A number when divided.................Part 2!

It was just a trial and error.If the number is 51,it satisfies both the conditions given above

Step 1: Let the dividing number to $a$ .

Step 2: After, let the divisor to $b$ .

Step 3: Then let the quotient to $c$ .

Step 4: Next, set a simultaneous equation that $\frac{a}{b} = c \cdots 27$ and $2( \frac{a}{b} ) = c \cdots 3$ .

Step 5: Solve the simultaneous equation.

Then $b$ must not be $0$ because dividing with $0$ cannot get the answer.

We knew that $b$ is not $0$ , then lets multiply $b$ at the both side.

Then it becomes $a = b \times c +27$ and $2a = b \times c +3$ .

Let's solve it.

$2a = 2b \times 2c + 54$ and $2a = b \times c + 3$ .

Then, $0 = b \times c + 51$ .

Switch sides, then it becomes $b \times c + 51 = 0$ .

Then let's substitute the given numbers to $b$ .

1 $34 \times c + 51 = 0$ .

$34c = -51$ .

$c = -\frac{3}{2}$ is no a solution because the quotient cannot be a fraction.

2 $54 \times c + 51 = 0$ .

$54c = -51$ .

$c = -\frac{17}{18}$ is not solution because the quotient cannot be a fraction.

3 $51 \times c + 51 = 0$ .

$51c = -51$ .

$c = -1$ is the solution because the quotient can be negative and it is not a fraction.

So $b = 51$ is the answer.

Answer : $\boxed{51}$ .

Let the number and divisor be $N$ and $d$ respectively. We are given that, for some integers $(x,y)$ , $N=d\cdot x+27$ $2N=d\cdot y+3\;\;\;\;$ Multiplying the first equation by $2$ , $2N=d\cdot (2x)+54=d\cdot y+3$ $\implies d(y-2x)=51$ Therefore, $d$ is a divisor of $51\implies d\in {(1,3,17,51)}$ . But $N$ leaves a remainder $27$ when divided by $d$ , which implies that $d>27$ .

Hence, $\boxed{d=51}$ is the only possibility.