Carelessly Careless

Since $819 = 2 \times 7^{2} \times 13$ and $1274 = 3^{2} \times 7 \times 13$ we can find all of the two digit factor pairings that make up each number. $819 = 63 \times 13, 21 \times 39$ and $1274 = 98 \times 13, 49 \times 26, 14 \times 91$ . We see that the only pair that fits the description is $819 = 21 \times 39$ and $1274 = 49 \times 26$ . Thus $x=21$ and $y=49$ . $49 + 21 = \boxed{70}$ .

Let's get the prime divisors of these two numbers to get the possible two digit number combinations:

$1274 = 2 \times 7 \times 7\times 13$ which makes the possible combinations: $(13,98), (14,91), (26,49)$

$819 = 3 \times 3\times 7\times 13$ which makes the possible combinations: $(13,63), (21,39)$

Now we look for an identity where the ones digit of one element of the first set altered and the tens digit of one element of the second set altered. With little inspection it is easy to see that $(26,49)$ and $(21,39)$ satisfies the above condition. Thus x and y are $(21, 49)$ which makes their sum $\boxed{70}$

1274 could be resolved to $(14*91)$ or $(13*98)$ or $(49*26)$

819 could be resolved to $(21*39)$ or $(13*63)$ or $(9*91)$ but the last one refused as 9 contains one digit only.

by comparing the resolutions of the two numbers to find two resolutions (one from each number) have the same tens place in two numbers and the same ones place in the other numbers we found $(26*49)$ and $(21*39)$

so x=21, y=49

then x+y= $\boxed{70}$

When you factorize both numbers, you get 1274= 26 x 49 and 819= 21 x 39. Observe that 26 and 21 differ in 1's place. 49 and 39 differ in 10's place and this is the only pair that does so.

Factorising 1274 , we get 1274 = 2\times7\times7\times13 , since 1274 is obtained by product of two 2 - digit numbers , probable factors of 1274 are 14\times91 or 98\times13 or 26\times49 Similarly factorising 819, we get 819 = 3\times3\times13\times7 , by above logic probable factors of 819 are 39\times21 or 63\times13 by observation we find that 49\times21 are correct values of x and y which got changed to 39 and 26 respectively giving 39\times21 =819 and 49\times26 = 1274

So, first we need to find the factor pairs of both 1274 and 819 which are both 2 digit numbers. These are 13,63 and 21,39 for 819 and 13,98 14,91 and 26,49 for 1274. This gives us six possible combinations but upon comparison, we realise that the 21,39 and 26,49 pair is the only pair that could have the misread numbers mentioned in the question. So, Mr Careless had the tens digits correct and Mrs Careless had the ones digits correct, giving 21 and 49, for a sum of 70.