GATE Electrical 2023 | GA | Question: 9

Question

The digit in the unit's place of the product $3^{999} \times 7^{1000}$ is _________.

• A. $7$
• B. $1$
• C. $3$
• D. $9$

Answer 1

The pattern of the unit's digits of powers of $3:$

1. $3^1 = 3$
2. $3^2 = 9$
3. $3^3 = 7$
4. $3^4 = 1$

The pattern repeats every $4$ powers of $3.$

The pattern of the unit's digits of powers of $7:$

1. $7^1 = 7$
2. $7^2 = 9$
3. $7^3 = 3$
4. $7^4 = 1$

The pattern repeats every $4$ powers of $7.$

Now, Unit digit of $3^{999} \times 7^{1000} = \underbrace{3^{4{(249)}}}_{\text{Unit digit = 1}} \times 3^3 \times \underbrace{7^{4(250)}}_{\text{Unit digit = 1}} = 1 \times 7 \times 1 = 7$

So, the unit's digit of $3^{999} \times 7^{1000}$ is $7.$

Correct Answer: A

Answer 2

$3^{999} * 7^{1000} = 3^{999} * 7^{999} * 7^1 = 21^{999} * 7$

The last Digit of 21 is 1. Therefore 21 power anything is 1.

1*7 = 7.

Therefore last digit is 7.

Comments

@SaiKo clever strategy.

Answer 3

Although with cyclicity this would be easy. Here is an alternative using binomial theorem only:

$3^{999} \equiv 3 (10 - 1)^{499} \equiv 3 (10 - 1) \equiv - 3 \equiv 7 \text{ mod } 10$

and $7^{1000} \equiv (10 -3)^{1000} \equiv (-1)^{1000} (3)^{1000} \equiv (10 - 1)^{500} \equiv 1 \text { mod 10}$

Now $(3^{999} \times 7^{1000}) \text { mod }  10 \equiv \left[(3^{999} \text{ mod } 10) \times (7^{1000} \text{ mod } 10)\right] \text { mod 10 } \equiv 7 \text { mod } 10 \equiv 7.$

Comments

$3^{999} (\text{mod} \, 10)$

$\equiv [3 \cdot 3^{998}] (\text{mod} \, 10)$

$\equiv [3 \cdot (3^{2})^{499}] (\text{mod} \, 10)$

$\equiv [3 \cdot (9)^{499}] (\text{mod} \, 10)$

$\equiv [3 \cdot (10-1)^{499}] (\text{mod} \, 10)$

$\equiv [\{3 (\text{mod} \, 10) \} \cdot \{(10-1)^{499} (\text{mod} \, 10)\}] (\text{mod} \, 10)$

$\equiv [3  \cdot \{(10-1) (\text{mod} \, 10)\}^{499}] (\text{mod} \, 10)$

$\equiv [3  \cdot \{10 (\text{mod} \, 10) - 1 (\text{mod} \, 10)\}^{499}](\text{mod} \, 10)$

$\equiv [3  \cdot \{0 - 1\}^{499}] (\text{mod} \, 10)$

$\equiv [3  \cdot (-1)^{499}](\text{mod} \, 10)$

$\equiv [3  \cdot (-1)](\text{mod} \, 10)$

$\equiv -3(\text{mod} \, 10)$

$\equiv 7(\text{mod} \, 10)$

$\equiv 7$

similarly, for $7^{1000}$ also.