In a two-digit number the tens digit is 3 more than the units digit, and the number itself is 7 times the sum of its two digits. Which number is it?

Let the units digit be $u$, so the tens digit is $u+3$ and the number is $10(u+3)+u=11u+30$. Seven times the digit sum is $7(u+3+u)=14u+21$. Setting them equal gives $11u+30=14u+21$, so $3u=9$ and $u=3$, making the number 63. Check: 6 is 3 more than 3, the digits add to 9, and $7\times 9=63$. The traps: 36 is 63 with the digits swapped, so its tens digit is 3 LESS than its units digit; 81 and 72 both have digits adding to 9, but 81 is nine times that sum and 72 is eight times it, not seven.