Question

10. The value of 5^2003 mod 7 is?
a) 3
b) 4
C) 8
d) 9​

Answer 1

Answer:

3 (mod 7)

Step-by-step explanation:

5^2003= (5^6)^334.5^-1 (mod 7)

1 . 5^-1 (mod 7)

3 (mod 7) (The inverse of 5 mod 7 is 3)

Answer 2

Answer:

Required value is 3.

Option a is the correct answer.

Step-by-step explanation:

Here we want to find ${5}^{2003}$mod 7.

By Fermat's Little Theorem, we can find this easily.

The Fermat's little theorem states that ${a}^{p - 1}$≡ 1 (mod p) if a and p are relatively prime. We take some questions as example which will clear your concept .

1.Find 331 mod 7.

By Fermat’s Little Theorem, 36 ≡ 1 mod 7. Thus, 331 ≡ 3

1 ≡ 3 mod 7.

2. Find 235 mod 7.

By Fermat’s Little Theorem, 26 ≡ 1 mod 7. Thus, 235 ≡ 2

5 ≡ 32 ≡ 4 mod 7.

3. Find 128129 mod 17.

By Fermat’s Little Theorem, 12816 ≡ 9

16 ≡ 1 mod 17. Thus, 128129 ≡ 9

1 ≡ 9 mod 17.

4. The number 21000 is divided by 13. What is the remainder?

By Fermat’s Little Theorem, 212 ≡ 1 mod 13. Thus, 21000 ≡ 2

400 ≡ 2

40 ≡ 2

4 ≡ 16 ≡ 3 mod 13.

Now question is what is the value of ${5}^{2003}$mod 7 ?

Now,

${5}^{2003} = {5}^{ {6}^{ {334.5}^{( - 1)} } } (mod \: \: 7)$1.

${5}^{( - 1)} (mod \: \: 7)$

→$3 \: (mod \: 7)$

(The inverse of 5 mod 7 is 3)