Stealthy Numbers

Call the divisors in question $\alpha$ and $\beta$ , so that $N - \alpha = 270$ and $N - \beta = 280$ . Then

$\begin{array}{rrlll} & \alpha &= \ \ N - 270 & & (1)\\ & \beta &= \ \ N - 280 & & (2)\\ \implies & \alpha - \beta &= \ \ 10 & & (3) \end{array}$

Since $\alpha \ \Big\rvert \ N$ , it follows from equation (1) that $\alpha \ \Big\rvert \ 270$ . Similarly, equation (2) implies that $\beta \ \Big\rvert \ 280$ . Combining that with equation (3) means that we are seeking divisors of $270$ that are exactly ten more than some divisors of $280$ .

Listing the divisors of $270$ and $280$ :

$\begin{array}{ll} & {\boldsymbol{270:}} \quad & 1, 2, 3, 5, 6, 9, 10, {\color{#D61F06}{15}}, {\color{#20A900}{18}}, 27, {\color{#3D99F6}{30}}, {\color{#EC7300}{45}}, 54, 90, 135, 270 \\ \\ & {\boldsymbol{280:}} \quad & 1, 2, 4, {\color{#D61F06}{5}}, 7, {\color{#20A900}{8}}, 10, 14, {\color{#3D99F6}{20}}, 28, {\color{#EC7300}{35}}, 40, 56, 70, 140, 280 \end{array}$

Thus $(\alpha, \beta)$ has four solutions: $(15, 5), (18, 8), (30, 20)$ and $(45, 35)$ ; these in turn give us four values for $N: \ 285, 288, 300$ and $315$ and our answer is $\boxed{4}$

Relevant wiki: Diophantine Equations - Solve by Factoring

We can let the two divisors be $d_{1},d_{2}$

If $\text{gcd}(d_{1},d_{2})=g$ ,

then we can let $d_{1}=gx,d_{2}=gy$ such that $x,y$ are co-prime.

We can deduce that $N=\text{lcm}(d_{1},d_{2})\cdot m$ for some positive integer $m$ .

$\text{lcm}(d_{1},d_{2})=gxy\Rightarrow N=gxym$

$N-d_{1}=270\Rightarrow gxym-gx=270\Rightarrow gx(ym-1)=270\Rightarrow gxym(1-\cfrac{1}{ym})=270----(1)$

$N-d_{2}=280\Rightarrow gxym-gy=280\Rightarrow gy(xm-1)=280\Rightarrow gxym(1-\cfrac{1}{xm})=280----(2)$

$(1)\div(2):\cfrac{gxym(1-\cfrac{1}{ym})}{gxym(1-\cfrac{1}{xm})}=\cfrac{270}{280}=\cfrac{27}{28}$

$\cfrac{1-\cfrac{1}{ym}}{1-\cfrac{1}{xm}}=\cfrac{27}{28}$

$\cfrac{\cfrac{ym-1}{ym}}{\cfrac{xm-1}{xm}}=\cfrac{27}{28}$

$\cfrac{xm(ym-1)}{ym(xm-1)}=\cfrac{27}{28}$

For simplicity, let $a=xm, b=ym$

$\Rightarrow \cfrac{a(b-1)}{b(a-1)}=\cfrac{27}{28}$

$28a(b-1)=27b(a-1)$

$28ab-28a=27ab-27b$

$ab-28a+27b=0$

$a(b-28)+27b=0$

$\Rightarrow a(b-28)+27(b-28)=-27\cdot 28$

$(a+27)(b-28)=-27\cdot 28$

$(27+a)(28-b)=27\cdot 28$

$\because b>0\Rightarrow 28-b<28$

and $\because a>0\Rightarrow 27+a>0$

$\therefore 28-b>0$

$0<28-b<28$

By prime factorization of $27\cdot28$ , we get $28-b=1,2,3,4,6,7,9,12,14,18,21,27$

$b=27,26,25,24,22,21,19,16,14,10,7,1$

$b-1=26,25,24,23,21,20,18,17,15,13,9,6,0$

$(1):gx(ym-1)=270\Rightarrow gx(b-1)=270$

$b-1$ is a factor of $270$

$\therefore b-1=18,15,9,6\Rightarrow b=19,16,10,7\Rightarrow ym=19,16,10,7$

$(1):gxym(1-\cfrac{1}{ym})=270$

Substituting the values of $ym$ , we have $gxym=285,288,300,315 \Rightarrow N=285,288,300,315$

$\therefore$ the number of possible values of $N=\boxed{4}$