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Last Digit

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Module 5 of 8

Overview

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

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.

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

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

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

2

.

Units digit is

4

for odd powers and

6

for even powers.

∴

\bm{U(4^{84})}

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

CAT 2025 Lesson : Divisibility - Last Digit

Divisibility

4. Last Digits

Example 13

Solution

Example 14

Solution

4.1 Units Digit Cyclicity in Powers

Example 15

Solution

Example 16

Solution

Example 17

Solution