4. Last Digits

To find the last $n$ digits in the product of certain numbers, we can simply multiply the last $n$ digits of each of these numbers.

In other words, to find the units digit of a product of numbers, we should multiply only their units digits. This is shown in the example below.

In this section we will use the following functions
1) $U$($x$), which provides the units digit or the last digit of $x$.
2) $T$($x$), which provides the tens and units digit or the last $2$ digits of $x$.

Example 13

What is the units digit of $303 \times 8456 \times 212$?

Solution

$U(303 \times 8456 \times 212)$

$= U (3 \times 6 \times 2) = U(36)$ = 6

Answer: $6$

Similarly, to find the last $2$ digits in a product, we find the product of the last $2$ digits.

Example 14

What are the last two digits of $303 \times 8456 \times 212$?

Solution

$T(303 \times 8456 \times 212) = T(3 \times 56 \times 12)$

$= T(168 \times 12) = T(68 \times 12) = T(816) = 16$

Answer: $16$

4.1 Units Digit Cyclicity in Powers

To find the last digit for a number expressed as an exponent, say $x^n$, we need to apply the power only to the last digit of $x$.

Example 15

What is the units digit of $164^{683}$?

Solution

To find the units digit of this exponent, we look at powers of the units digit alone. Units digit of $164$ is $4$.

$4^{1} = 4$ ; $4^{2} = 16$ ;
$4^{3} = 64$ ; $4^{4} = 256$

The units digits are alternating between $4$ and $6$. In fact, the units digits are $\bm{4}$ for odd powers and 6 for even powers of $\bm{4}$.

In $164^{683}$, the power of $683$ is an odd number. ∴ The units digit is $4$.

Answer: $4$

In the above example, the units digit is repeating for every second power of $4$. This repetition or cyclicity varies across the $10$ different digits of our decimal number system.

For instance, $3$ raised to the powers of $1$, $2$, $3$, $4$ and $5$ are $3$, $9$, $27$, $81$ and $243$ respectively. The last digits of these powers, i.e. $\bm{3, 9, 7, 1 \text{and} 3}$, are filled where $x = \bm{3}$ in the table below.

The last digits of consecutive powers of $3$ are $3$, $9$, $7$, $1$, $3$, $9$, $7$, $1$, $3$, $9$, ... Note that the last digit for every fourth power is repeated. Therefore, when the last digit is $3$, the cyclicity is $4$.

	Last digit of $x^{n}$ where $x = $
	0	1	2	3	4	5	6	7	8	9
$x^1$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$x^2$	$0$	$1$	$4$	$9$	$6$	$5$	$6$	$9$	$4$	$1$
$x^3$	$0$	$1$	$8$	$7$	$4$	$5$	$6$	$3$	$2$	$9$
$x^4$	$0$	$1$	$6$	$1$	$6$	$5$	$6$	$1$	$6$	$1$
$x^5$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$

Cyclicity	Digits
$1$	$0, 1, 5, 6$
$2$	$4, 9$
$4$	$2, 3, 7, 8$

The above tables need not be memorised. All you need to note here is that the cyclicity is a maximum of 4 and that the last digits can be quickly found. The same is shown in the example below.

Example 16

What is the units digit of $36^{83} \times 77^{459} + 88^{364} \times 64^{84}$?

Solution

We need to find the units digit of $6^{83} \times 7^{459} + 8^{364} \times 4^{84}$

$6^{1} = 6$ ; $6^{2} = 36$.
Cyclicity of $6$ is $1$. ∴ $\bm{U(6^{83})}$ is 6.

$U(7^1) = 7$
$U(7^2) = U(7 \times 7) = 9$
$U(7^{3}) = U(49 \times 7) = 3$
$U(7^4) = U(3 \times 7) = 1$
$U(7^5) = U(1 \times 7) = 7$ (Starts repeating here)

Cyclicity for $7$ is $4$. When the powers of $7$ are of the form $4k + 1$, $4k + 2$, $4k + 3$ and $4k$, where $k$ is a natural number, the units digits are $7$, $9$, $3$ and $1$ respectively.
When $459$ is divided by $4$, the remainder is $\bm{3}$. So, $459$ is of the form $\bm{4}$k + $\bm{3}$.

∴ $\bm{U(7^{459})}$ is that of $7^3$, which is 3.

Units digits of $8^1$, $8^2$, $8^3$ and $8^4$ are $8$, $4$, $2$ and $6$ respectively. Cyclicity for $8$ is also $4$.

When $364$ is divided by $4$, the remainder is $\bm{0}$. So, $364$ is of the form $\bm{4}$k.

∴ $\bm{U(8^{364})}$ is that of $8^4$, which is $\bm{6}$.
(Note: When the remainder is $0$, we have to take units digit of $8^4$ and not $8^0$.)

Units digits of $4^1$, $4^2$, $4^3$ and $4^4$ are $4$, $6$, $4$ and $6$ respectively. Cyclicity for $4$ is $2$.

Units digit is $4$ for odd powers and $6$ for even powers.

∴ $\bm{U(4^{84})}$ is $6$.

∴ $U(6^{83} \times 7^{459} + 8^{364} \times 4^{84}) = U(6 \times 3 + 6 \times 6) = U(54) = \bm{4}$

Answer: $4$

Example 17

What is the last digit of $32^{31^{31}}$?

Solution

The cyclicity for $2$ is also $4$. The units digits of $2^1$, $2^2$, $2^3$, $2^4$ are $2$, $4$, $8$ and $6$ respectively.

We need to find the remainder when $31^{31}$ is divided by $4$. The following is covered under Factors & Remainders lessson.

$\text{Rem}\left(\dfrac{31^{31}}{4} \right) = \text{Rem}\left(\dfrac{(32 - 1)^{31}}{4} \right) = \text{Rem} \left(\dfrac{(-1)^{31}}{4} \right) = \text{Rem} \left(\dfrac{-1}{4} \right)$

$= -1 + 4 = 3$

∴ $31^{31}$ is of the form $4k + 3$.

Units digit of $\bm{U(32^{31^{31}}) = U(2^{4k + 3}) = U(2^{3}) = 8}$

Answer: $8$