The tens digit of a two-digit number is 2 more than its units digit. The number itself is 3 more than six times the sum of its digits. Find the number.

Let the units digit be $u$, so the tens digit is $u+2$ and the number is $10(u+2)+u=11u+20$. Its digit sum is $2u+2$, so $11u+20=6(2u+2)+3=12u+15$, giving $u=5$. The number is 75. Check: 7 + 5 = 12 and $6\times 12+3=75$. The trap 57 has the two digits swapped; 53 satisfies only the digit condition $(6\times 8+3=51$, not 53).