2. Find all four-digit numbers A that satisfy the following three conditions:

a) the first digit of A is twice as small as the last;

b) the second and third digits of A are the same;

c) if the number A is decreased by 2, the result is divisible by 143.
Solution. Let $A=\overline{a b b d}$. Then $\overline{a b b d}=1000 a+110 b+d$. From condition a) we have $\mathrm{d}=2 \mathrm{a}$, so accordingly $\overline{\mathrm{abbd}}=1002 \mathrm{a}+110 \mathrm{~b}$. Therefore, $\mathrm{A}-2=1002 \mathrm{a}+110 \mathrm{~b}-2=1001 \mathrm{a}+110 \mathrm{~b}+\mathrm{a}-2$. Since $143=11 \cdot 13$, we have that $11 \mid(1001 \mathrm{a}+110 \mathrm{~b}+\mathrm{a}-2)$. However, 11|1001, 11|110, so 11|(a-2), i.e., $a=2$, then $d=2 a=4$. Now, $A=\overline{2 b b 4}=2004+110 b$. Therefore, $A-2=2002+110 b$. Since $13|2002,13|(A-2)$, we have that $13 \mid 110$ b. However, GCD $(13,110)=1$ so 13|b. Since b is a digit, it follows that $b=0$. Finally, the desired number is $\mathrm{A}=2004$.