Which one of the following is the remainder when 10^20 is divided by

Which one of the following is the remainder when 10^20 is divided by 7?

1
2
4
6
This question was previously asked in
UPSC CAPF – 2018
The remainder when 10^20 is divided by 7 is 2.
– We need to compute 10^20 mod 7.
– First, simplify the base: 10 mod 7 = 3. So, 10 ≡ 3 (mod 7).
– Then, 10^20 ≡ 3^20 (mod 7).
– Now, we compute powers of 3 modulo 7 to find a pattern (cycle):
– 3^1 ≡ 3 (mod 7)
– 3^2 ≡ 9 ≡ 2 (mod 7)
– 3^3 ≡ 3 * 2 = 6 (mod 7)
– 3^4 ≡ 3 * 6 = 18 ≡ 4 (mod 7)
– 3^5 ≡ 3 * 4 = 12 ≡ 5 (mod 7)
– 3^6 ≡ 3 * 5 = 15 ≡ 1 (mod 7)
– The powers of 3 modulo 7 repeat with a cycle length of 6 (3, 2, 6, 4, 5, 1).
– To find 3^20 mod 7, we need to find the position in the cycle corresponding to the exponent 20. This is done by finding the remainder of 20 when divided by the cycle length 6.
– 20 ÷ 6 = 3 with a remainder of 2. (20 = 3 * 6 + 2).
– So, 3^20 ≡ 3^(6*3 + 2) ≡ (3^6)^3 * 3^2 (mod 7).
– Since 3^6 ≡ 1 (mod 7), we have:
– 3^20 ≡ (1)^3 * 3^2 (mod 7)
– 3^20 ≡ 1 * 9 (mod 7)
– 3^20 ≡ 9 ≡ 2 (mod 7).
– The remainder is 2.
This method uses modular arithmetic properties, specifically finding the cyclic nature of powers modulo a number. Fermat’s Little Theorem could also be applied here since 7 is a prime number and 3 is not divisible by 7. Fermat’s Little Theorem states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) ≡ 1 (mod p). Here, a=3, p=7. So, 3^(7-1) = 3^6 ≡ 1 (mod 7). This confirms our calculated cycle length. Then we proceed as shown.
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