It is known that \( x \) and \( y \) are two-digit numbers, according to the table below:
| 26 | 12 | 48 | 55 | 24 | \(x\) |
| 21 | 62 | 55 | 84 | \(y\) | 42 |
What is the most likely minimum result of \( x + y \)?
-
110
-
154
-
160
-
120
-
144
Try to solve it before checking the answer.
-
110
We will solve this problem by studying all possible answers, but first we are going to divide the table into the following three subtables:
| 26 | 12 |
| 21 | 62 |
| 48 | 55 |
| 55 | 84 |
| 24 | \(x\) |
| \(y\) | 42 |
In the first sub-table, the numbers in the first column are 26 and 21. If we swap the tens and ones digits of 26 and 21, we get the numbers 62 and 12, which are the same numbers of the second column, except for the rows being switched. This pattern is repeated in the second subtable. Therefore, the same thing should happen in the third subtable. That is:
If \( y = ab = 10a + b \), then \( x = ba = 10b + a \)
If \( y = ab = 10a + b \),
then;
\( x = ba = 10b + a \)
Now, we have:
\[ \begin{aligned} x + y &= 10b + a + 10a + b \\[2em] &= 11(a + b) \end{aligned} \]This means that \( x + y \) is a multiple of 11.
Of the five possible answers in the statement (110, 154, 160, 120, and 144), the only multiples of 11 are 110 and 154. Of these two, the minimum is 110.