K and m are integers > 0
Target question => What is the remainder when x is divided by m + 1?
The remainder r can be represented as => a(m+1) +1
where a is an integer > 0 and r will definitely be a number within the range 0 < or = r < (m+1)
Statement 1 => k - m when divided by m + 1 leaves a remainder 1
$$k-m=b(m+1)+1;where\ b\ is\ an\ integer>or=0$$
$$k=b(m+1)+1+m$$
$$k=bm+b+m+1$$
$$\frac{k}{\left(m+1\right)}=\frac{\left(m+1\right)\left(b+1\right)}{\left(m+1\right)}$$
$$\frac{k}{\left(m+1\right)}=\left(b+1\right)+0$$
The remainder is 0, statement 1 is SUFFICIENT
Statement 2 => k/3 and k/2, when divided by m + 1, leaves a remainder equal to 4 and 3 respectively
$$\frac{k}{3}=c\left(m+1\right)+4\ where\ c\ \ge0$$
$$k=3c\left(m+1\right)+12.....eqn\ 1$$
$$\frac{k}{2}=d\left(m+1\right)+3\ where\ d\ \ge0$$
$$k=2d\left(m+1\right)+6......eqn\ 2$$
$$multiply\ eqn\ 2\ by\ 2$$
$$2k=4d\left(m+1\right)+12.....eqn\ 3$$
$$subtract\ eqn\ 1\ from\ eqn\ 3$$
$$2k=4d\left(m+1\right)+12$$
$$k=3c\left(m+1\right)+12$$
$$k\left[4d\left(\ m+1\right)\right]-\left[3c\left(m+1\right)\right]+0$$
$$\frac{k}{\left(m+1\right)}=\frac{\left(m+1\right)\left(4d-3c\right)+0}{\left(m+1\right)}$$
$$\frac{k}{\left(m+1\right)}=\left(4d-3c\right)+0\ the\ remainder\ is\ 0,\ statement\ 2\ is\ SUFFICIENT$$
Each statement alone is SUFFICIENT,
Answer = D
