This is a tough one! This strikes me as a little tougher than official questions tend to be, but let's break it down. Imagine the numbers all stacked together like a pyramid.
_____2
____22
___222
__2222
and so on...
So let's reason column by column.
In the first column, the units digits, we'll have a total of 30 2's. 30*2 = 60, the units digit of the sum will be 0, and we'll carry the 6.
When we move to the column to the left, the tens digits, we'll have 29 2's (all but the first term). With the 6 we carried over, we'll have 6 + 29*2 = 64. So 4 is the digit for the tens column, and again, we'll carry a 6.
In the third column, there will be 28 2's. Now we'll have 28*2 + 6 = 62. 2 is the digit for the hundreds column, carry the six...
In the fourth column, there will be 27 2's. Now we'll have 27*2 + 6 = 60. 0 is the digit, carry the six.
Hmm. This is time consuming.
So we started with 30 2's in the units column, and each time we shift over a column, we're losing a 2. The number of 2 's we'll have in the Nth column can be summarized as: 30 - (N - 1). Notice also that, after the first column, the sum should be decreasing by 2 each time we shift over a column.
Let's skip to the 10th column and use some logic. Here, before we consider what should have been carried over from the ninth column, we'll have 21 2's, so that gives us 42. Even if we'd continued carrying 6's, (it's not actually a 6 - as noted earlier, the sum is getting smaller and smaller) the number we're carrying to the 11th column would HAVE to be a 4, because 42+ 6 = 48, and our actual number should be between 42 and 48. So let's carry a 4 over to the 11th column.
In the 11th column, we'll have 20 2's. And we're carrying a 4. That leaves us with 20*2 +4, and a units of 4.
Answer is C.
Now I need a nap.
