A Division has different results (1)

We want $N = 51x = 25y + 9$ for some integers $x,y$ . But then $\begin{aligned} 51x - 25y & = \; 9 \; = \; 9(51 - 2\times25) \\ 51(x-9) & = \; 25(y -18) \end{aligned}$ SInce $51$ and $25$ have no common factor, we deduce that $x-9 = 25z$ and $y-18=51z$ for some integer $z$ . Thus $N = 51x = 459 + 1275z$ for any integer $z$ . Thus $\boxed{459}$ is a solution, but so is $1734$ and infinitely many other numbers also work

Assuming that $P$ is the integer solution to $N/25.5$ , here’s what we know:

$N/25.5=P$

^this simplifies to $N=25.5P$ , and

$N/25=P\text{ }9/25$

For a fraction of the form A $^B/_C$ , the simplified form would be:

$^{(A*C)+B}/_C$

So $P\text{ }9/25$ is the same as:

$^{25P+9}/_{25}$

So now,

$N/25= (25P+9)/25$

which simplifies to

$N=25P+9$

Now, we have two equations for $N$ :

$N=25.5P$

$N=25P+9$

Setting these equations equal to each other, we find that $P=18$ .

So if $P=18$ , then plugging 18 into the original equations shows that $N=459$ .

Using the formula:Divisor×Quotient+Remainder=Dividend

Taking Divisor as 25,Quotient as X,Remainder as 9,Dividend as N

25X+9=N ...eq.1

Taking Divisor as 25.5,Quotient as X,Remainder as 0,Dividend as N

25.5X+0=N ...eq.2

Solving both equation we get

X=18 Putting X in eq.2 we get N as 459.

The 2nd expression indicates the quotient must be even, so we can use Chinese Remainder Theorem to solve.

We set the system of congruences $\begin{cases} N\equiv 9\mod{25} \\ N\equiv 0\mod{51} \end{cases}$ ,

then we get $N=459+1275k$ .

When $k=0$ , the minimum $N$ is $\boxed{459}$ .