Revision Notes
1. What determines the unit digit of an?
Unit digit of any an depends on just the ‘unit digit of a’ (and the rest of the digits of a can be safely ignored) and on n.
As n assumes values 1, 2, 3, 4, ……. , the unit digit of an follows a cycle of specific values. And using this cycle, we can identify the unit digit for any value of a & n.
2. Find the cycle of unit digit of an
Say you want to find the cycle of the unit digit of 7n as n assumes values 1, 2, 3, 4, ……
Start with unit digit of 71 i.e. 7 will be the first term of the cycle. To find the next term in the cycle, multiply the last term of the cycle with 7. Just focus on the unit digit of the product, that will be the next term in the cycle. Continue this till you again reach the same value as the 1st term i.e. 7.
First in the cycle will be 7 itself;
Next will be the unit digit of 7 × 7 i.e. 9;
Next will be the unit digit of 9 × 7 i.e. 3;
Next will be unit digit of 3 × 7 i.e. 1;
Next will be the unit digit of 1 × 7 i.e. 7, the same value which was the first term in the cycle. And here-on, the same set of calculations will ensue i.e. the cycle repeats all over again. So the cycle of unit digit of 7n will be 7, 9, 3, 1.
Cycle of unit digit of 8n?
First in the cycle will be 8 itself;
Next will be the unit digit of 8 × 8 i.e. 4;
Next will be the unit digit of 4 × 8 i.e. 2;
Next will be unit digit of 2 × 8 i.e. 6;
Next will be the unit digit of 6 × 8 i.e. 8, the first term and here-on the cycle starts all over again. So the cycle of unit digit of 8n will be 8, 4, 2, 6.
Using the above procedure, find the cycle for a = 0 to 9. These can be the only possible values for unit digit of a.
3. The cycle of values, for different unit digit of a is as follows:
| Unit digit of a | Cycle of the unit digit of an | Cycle has ? terms |
| 0 | 0 | 1 |
| 1 | 1 | 1 |
| 2 | 2, 4, 8, 6 | 4 |
| 3 | 3, 9, 7, 1 | 4 |
| 4 | 4, 6 | 2 |
| 5 | 5 | 1 |
| 6 | 6 | 1 |
| 7 | 7, 9, 3, 1 | 4 |
| 8 | 8, 4, 2, 6 | 4 |
| 9 | 9, 1 | 2 |
4. Finding the unit digit of an, for any n.
When a is such that the cycle of unit digits has just 1 term
- Unit digit of 0n, 1n, 5n and 6n, will be 0, 1, 5 and 6 respectively, for any value of n.
Next, we turn to values of a which have 2 terms in the cycle
- Unit digit of 4(any odd natural power) will be 4;
and unit digit of 4(any even natural power) will be 6.
- Unit digit of 9(any odd natural power) will be 9;
and unit digit of 9(any even natural power) will be 1.
Lastly, values of a which have 4 terms in the cycle
- To find unit digit of an, where a = 2, 3, 7 or 8
divide n by 4 and find the remainder
if remainder is 1, the unit digit is 1st term in the cycle; if remainder is 2, the unit digit is 2nd term in the cycle; if the remainder if 3, the unit digit is 3rd in the cycle; and if remainder is 0, the unit digit is last term of the cycle.
5. Unit digit of any expression depends on just the unit digit of the terms in the expression.
E.g. Unit digit of 14334 + 357845 × 9475 – 2914
Will be same as unit digit of
334 + 845 × 475 – 914
because unit digit of an depends on just the unit digit of a.
And finding the unit digit of each term as explained in point #4, the unit digit will be the unit digit of
1 + 8 × 4 – 1
Please do follow the BODMAS rule and focus only on the unit digit, ignoring all other digits to get the answer as 1 + 2 – 1 i.e. 2.
ERROR-PRONE
Just be careful when you subtract, because you may also get something of the type …
Case 1: Find unit digit of 2819 – 1211
You should reach
2 – 8
But the unit digit can never be −6. So how to proceed?
It may not be very difficult for many to figure out that there must be a ‘borrow’ from the ten’s position and the answer will be 12 – 8 i.e. the unit digit of 2819 – 1211 will be 4. And this is correct. Contrast this with …
Case 2: Find unit digit of 1811 – 2219
Here too, we reach the same step,
2 – 8
Will the answer in this case also be the same 12 – 8 = 4?
No, it wont be the same for this question.
In the former case 1, focus on the values in the original question, 2819 – 1211. Here 2819 is greater than 1211. And hence 12 – 8 = 4 was perfect.
But in case 2, 1811 – 2219, 1811 is less than 2219 and thus the actual answer will be a negative number i.e.
− (2219 – 1811)
So the result will a negative number. And the unit digit will be 8 – 2 i.e. 6.
6. Unit digit of \( {a}^{{n}^{m}}\)
Here where most student err is in confusing \( {2}^{{3}^{4}}\) with \({\left({2}^{3}\right)}^{4}\).
Using rules of indices, most of us will be knowing that
\({\left({2}^{3}\right)}^{4}\)
= 23×4
= 212
And the unit digit of 212 can easily be found out using whatever is learnt already.
But \( {2}^{{3}^{4}}\) is NOT same as \({\left({2}^{3}\right)}^{4}\).
In an expression like \( {2}^{{3}^{4}}\), we first need to compute starting from the top-most index-relation i.e. first we will need to do 34 = 81 and then it becomes 281, which obviously is not same as earlier 212.
However here we knew 34 and hence could manage this. But in the values used in questions, we would not be knowing the value of the index nor can we compute it easily. E.g. \( {17}^{{18}^{19}}\). How will we compute the value 1819?
Then there are a lot of students, who, when faced with \( {17}^{{18}^{19}}\) consider this as just 1718 to start with, and find the remainder when 18 is divided by 4. However this is also wrong.
Since we first need to do 1819, the index of 17 is not 18 but is 1819. Representing \( {17}^{{18}^{19}}\) in the form an, the correct form is a = 17 and n = 1819. And since, to find unit digit of 17n, we need to find remainder when n is divided by 4, in this question we will need to find the remainder when 1819 is divided by 4.
Now finding the remainder when an is divided by a number is another question itself, which we will learn in the topic of Remainders. But in this specific case, it is easy. While we cannot compute 1819, it is 18 × 18 × 18 × … × 18, which is enough to realise that it is a multiple of 4. And being a multiple of 4, 1819 divided by 4 will leave a remainder of 0. And the unit digit of \( {17}^{{18}^{19}}\) will be the last in the cycle of (7, 9, 3, 1) i.e. 1
Thus to find the unit digit of \( {a}^{{n}^{m}}\), we divide nm with the number of terms in the cycle of unit digit of a, find the remainder and then proceed as learnt in point 4.
7. The reverse … for what values of n will the unit digit of an be a given value.
Though this appears tough, this is pretty easy. Starting with the easier cases to get a hang of the process ….
When unit digit of 4n or 9n is given
If unit digit of 4n is 4, then n can be ANY odd natural number; and if unit digit of 4n is 6, then n can be ANY even natural number.
Similarly, if unit digit of 9n is 9, then n can be ANY odd natural number; and if unit digit of 9n is 1, then n can be ANY even natural number.
In this case, n can be odd or even i.e. divisible by 2 or not (remainder = 1 when divided by 2) …. just two possibilities because the number of terms in cycle of 4n and 9n is just 2.
With 2n, 3n, 7n, 8n, since the number of terms in the cycle is 4, and when index is divided by 4, the remainders could be ANY of 1, 2, 3 or 0. For these we have …
If unit digit is first in the cycle, then n could be ANY (multiple of 4 + 1) i.e. of the form 4k+1
If unit digit is second in the cycle, then n could be ANY (multiple of 4 + 2) i.e. of the form 4k+2
If unit digit is third in the cycle, then n could be ANY (multiple of 4 + 3) i.e. of the form 4k+3
If unit digit is last in the cycle, then n could be ANY (multiple of 4) i.e. of the form 4k.
E.g. Thus, if unit digit of 3n is given to be 1 …. since 1 is the last in the cycle (3, 9, 7, 1), hence n has to be of the form 4k.
8. Some general expressions.
8.1 Unit digit of: 1n + 2n + 3n + 4n + …… + 9n
If n is any number of the form 4k+1, or 4k+2 or 4k+3, the unit digit of this expression will be 5; whereas if n is of the form 4k, the unit digit will be 1.
You could assume any value of n satisfying the form and check for your yourself.
CAUTION: Be careful that if n is a whole number, then n = 0 is a completely different case since (anything)0 = 1; and for this one case, n = 0, the unit digit will be 9.
8.2 Unit digit of: a1 + a2 + a3 + a4 + …. + an
For a = 0, 1, 5 or 6, the unit digit of each term in the series will be the same, will be 0, 1, 5, or 6 respectively and the required unit digit will just be the sum of these values, n times i.e. the unit digit of a × n.
For a = 4 or 9, the sum of two successive terms will be constant, e.g. for case of a = 4, it will be
(4 + 6) + (4 + 6) + (4 + 6) + ….
Since each bracket results in unit digit being 0, the unit digit of the expression will thus be 4 if n is odd; or 0 if n is even.
Very similar workings follow for a = 9.
For a = 2, 3, 7 or 8, the sum of four successive terms will be constant e.g. for a = 2, it will be
(2+4+8+6) + (2+4+8+6) + (2+4+8+6) + …
Here too, the unit digit of each bracket becomes 0 and thus, the unit digit of the expression will depend on how many terms are there in the last bracket.
For n = 4k+1, the unit digit will be just 2; for n = 4k+2, the unit digit will be 2+4 = 6; for n = 4k+3, the unit digit will be 2+4+8 = 4; and for n = 4k, it will be 0.
Very similar working follow for a = 3 or 7 or 8.
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