CAT 2025 Lesson : Divisibility - Last Digit

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 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.