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

COnsec Integers 1

n = 2

COnsec Integers 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.
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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