A Countryside Problem

Suppose that $x$ days before this particular day is the day whence the mailman is to visit in three days time, the grocer in seven days time and the physician in ten. This day will be known as the starting day, and our particular day given in the question is the $(x-1)^{st}$ day after this.

We then have $x \equiv 1\pmod{3},x \equiv 3\pmod{7},x \equiv 2\pmod{10}$ . Applying Chinese Remainder Theorem on these 3 modular congruences, we attain $x \equiv 52\pmod{210}$ . So this particular day has been the $52-1=51$ days since the first day we have chosen.

We know the three men will first visit on the same day every $*lcm(3,7,10)-1=210-1=209$ days from the first day and this duration from our particular day is thus $209-51=\boxed{158}$ days.

Let the first common visit come after m days.then m = -1(mod 3) , m= -3(mod 7) and m= -2(mod 10). The third condition implies m= 10k+8 for some k. Reducing modulo 3 , the first condition implies that k=3r for some r. So m =3r +8 . Reduce modulo 7 and use the second condition to get 2r= -4 and hence r=-2modulo 7. So r may be taken as 5 , which gives m equal to 158