_1 + n_2 + n_3 = 23 \times 3 = 69$.
The average of the last three numbers is $42$,so their sum is $n_5 + n_6 + n_7 = 42 \times 3 = 126$.
Substituting these sums into the total sum equation:
$69 + n_4 + 126 = 235$
$195 + n_4 = 235$
$n_4 = 235 - 195 = 40$.
Therefore,the fourth number is $40$.

(A) Let the two numbers be $a$ and $b$.
Given that the sum of their squares is $a^2 + b^2 = 68$.
Also,the square of their difference is $(a - b)^2 = 36$.
We know the algebraic identity: $(a - b)^2 = a^2 + b^2 - 2ab$.
Substituting the given values into the identity:
$36 = 68 - 2ab$.
Rearranging the terms to solve for $2ab$:
$2ab = 68 - 36$.
$2ab = 32$.
Dividing by $2$:
$ab = 16$.
Therefore,the product of the two numbers is $16$.

(A) number is divisible by $9$ if the sum of its digits is divisible by $9$.
Sum of the digits of $6735K1 = 6 + 7 + 3 + 5 + K + 1 = 22 + K$.
For the number to be divisible by $9$,the sum $22 + K$ must be a multiple of $9$.
The multiples of $9$ are $9, 18, 27, 36, \dots$.
Since $K$ is a single digit $(0-9)$,$22 + K$ must be at least $22$ and at most $22 + 9 = 31$.
The only multiple of $9$ in the range $[22, 31]$ is $27$.
Therefore,$22 + K = 27$.
$K = 27 - 22 = 5$.
Thus,the least value of $K$ is $5$.

(A) number is divisible by $8$ if the number formed by its last three digits is divisible by $8$.
In the given number $7236K2$,the last three digits are $6K2$.
We need to check for which value of $K$ the number $6K2$ is divisible by $8$.
If $K = 3$,the number is $632$. Since $632 / 8 = 79$,it is divisible by $8$.
If $K = 7$,the number is $672$. Since $672 / 8 = 84$,it is divisible by $8$.
Comparing this with the given options,$7$ is present as an option.

(A) number is divisible by $88$ if it is divisible by both $8$ and $11$,since $88 = 8 \times 11$ and $\gcd(8, 11) = 1$.
First,apply the divisibility test for $8$. $A$ number is divisible by $8$ if its last three digits are divisible by $8$. The last three digits are $23y$.
For $23y$ to be divisible by $8$,we check $230 \div 8 = 28$ with remainder $6$. Adding $2$ to $230$ gives $232$,which is $8 \times 29$. Thus,$y = 2$ is the smallest value.
Now the number is $5x4232$. Apply the divisibility test for $11$. $A$ 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 $11$.
Sum of digits at odd places (from right): $2 + 2 + x = x + 4$.
Sum of digits at even places (from right): $3 + 4 + 5 = 12$.
Difference: $(x + 4) - 12 = x - 8$.
For the number to be divisible by $11$,$x - 8 = 0$,which gives $x = 8$.
Therefore,the least values are $x = 8$ and $y = 2$.

(A) Let the first part be $x$. Then the second part is $(24 - x)$.
According to the problem,$7$ times the first part added to $5$ times the second part equals $146$.
So,$7x + 5(24 - x) = 146$.
Expanding the equation: $7x + 120 - 5x = 146$.
Simplifying: $2x + 120 = 146$.
Subtracting $120$ from both sides: $2x = 26$.
Dividing by $2$: $x = 13$.
Therefore,the first part is $13$.

(B) Let the second number be $x$.
Then,the first number is $2x$.
The third number is one-third of the first,which is $\frac{1}{3} \times 2x = \frac{2x}{3}$.
According to the problem,the sum of these three numbers is $132$:
$2x + x + \frac{2x}{3} = 132$
Multiply the entire equation by $3$ to clear the fraction:
$6x + 3x + 2x = 396$
$11x = 396$
$x = \frac{396}{11} = 36$.
Thus,the second number is $36$.

(C) To be divisible by both $5$ and $9,$ the number must be divisible by their least common multiple $(LCM)$.
Since $5$ and $9$ are coprime,their $LCM = 5 \times 9 = 45.$
Now,divide $7231$ by $45$ to find the remainder:
$7231 \div 45 = 160$ with a remainder of $31.$
To make the number exactly divisible by $45,$ we need to add the difference between the divisor and the remainder to the original number.
Required number $= 45 - 31 = 14.$

(A) number divisible by both $3$ and $11$ must be divisible by their least common multiple,which is $33.$
To find the nearest number,we divide $9231$ by $33.$
$9231 \div 33 = 279$ with a remainder of $24.$
Since the remainder $24$ is greater than half of the divisor $(33 / 2 = 16.5)$,the next multiple of $33$ is closer to $9231$ than the current quotient's multiple.
The nearest number is $9231 + (33 - 24) = 9231 + 9 = 9240.$

(A) The first four prime numbers are $2, 3, 5,$ and $7$.
Their product is $2 \times 3 \times 5 \times 7 = 210$.
To find the nearest number to $12199$ divisible by $210$,we divide $12199$ by $210$:
$12199 \div 210 = 58$ with a remainder of $19$.
Since the remainder $19$ is less than half of the divisor $(210 / 2 = 105)$,the nearest multiple is obtained by subtracting the remainder from the dividend.
Therefore,the required number is $12199 - 19 = 12180$.

(A) Let the two numbers be $x$ and $y$.
According to the problem:
$x^{2} + y^{2} = 90$ $.....(1)$
$(x - y)^{2} = 46$ $.....(2)$
We know that $(x - y)^{2} = x^{2} + y^{2} - 2xy$.
Substituting the values from equation $(1)$ and $(2)$ into the identity:
$46 = 90 - 2xy$
Rearranging the terms to solve for $xy$:
$2xy = 90 - 46$
$2xy = 44$
$xy = 22$
Therefore,the product of the two numbers is $22$.

(A) Let the number be $x$.
Given that $40 \%$ of $x = 360$.
Therefore,$\frac{40}{100} \times x = 360$.
$x = \frac{360 \times 100}{40} = 9 \times 100 = 900$.
Now,we need to find $15 \%$ of this number $x$.
$15 \%$ of $900 = \frac{15}{100} \times 900 = 15 \times 9 = 135$.

(A) Let the two-digit number be $10x + y$,where $x$ is the tens digit and $y$ is the units digit.
According to the problem,the sum of the digits is $x + y = 8$ $...(1)$.
When the digits are reversed,the new number is $10y + x$.
Given that the new number is $54$ more than the original number: $10y + x = (10x + y) + 54$.
Simplifying this equation: $9y - 9x = 54$,which gives $y - x = 6$ $...(2)$.
Adding equations $(1)$ and $(2)$: $(x + y) + (y - x) = 8 + 6$,so $2y = 14$,which means $y = 7$.
Substituting $y = 7$ into equation $(1)$: $x + 7 = 8$,so $x = 1$.
The original number is $10x + y = 10(1) + 7 = 17$.

(A) Let the number be $N$. The divisors are $3, 4, 6$ and the respective remainders are $1, 2, 4$.
Observe the difference between the divisor and the remainder:
$3 - 1 = 2$
$4 - 2 = 2$
$6 - 4 = 2$
The constant difference is $2$.
To find $N$,we first find the $LCM$ of $(3, 4, 6)$,which is $12$.
The general form of the number is $N = 12k - 2$,where $k$ is an integer.
For the largest two-digit number,we set $12k - 2 < 100$,which gives $12k < 102$,so $k < 8.5$.
Taking $k = 8$,we get $N = 12(8) - 2 = 96 - 2 = 94$.
Now,we find the remainder when $N = 94$ is divided by $5$.
$94 = 5 \times 18 + 4$.
Therefore,the remainder is $4$.

(C) Given the equation: $A + B + AB = 65$.
Add $1$ to both sides to factorize the expression:
$A + B + AB + 1 = 65 + 1$
$A(1 + B) + 1(1 + B) = 66$
$(A + 1)(B + 1) = 66$
Since $A$ and $B$ are positive integers,$(A + 1)$ and $(B + 1)$ must be factors of $66$ greater than $1$.
The pairs of factors of $66$ are $(1, 66), (2, 33), (3, 22), (6, 11)$.
Since $A, B \leq 15$,then $(A + 1) \leq 16$ and $(B + 1) \leq 16$.
The only pair satisfying this condition is $(6, 11)$.
Thus,$A + 1 = 6$ and $B + 1 = 11$ (or vice versa).
$A = 5$ and $B = 10$.
The difference between $A$ and $B$ is $|10 - 5| = 5$.

(D) To compare fractions,we can find the reciprocal of each fraction and compare their values. The fraction with the smallest reciprocal will be the largest fraction.
$1$. Reciprocal of $\frac{5}{113}$ is $\frac{113}{5} = 22.6$
$2$. Reciprocal of $\frac{7}{120}$ is $\frac{120}{7} \approx 17.14$
$3$. Reciprocal of $\frac{13}{145}$ is $\frac{145}{13} \approx 11.15$
$4$. Reciprocal of $\frac{17}{160}$ is $\frac{160}{17} \approx 9.41$
Comparing the results: $22.6 > 17.14 > 11.15 > 9.41$.
Since $\frac{160}{17}$ is the smallest value among the reciprocals,$\frac{17}{160}$ is the largest fraction.