Famous comedic duo Abbott and Costello have a routine in which Costello performs clearly incorrect math, but somehow continues to arrive at the same answer. Abbott is baffled to see how Costello’s outrageous claim proves to be true time and time again. Before going any further take a look at their skits, the first showing how 25 divided by 5 is 14, and the second how 7 x 13 = 28.
Costello shows that 25 divided by 5 is 14 by addition, multiplication, and by division. However, he performs each of these operations incorrectly, because of how he handles the ones and tens places. When adding the five 14s together, he counts each number individually: 1 + 4 + 1 + 4 + ….. instead of using their place values ( 10 + 4 + 10 + 4…). When multiplying Costello similarly does 14 x 5 as 4 x 5 added with 1 x 5. The correct way is 4 x 5 added to 10 x 5. The same holds for division, where his method disregards place values.
What is astounding about this incorrectness is that he always arrives back at the same conclusion that 5 x 14 = 25, by those three operations. However, he makes the same mistake in each operation, which is the reason why it always works out the same way. In the second skit, Costello uses the same flawed methodology to show that 7 x 13 = 28. This led me to wonder if these select examples were special and unique, if other numbers would have a similar outcome, or if all numbers would behave this way.
After some experimenting with 16 x 6 (which equals 42 by the Costello Method) among other number combinations, the results continued to hold true. The numbers selected for the skit are more impressive than some that could be chosen. Evidence shows that if you commit to the Costello Method for addition, multiplication, and division then you can usually manipulate your way to the same answer each time. Two exceptions would be if there is a 0 in the ones place, or if you are multiplying by 1.
Example: Consider 3 x 20. By ignoring place value this adds as 2 + 0 + 2 + 0 + 2 + 0 = 6, which is so wrong that it is just goofy. It would take clever misdirection to even try to pull off that 6 divided by 3 is 20.
Example: 1 x 14 by the Costello Method = 5. The addition step for 1 x 14 is adding 14 to nothing, and I would love to see Costello try to explain that one.
There are certainly more exceptions that could be considered, such as if both numbers were each one digit, if it was being multiplied by zero, etc.
So – why does this happen?
Consider the case of 29 x 3.
29 x 3 would be 2 x 3 + 9 x 3 = 33. By adding this it would be 2 + 2 + 2 + 9 + 9 + 9 = 33. If I rewrite the addition step in this way 3(2) + 3(9) then it becomes clear how the mistake in the multiplication example leads you astray in both cases. Again, you are multiplying 3 to both digits, regardless of their place value.
As a teacher, these videos are among my favorites for the shock value and then the discussion about where Abbott and Costello made their mistakes.
All the Calculations 
What about 27 / 3 = 27? And 27 * 3 = 27…