A number when divided by............Part 3!

Using remainder theorem i.e. p(x)=q(x)*g(x)+r(x).

Let p(x) be X and 'a' = quotient

It can be writen in the form of

X=237*a+192,where a = q(x) , 237 = g(x) and r(x)=193

similarly consider p(x)=Y

So,Y= 117*a+108 (given that quotient is equal)

now add X and Y ,

X+Y=354a + 300

Now divide X+Y by 118

354a+300÷118 gives a quotient of 3a +2 leaving a remainder of 64

THEREFORE THE REMAINDER =64

It is given that, $X = 237k+ 192$ $\ldots (1)$ . And $Y= 117k + 108$ $\ldots (2)$

Now $(1)+(2)$

$X + Y = k(237+117) + 300$ $X+Y= 118 \cdot 3k + 300$

And $300 \equiv 64 \pmod{118}$

Therefore $\boxed{64}$ is the answer

For the divisor $237$ we have $237 \equiv 1 \quad \textrm{mod}\ 118$ while the divisor $117$ we have $117 \equiv -1\quad \textrm{mod}\ 118.$ Since the quotients of the divisions are the same, $X+Y \equiv 192 + 108 \equiv 74 - 10 = 64 \quad \textrm{mod}\ 118.$

In other words (using less words and more formulas)

$X = 237q + 192 = 236q + q + 192 = (2q)118 + q + 192$ $Y = 117q + 108 = 118q - q + 108$

$X + Y = (3q)118 + 300 = (3q)118 + (2)118 + 64 = (3q +2)118 + 64$

Hence the answer is $64$ .