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Your first reaction to the title of this post is probably, “I know my subtraction well, thank you!” No surprise there. But what is surprising is that our statistics tell us that the following question – which is nothing extraordinary, but does involve some tricky subtraction – is a 700-level question. That made us decide to write this post. We will discuss this concept along with the question:
Question: The last digit of 
(A) 0
(B) 1
(C) 5
(D) 8
(E) 9
This is a simple question based on the cyclicity of units digits.
There are 3 terms here: 
, 
and 
. Let’s find the last digit of each of these terms:
The units digit of 12 is 2.
2 has a cyclicity of 2 – 4 – 8 – 6.
The cycles end at the powers 4, 8, 12 … etc. So, twelve 2’s will end in a units digit of 6.
The units digit of 13 is 3.
3 has a cyclicity of 3 – 9 – 7 – 1.
A new cycle starts at the powers 1, 5, 9, 13 … etc. So, thirteen 3’s will end in a units digit of 3.
This term is actually the most simple to manage in the case of its units digit – an even number multiplied by a multiple of 5 will end in 0. Also, note that this will be a huge term compared to the other two terms.
This is what our expression looks like when we consider just the units digits of these terms:
(A number ending in 6) + (A number ending in 3) – (A much greater number ending in 0)
Looking at our most basic options, a number ending in 6 added to a number ending in 3 will give us a number ending in 9 (as 3 + 6 = 9). So, the expression now looks like this:
(A number ending in 9) – (A much greater number ending in 0)
It is at this point that many people mess up. They deduce that 9-0 will end in a 9, and hence, the answer will be E. All their effort goes to waste when they do this. Let’s see why:
How do you subtract one number out of another? Take, for example, 10 – 7 = 3
This can also be written as 7 – 10 = – 3. (Here, you are still subtracting the number with a lower absolute value from the number with a greater absolute value but giving it a negative sign.)
Let’s try to look at this in tabular form. The number with the greater absolute value goes on the top and the number with the smaller absolute value goes under it. You then subtract and the result gets the sign of the number with the greater absolute value.
(i) 100 – 29
100
– 29
071
(ii) 29 – 100
100
– 29
071
(But since the sign of 100 is negative, your answer is actually – 71.)
So, the number with greater absolute value is always on top. Going back to our original question now, (A number ending in 9) – (A much greater number ending in 0) will look like:
abcd0
– pq9
ghjk1
Ignoring the letter variables (these are simply placeholders), note that the greater number ending in 0 will be on the top and the smaller one ending in 9 will be below it. This means the answer will be a negative number ending in a units digit of 1. Therefore, our answer is (B).
As we learn more advanced concepts, make sure you are not taking your basic principles for granted!
