If you add the digits in a two-digit number and multiply the sum by 7, you get the original number. if you reverse the digits in the two-digit number, the new number is 18 more than the sum of its two digits. what is the original number?

Hello,

Let's assume a the digit of tenth
b the digit of unities.
7*(a+b)=10a+b ==>3a=6b==>a=2b
b*10+a=18+a+b==>9b=18==>b=2 and a=4
The original number is 42.
7*(4+2)=42
24=18+4+2

The first one, i believe, would be correct. all the others have one thing in common, and that's that they have repeating numbers. in the second one, x and y have repeating 6's and 3's, and in the third, x and y have repeating 6's as well. in the last one, there are repeating 3's. the first one has no repeating numbers.