## Consecutive integers; a(10^n * b + c) + 1 = (a + l)*b*c — Part 1

$x \; = \; a \,(10^{n} \; b \; + \; c)$,        $x + 1 \; = \; (a + 1) \, b \, c$

meaning,

$a(10^n \; b \; + \; c) \; + \; 1 \; = \; (a + 1) \, b \, c$

where    $5 \; \times \; 10^{n-1} \; < \; c \; < \; 10^{n} \; - \; 1$

n = 1

n = 2

n = 3

$a \,(10^3 \; b \; + \; c) \; + \; 1 \; = \; (a + 1) \, b \, c$        $500 < c < 999$

Here are the first few solutions:

$2667 \; = \; 1 \; \times \; 2667$
$2668 \; = \; 2 \; \times \; 2 \; \times \; 667$

$84503 \; = \; 1 \; \times \; 84503$
$84504 \; = \; 2 \; \times \; 84 \; \times \; 503$

$251501 \; = \; 1 \; \times \; 251501$
$251502 \; = \; 2 \; \times \; 251 \; \times \; 501$

$2671334 \; = \; 2 \; \times \; 1335667$
$2671335 \; = \; 3 \; \times \; 1335 \; \times \; 667$

$239244 \; = \; 4 \; \times \; 59811$
$239245 \; = \; 5 \; \times \; 59 \; \times \; 811$

$2567204 \; = \; 4 \; \times \; 641801$
$2567205 \; = \; 5 \; \times \; 641 \; \times \; 801$

Find more solutions.